Target Publications Pvt. Ltd.

Chapter 01: Mathematical Logic

01 Mathematical Logic
1. x + 3 = 10 is an open sentence. 8.
 It is not a statement. 1 2 3 4 5 6
 option (C) is correct. p q ~q p  q p  ~q ~(p  ~q)
T T F T F T
2. Since p  q is false, when p is true and q is
T F T F T F
false.
F T F F T F
p  (q  r) is false, F F T T F T
 p is true and q  r is false
 p is true and both q and r are false. The entries in the columns 4 and 6 are
identical.
3. Since, contrapositive of p  q is ~q  ~p.  ~(p  ~q)  p  q
 contrapositive of (~p  q)  ~r is  statement-l is true.
~(~r)  ~(~p  q)  r  (p  ~q) Also, all the entries in the last column of the
above truth table are not T.
4. ~p: Rohit is short.  ~(p  ~q) is not a tautology.
The given statement can be written  statement-2 is false.
symbolically as p  (~p  q).  option (B) is correct.
5. Let p: x is a complex number 9. Consider option (C)
q: x is a negative number (p  q)  (p  r)  (T  T)  (T  F)
 Logical statement is p  q TT
 converse of p  q is q  p T
 option (C) is correct.
 option (B) is correct.
11. The statement “Suman is brilliant and
6. Consider option (C)
dishonest iff suman is rich” can be expressed
p q r ~q p  ~q (p  ~q)  r as Q  (P  ~R)
T T T F F T The negation of this statement is
T T F F F T ~(Q  (P  ~R))
T F T T T T
12. (q)  (p) is contrapositive of p  q.
T F F T T F
F T T F F T  p  q  (~q)  (~p)
F T F F F T  option (D) is true.
F F T T F T 13. (~p  ~q)  (p  q)  (~p  q)
F F F T F T  ~p  (~q  q)  (p  q)
 (p  ~q)  r is a contingency  (~p  T)  (p  q)
 option (C) is correct.  ~p  (p  q)
 (~p  p)  (~p  q)
7. Consider option (A)  T  (~p  q)
p q p  q p  q ~(p q) (p  q) (p  q))  ~p  q
T T T T F F  option (B) is correct.
T F F T F F 14. Since, inverse of p  q is ~p  ~q.
F T F T F F  inverse of (p  ~q)  r
F F F F T F is ~(p  ~q)  ~r
 (p  q)  (~(p  q)) is a contradiction. i.e., ~p  q  ~r
 option (A) is correct.
1
Target Publications Pvt. Ltd. Chapter 02: Matrices

02 Matrices

1 2 2  1  tan  1  1  tan 
  tan    2  tan 
2. |A| = 2 1 1 = 1(1)  2(7) + 2(9)  1  sec   1 
1 4 3 1 1  tan 2  2 tan  
=  
=30 sec2   2 tan  1  tan 2  
1
 A exists.  cos 2  sin 2 
T =  
 1 7 9   1 2 0   sin 2 cos 2 
  
adj A =  2 5 6 =  7 5 3 By equality of matrices, we get
 0 3 3  9 6 3  a = cos 2, b = sin 2
 1 2 0 
1  1 2 3  x   6 
 A =  7 5 3
1
2 4 1  y    7 
3 6.     
 9 6 3 
 3 2 9   z  14 
 sum of the elements of A1
R2  R2  2R1, R3  R3  3R1
1
= ( 1  2 + 0  7  5  3 + 9 + 6 + 3) = 0 1 2 3   x   6 
3  0 0 5   y  =  5 
      
3. (adj A) A = |A| In
 0 4 0   z   4 
2 0 3 1 0 0 
 5z = 5  z = 1
A = 1 1 2 0 1 0 
4y = 4  y = 1
3 2 0 0 0 1  x + 2y + 3z = 6  x = 1
1 0 0 
= 11 0 1 0   1 2 1
7. Let A =  2  3 
0 0 1 
 1 0 3 
11 0 0 
Matrix will not be invertible if |A| = 0
=  0 11 0 
1 2 1
 0 0 11
 2  3 =0
1 2 4 1 0 3
4. Let A =  3 19 7   1(3) + 2(9)  1() = 0
 2 4 8  =9
 |A| = 0 8. Given, |A|  0 and |B| = 0
 A1 does not exist.
 |AB| = |A| |B| = 0
 1 tan  
1
1  1 tan  
T and |A1 B| = |A1| |B|
5.   tan  =
 1  
sec    tan 
2
1  =
1
|B|
 1
....  | A | 
1 
|A|  | A | 
1  1  tan  =0
= 
sec   tan 
2
1   1
AB and A B are singular.
1
Std. XII : Triumph Maths 
9. (AB)1 = B1 A1 13. Since, A(adj A) = |A|.I
 1 1  Replacing A by adj A, we get
  adj A (adj(adj A)) = |adj A|I
 B1 A1 =  2 2 
 1 0  A1.|A| (adj(adj A)) = |adj A|I
 4   1 
….  A 1  (adjA) 
 |A| 
10. (A2  8A)A1 = A.A.A1  8A.A1 1 2
  A (adj (adj A)) = |A| .I
= A  8I
….[ |adj A| = |A|n1]
 1 4 4  8 0 0 
=  4 1 4   0 8 0    A1(adj (adj A)) = 2I
 4 4 1  0 0 8   A1 (adj (adj A)) = I
Given, A1(adj (adj A)) = kI
 7 4 4
 k=
=  4 7 4 

 4 4 7 

1 2 0
11. det A = 1 1 2
2 1 1
= 13
2
 det (adj (adj A)) = (det A)(3 1)

….   adj(adjA)  A
(n 1)2 

 
= (det A)4 = (13)4

4 0 0
12. A. (adj A) =  0 4 0  ….(i)
 0 0 4 
1 0 0 
= 4 0 1 0 
0 0 1 
= 4.I
Since, A(adj A) = |A|.I
 |A| = 4
From (i), |A| . |adj A| = 64
64
 |adj A| = = 16
4
(n 1) 2
Also, |adj (adj A)| = A
(31)2
= A
= (4)4 = 256
adj(adjA) 256
 = = 16
adjA 16

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Target Publications Pvt. Ltd. Chapter 03: Trigonometric Functions

03 Trigonometric Functions

1. sin 3x = 3 sin x  4 sin3x But cos x > 0 (x must be in 1st or 4th Quadrant)
1  3
 sin3x = (3sin x  sin 3 x)  the possible values are , .
4 8 8
cos 3x = 4 cos3x  3 cos x Case II:
1 If cos x < 0,
 cos3x = (cos3 x  3cos x) 1 1
4 sin x(cos x) =  sin 2x = 
Given, cos 3x cos3x + sin 3x sin3 x = 0 2 2 2
1 5 7 
 cos 3x (cos 3x + 3 cos x)  2x = ,
4 4 4
1 5 7 
+ sin 3x (3 sin x  sin 3x) = 0  x= ,
4 8 8
1  The values of x satisfying the given equation
 cos 2 3x+ 3cos x cos3x +3sin x sin 3x  sin 2 3x   0
4  3 5 7 
between 0 and 2 are , , , .
 cos2 3x  sin2 3x + 3(cos 3x cos x 8 8 8 8
+ sin 3x sin x) = 0 
 cos 6x + 3 cos 2x = 0 These are in A.P. with common difference .
4
 4 cos3 2x  3 cos 2x + 3 cos 2x = 0
2 2x
 4 cos3 2x = 0 3. 16sin x  16cos = 10
 cos 2x = 0 sin 2 x 2
 16 161 sin x = 10

 2x = (2n + 1) 2 16
2  16sin x  2
= 10
 16sin x
 x = (2n + 1) 2
4 Let t = 16sin x
16
2. sin x 8cos 2 x = 1 t+ = 10
t
 sin x 2 2 cos x = 1 ….  8  2 2   t2 + 16 = 10t
 t2  10t + 16 = 0
1  (t  2) (t  8) = 0
 sin x |cos x| =
2 2  t = 2 or t = 8
Case I: 2 2
 16sin x = 2 or 16sin x = 8
1 2 2
If cos x > 0, sin x cos x =  24sin x = 21 or 24sin x = 23
2 2
 4 sin2x = 1 or 4 sin2x = 3
1 1
 sin 2x = 1 3
2 2 2  sin2 x = or sin2 x =
4 4
1
 sin 2x =  
2  sin2x = sin2   or sin2x = sin2  
6
  3
 3 9 11  
 2x = , , ,  x = n ± or x = n ±
4 4 4 4 6 3
….[ x  (0, 2),  2x  (0, 4)]  7  5 11  4  2  5
x= , , , or x = , , ,
 3 9 11 6 6 6 6 3 3 3 3
 x= , , ,
8 8 8 8  There are 8 solutions in [0, 2].
1
Std. XII : Triumph Maths 
4. The maximum value of a sin x + b cos x is 6. tan4x  2 sec2x + a2 = 0
a 2  b2 .  tan4x  2(1 + tan2x) + a2 = 0
 tan4x  2 tan2x  2 + a2 = 0
 Maximum value of sin x + cos x is 2 and the
maximum value of 1 + sin 2x is 2.  tan4x  2 tan2x + 1  3 + a2 = 0
 The given equation will be true only when  (tan2 x  1)2 = 3  a2
 3  a2  0
sin x + cos x = 2 and 1 + sin 2x = 2
 a2  3
If sin x + cos x = 2
 |a|  3
1 1
 cos x  + sin x  =1
2 2 7. 3 cos x + 4 sin x = 5
   2 x   x 
 cos x cos + sin x sin = 1  1  tan 2   2 tan 2 
4 4  3  + 4 = 5
   1  tan 2 x   1  tan 2 x 
 cos  x   = 1  2  2
 4
x
 Let tan = t
 x  = 2n, 2
4  3  3t + 8t = 5 + 5t2
2
  8t2  8t + 2 = 0
 x = 2n + ….(i)
4  4t2  4t + 1 = 0
1 + sin 2x = 2  sin 2x = 1  (2t  1)2 = 0
 1
 sin 2x = sin t=
2 2
 x
 2x = n + (1)n.  tan = tan 
2 2
n  x
x= + (1)n. ….(ii)  = n + 
2 4 2
The value of x [, ] which satisfies both  x = 2n + 2

(i) and (ii) is .  2 
4  
8. tan  + tan     + tan    = 3
 3  3 
5. sin4 x + cos4 x = sin x cos x
 (sin2x + cos2x)2  2 sin2x cos2x = sin x cos x tan   3 tan   3
 tan  + + =3
1 1 1  3 tan  1  3 tan 
 1  (2 sin x cos x)2 = .2sin x cos x
2 2 tan (1  3 tan 2 )  (tan   3) (1  3 tan )
1 1
 1  sin2 2x = sin 2 x  (tan   3) (1  3 tan )
2 2  =3
1  3 tan 2 
 sin22x + sin 2x  2 = 0
9 tan   3tan 3 
 (sin 2x + 2) (sin 2x  1) = 0  =3
1  3tan 2 
 sin 2x = 1 ….[sin 2x  2]
  3tan   tan 3  
 sin 2x = sin  3 =3
 1  3tan  
2
2
  3 tan 3 = 3
 2x = n + (1)n
2 
 tan 3 = 1 = tan
n  4
x= + (1)n
2 4 
 3 = n +
 5 4
 The value of x in [0, 2] are and .
4 4 
  = (4n + 1)
 There are 2 solutions. 12
2
 Chapter 03: Trigonometric Functions
5 20 2 (b  c) 2  2bc  a 2
9.   cos =
6 37 3 2bc
A B 1 (3 3) 2  2 18  a 2
 tan > tan  =
2 2 2 2 18
A B
    18 = 27 + 36  a2
2 2
 a2 = 27 + 36 + 18 = 81
 A>B
A B  a = 9 cm
tan  tan
 A B 2 2
 tan    = 11.
 2 2  1  tan A tan B B
2 2
30
5 20 3 1

 A  B  6 37
 tan   =
 2  1  5  20
6 37 105 45
A C
   C  185  120
 tan  =
 2  222  100
Let B = 30, C = 45  A = 105
  C  305
 tan    = sin A sin B sin C
 2 2  122  
a b c
C 305 sin105 sin 30 sin 45
 cot =   
2 122 3 1 b c
C 122
 tan =
2 305
 b=
 
3 1 sin 30

3 1
20 122 sin105 2sin105
Since, >

B
37
C
305
c=
 
3 1 sin 45
=
3 1
 tan  tan sin105 2 sin105
2 2
B C 1
  A(ABC) = bc sin A
2 2 2
B>C 1 3 1 3 1
=    sin105
A>B>C 2 2sin105 2 sin105
a>b>c
 
2
3 1
9 3 =
10. A(ABC) = 4 2 sin (60  45)
2
 
2

1 9 3 3 1
 bcsin A = =
2 2  3 1 1 1 
4 2    

1
 bc 
3 9 3
=  2 2 2 2
2 2 2
 
2
3 1
 2 3 =
….  sin A  sin    3 1 
 3 2  4 2 
 bc = 18  2 2 
b2  c2  a 2 3 1
cos A = =
2bc 2
3
Std. XII : Triumph Maths 
bc ca ab 2(a  b  c)  n2  3n  4 = 0
12. Let   k =
11 12 13 36  (n + 1) (n  4) = 0
abc  n = 1 or n = 4
= But n cannot be negative.
18
….(By property of equal ratio)  n=4
 b + c = 11k, c + a = 12k, a + b = 13k,  The sides of the  are 4, 5, 6.
a + b + c = 18 k
14.
 a = 7k, b = 6k, c = 5k A
b 2  c2  a 2
cos A =
2bc
E O B
36k  25k 2  49k 2
2
=
2(6k)(5k) r 72 r
12k 2 1
= 2
=
60k 5 D C
1 360
 cos A = In ODC, OD = OC = r, DOC = = 72
5 5
13. A 1 1
 A(ODC) = r.r. sin 72 = r2 sin 72
2 2
n+1 n 5
 A2 = Area of pentagon = r2 sin 72
2
B A1 = Area of circle = r2
C
n+2 A1 r 2
 =
Let AC = n, AB = n + 1, BC = n + 2 A2 5 2
r sin 72
 Largest angle is A and smallest angle is B. 2
 A = 2B 2 2 2 
Since, A + B + C = 180 = = sec 18 = sec
5cos18 5 5 10
 3B + C = 180
 C = 180  3B 15. Let a = 4k, b = 5k, c = 6k
 sin C = sin(180  3B) = sin 3B abc 4k  5k  6k 15k
Now, s = = =
sin A sin B sin C 2 2 2
 = =
n2 n n 1  = s(s  a) (s  b) (s  c)
sin 2B sin B sin 3B
 = = 15k  15k  15k  15k 
n2 n n 1 =   4k   5k   6k 
2  2  2  2 
2sin Bcos B sin B 3sin B  4sin 3 B
 = =
n2 n n 1 15k 7k 5k 3k 15 7 2
=    = k
2cos B 1 3  4sin B2
2 2 2 2 4
 = =
n2 n n 1 a a
By sine Rule, = 2R  sin A =
n2 n 1 sin A 2R
 cos B = , 3  4 sin2B =
2n n 1
n  1 = b csin A
 3  4(1  cos2 B) = 2
n 1 a abc
2   = bc =
n2 n 1 2 2R 4R
 3  4 + 4  =
 2n  n abc 4k.5k.6k 8
 R= = = k
n  4n  4
2
n 1 4 15 7k 2
7
1+ =
n 2
n Also  = rs, where r = Radius of incircle of
  n2 + n2 + 4n + 4 = n2 + n ABC
4
 Chapter 03: Trigonometric Functions

15 7 2 Let length of altitude = p

 k 7
 r= = 4 = k Since, A + B + C = 
s 15k 2
 5
2  A+ + =
R 8 2 16 8 8
 = k =
r 7 7k 7  5 
 A=  =
R 16 8 8 4
 =
r 7 1 1
Area of  = ap = bc sin A
b c a
2 2 2 2 2
16. cos A =
2bc 
 ap = bc sin
43a 2
4
 cos 30 =
4 3 1
 ap = bc 
3 7a 2
2
 =
2 4 3 bc
7a =62  p= ….(i)
2a
 a2 = 1
a=1 ….[ a  1] By sine rule,
1 1 a b c
= bcsin A =  2  3  sin 30 = =
2 2   5
sin sin sin
1 3 4 8 8
= 3 =
2 2 
a sin
a  b  c 1 2  3 3 3 8 = 
s = = =  b= 2a sin
2 2 2 1 8
 = rs 2
 3 2 5
 r = =  a sin
s 2 3 3 8 = 5
c= 2 a sin
3(3  3) 3 3 3 3 1 1 8
= = =
93 6 2 2
17. a4 + b4 + c4 = 2a2(b2 + c2)  From (i),
 a4 + b4 + c4  2a2b2  2a2c2 = 0
 5
 a4 + b4 + c4  2a2b2 + 2b2c2  2a2c2 = 2b2c2 2a sin . 2a sin
5 
8 8 =
 b 2  c2  a 2  =   p= 2a sin sin
2 2
 2bc 2a 8 8
 b2 + c2  a2 = 2bc 2a  5 
=  2sin sin 
b c a
2 2
2bc2
1 2  8 8
 cos A = = =
2bc 2bc 2
a   5    5   
 A = 45 =  cos  8  8   cos  8  8  
 2     
18. A
4 a   3 
=  cos  cos 
2  2 4
p
 a   1 
8
= 0    
2  2 
B a C
a
5  P=
8 2
5
Std. XII : Triumph Maths 
A B 22. A, B, C are in A.P.
19. tan and tan are the roots of the quadratic
2 2  A + C = 2B
2
equation 6x  5x + 1 = 0 Also, A + B + C = 180
 B = 60
A B 5 A B 1
 tan + tan = , tan . tan = sin A sin B sin C
2 2 6 2 2 6   k
a b c
A B 5 5  sin A = ak, sin B = bk, sin C = ck
tan  tan
A  2 2 = 6 = 6 =1
 tan    = a c
 2 2  1  tan A tan B 1  1 5  sin 2C + sin 2A
c a
2 2 6 6
a c
= (2 sin C cos C) + (2sin A cos A)
 AB c a
 tan   1
 2  a c
= (2 ck cos C) + (2ak cos A)
AB  c a
 = = 2ka cos C + 2kc cos A
2 4
= 2k(a cos C + c cos A)
 
 A+B=  C= = 2kb ….[ b = a cos C + c cos A]
2 2
= 2 sin B
 ABC is a right angled triangle.
3
=2 ….[ B = 60]
1 2
acsin B
 2 ac = 3
20. r= = =
s 1 a  bc
(a  b  c)
2 1 1 1
23. 2 cot1 3 = 2 tan1   = tan1 + tan1
….[ sin B = sin 90 = 1]  3 3 3

ac acb  1 1 
 r =  1
 33 
acb acb = tan  
ac(a  c  b) ac(a  c  b) 1 1  1 
= = 2  3 3
(a  c)  b
2 2
a  c 2  2ac  b 2
 33
acb = tan1  
= 2 2
….[ a + c = b ] 2
 9 1 
2
 Diameter = a + c  b 6
= tan1  
8
21. A = 55, B = 15, C = 110 3
a b c = tan1
 = = =k 4
sin 55 sin15 sin110
  1
 cot   2cot 1 3  =
 a = k sin 55, b = k sin 15, c = k sin 110  4    3
tan   tan 1 
 c2  a2 = k2 sin2 110  k2 sin2 55 4 4
= k2(sin 110 + sin 55) (sin 110  sin 55)   3
1  tan tan  tan 1 
 165 55   55 165  4  4
= k2  2sin cos =
  2sin cos    1  
 2 2   2 2  tan  tan  tan 
4  4
= k2 sin 165 sin 55
3
= k2 sin 15 sin 55 1  1
= 4  43 = 7
= (k sin 55) (k sin 15) 3 43
1
= ab 4
6
 Chapter 03: Trigonometric Functions

1 a  
24. Let cos 1   =   2x = sin   sin 1 x 
2 b 3 
a  
 cos1   = 2 = sin cos (sin1 x) – cos sin (sin1 x)
b 3 3
a 3 1
 cos 2 =  2x = cos (sin1 x)   x …. (i)
b 2 2
 1  a   1 Let sin1 x = 
1  a  
 tan   cos 1    + tan  4  2 cos  b    sin  = x
4 2  b    
cos  = 1  x 2
   
= tan     + tan      cos (sin1 x) = 1  x 2 ….(ii)
4  4 
1  tan  1  tan  From (i) and (ii), we get
=  3 1
1  tan  1  tan  2x =  1  x2  x
2 2
1  tan    1  tan  
2 2

=  4x = 3 1  x2  x
1  tan  2

2 1  tan 2    5x = 3 1  x 2
=  25x2 = 3  3x2 (squaring both sides)
1  tan 2 
 28x2 = 3
2 2 2 2b
= = = = 3
1  tan  cos 2
2
a a  x2 =
28
1  tan 
2
b
3 1 3 1 3
 x= =  =
25. cos1   cos1  = cos1   1   2 1  2  28 4 7 2 7
  (From the given relation it can be seen that x is
y positive)
Given, cos1 x – cos1 =
2
 33  1   
 xy  y2   27. L.H.S. = sin1  sin  + cos  cos 
 cos1   1  x 2  1    =   7   7 
 2  4 
  13  1  19 
+ tan1   tan  + cot   cot 
xy  2
  8   8 
 cos  =
2
 1  x  1  y4 
2
  2   1     
  = sin1 sin  5    + cos  cos     
  7    7 
 2
 xy
 1  x  1  y4 
2
= cos  
2     
+ tan1   tan     
 
  8 
 2

 2 1  x  1  y4 
2
= 2 cos   xy     
+ cot1   cot     
  8 
 
Squaring on both sides, we get
 2   3 
 y2  = sin1  sin   cos 1   cos 
4(1  x2) 1   = 4 cos2   4xy cos  + x2y2  7   7 
 4 
     
 4  y2  4x2 + x2y2 = 4 cos2   4xy cos  + x2y2 + tan1  tan  + cot1  cot 
 8   8 
 4x2 + y2  4xy cos  = 4  4 cos2 
 4x2 + y2  4xy cos  = 4 sin2  2 3 3 5
=   
7 7 8 8
 ….[ cos1 (x) =   cos1 x]
26. sin1 x + sin1 2x =
3
  13
 =   = 2  
 sin1 2x =  sin1 x 7 7 7
3
7
Std. XII : Triumph Maths 
13 a 1  1 1 
3
   3 sin1   = sin1 3   4   
7 b  3  3  3  
 a = 13, b = 7
 4 
 a + b = 13 + 7 = 20 = sin1 1  
 27 
4 5 16  23 
28. sin1 + sin1 + sin1 = sin1   = sin1 (0.852)
5 13 65  27 
4 2
 4  
2
3 1.732
 1    
1 5 5  = 0.866, 0.852 < 0.866
= sin 1  
5  13  13  5   2 2

 sin1 (0.852) < sin1 (0.866)
1 16
+ sin ….[ sin1 x is also an increasing function]
65
 3
 4 12 5 3  1 16 1
1  3 sin1   < sin1  2 
= sin  5  13  13  5  + sin 65  3  
 
 48  15  1  16  1 
= sin1   + sin    3 sin1   < ....(ii)
 65   65   3 3
 63   16   3
= sin1   + sin1 3
  sin1   = sin1 (0.6) < sin1  2 
 65   65  5  
  63  
2
 16 
= cos1  1     + sin1 3 
   sin1   < ….(iii)
  65    65  5 3

From (ii) and (iii), we get
 16   16 
= cos1   + sin1   1 3   
 65   65  B = 3 sin1   + sin1   < + =
 3 5 3 3 3

= 
2  B< ….(iv)
3
29. 2 = 1.414 From (i) and (iv), A > B
 2 2  1 = 2  1.414  1 = 2.828  1 = 1.828 
30. cot1 x + cot1 y + cot1 z =
 2 21> 3 ….[ 3  1.732 ] 2

 tan1 (2 2  1) > tan1  3 


2
 tan1 x +

2
 tan1 y +

2
 tan1 z =

2
….[ tan1 x is an increasing function]  1 1 1
tan x + tan y + tan z = 
  tan (tan1 x + tan1 y + tan1 z) = tan  = 0
 2 tan1 (2 2  1) > 2 
3 Let A = tan1 x, B = tan1 y, C = tan1 z
 tan  A  B   tan C
 A> ….(i)  tan (A + B + C) =
3 1  tan (A  B) tan C
sin 3  = 3 sin   4 sin3  tan A + tan B
 tan C
 3 = sin1 (3 sin   4 sin3 ) 1  tan A tan B
=
1 tan A  tan B
Put sin  = 1  tan C
3 1  tan A tan B
1 tan A  tan B  tan C  tan A tan B tan C
  = sin1   =
 3 1  tan A tan B  tan B tan C  tan C tan A

8
 Chapter 03: Trigonometric Functions
 tan (A + B + C) = 0  x2 (6x + 2) = 2(8x2 + 6x)
 tan A + tan B + tan C = tan A tan B tan C  6x3 + 2x2 – 16x2  12x = 0
 6x3  14x2  12x = 0
 tan (tan1 x) + tan(tan1 y) + tan(tan1z)
 3x3  7x2  6x = 0
= tan(tan1 x) tan(tan1 y) tan(tan1 z)  x(3x2  7x  6) = 0
 x + y + z = xyz  x(x  3) (3x + 2) = 0
2
 1  9 9    x = 0, 3, 
31. cos1   cos  sin   3
 2 10 10   But x > 0,  x=3
  9  9  1 
= cos1 cos cos  sin sin  34. cot1 x + sin1 =
 4 10 4 10  5 4
   9   1
= cos1  cos    1 
  4 10    tan1 + tan1 5 =
x 1 4
1   5  18   1
= cos cos  20   5
  
 x 
….  sin 1 x  tan 1 
  23   1  x2 
= cos1  cos   
  20   1 1 
 tan1  tan 1 
1   23   x 2 4
= cos  cos  2  20    1 1 
  
   
  17    17  tan1  x 2  =
= cos1 cos    and 0 ≤ ≤ 1 1  1  4
  20   20  x 2 
17 2 x 
=  = tan = 1
20 2x 1 4
17  2 + x = 2x – 1
 Principal value is .  x=3
20

 23 
32. tan1 2 + tan1 3 =  + tan1  
1 23 
….[ 2  3 > 1]
=  + tan1 (1)
=   tan1 1
 tan1 1 + tan1 2 + tan1 3 = 
1 1 2
33. tan1 + tan1 = tan1 2
1  2x 4x  1 x
 1 1 
   2
 tan1  1  2 x 4 x  1  = tan1 2
1 1  1  x
 1  2x 4x  1 
4x  1  2x  1 2
 = 2
1  2 x  4 x  1  1 x
6x  2 2
 = 2
4 x  8x  1  2 x  1 x
2

9
Target Publications Pvt. Ltd. Chapter 04: Pair of Straight Lines

04 Pair of Straight Lines

1. L1: ax2 + 2hxy + by2 = 0 2 1

2  =0
Equation of any line passing through origin 4 4
and perpendicular to L1 is given by 2 9
 =  2 = 9   =  3
bx2  2hxy + ay2 = 0 4 4
….(interchanging coefficients of x2 and y2 and
change of sign for xy term) 5. The given equation of pair of lines is
 The required equation of pair of lines is x2 + 2 2 xy – y2 = 0
–15x2 + 7xy + 2y2 = 0  a = 1, b = 1, h = 2
i.e. 15x2 – 7xy – 2y2 = 0 Now, a + b = 1 + (1) = 0
2h  The lines are perpendicular
2. Here, m1  m2  .....(i)
b 6. The joint equation of the lines through the
a point (x1, y1) and at right angles to the lines
and m1m2  ax2 + 2hxy + by2 = 0 is
b
b(x – x1)2 – 2h(x – x1)(y – y1) + a(y – y1)2 = 0
 (m1 – m2)2 = (m1 + m2)2 – 4m1m2
 joint equation of pair of lines drawn through
4h 2  4ab (1, 1) and perpendicular to the pair of lines
=
b2 3x2 – 7xy + 2y2 = 0 is
4h 2  3h 2 2(x – 1)2 + 7(x – 1)(y – 1) + 3(y – 1)2 = 0
= 2
….[ 4ab = 3h2 (given)]
b 7. The given equations are x – y – 1 = 0 and
h2 2x + y – 6 = 0
  The joint equation is given by
b2
(x – y – 1) (2x + y – 6) = 0
h
 m1  m2  .....(ii)  2x2 + xy – 6x – 2xy – y2 + 6y – 2x – y + 6 =0
b  2x2 – y2 – xy – 8x + 5y + 6 =0
On solving (i) and (ii), we get
h 3h 8. Let the equation of one of the angle bisector of
m1  and m2  the co-ordinate axes be x + y = 0  m1 = –1
2b 2b
Given equation of pair of lines is
 m1 : m2 = 1 : 3 2x2 + 2hxy + 3y2 = 0
3. The lines are parallel, if af2 = bg2  A = 2, H = h, B = 3
 4f2 = 9g2 a 2
Now, m1m2 =  m2 =
3 b 3
f= g
2 2h 2 2h
Also m1 + m2 =  –1 – =
Let g = 2 and f = 3 b 3 3
 abc + 2fgh – af2 – bg2 – ch2 5
h=
= 4 (9) (c) + 2 (3) (2) (6) – 4(3)2 – 9(2)2 – c (6)2 2
=0 9. The given equation of pair of lines is
 c is any number. 3x2 – 2y2 + xy – x + 5y – 2 = 0
4. Given equation is x2  y2  x  y  2 = 0. 5 1 
 a = 3, b = –2, c = –2, f = , g = ,h=
 1 2 2 2
 a = 1, b = 1, c = 2, f = ,g= ,h=0 Now abc + 2fgh – af2 – bg2 – ch2 = 0
2 2
This equation represents a pair of straight 5 75 1  2
12 –    =0
lines, if abc + 2fgh  af2  bg2  ch2 = 0 4 4 2 2
1
Std. XII : Triumph Maths 
 22 – 5  25 = 0
 ( 5)(2 + 5) = 0
5
  = 5 or
2
10. Let y = mx be the common line and let y = m1x
and y = m2x be the other lines given by
2x2 + axy + 3y2 = 0 and 2x2 + bxy  3y2 = 0
respectively. Then,
a 2
m + m1 =  , mm1 = , and
3 3
b 2
m + m2 = , mm2 = 
3 3
2  2
 (mm1) (mm2) =   
3  3
4
 m2(m1m2) = 
9
4
 m2 = ….[ m1m2 = 1 (given)]
9
2
m=
3
2
When m = ,
3
2 2
mm1 = and mm2 =   m1 = 1 and m2 = 1
3 3
a b
 m + m1 =  and m + m2 =
3 3
 a = 5 and b = 1
2
When m =  ,
3
2 2
mm1 = and mm2 =  m1 = 1 and m2 = 1
3 3
a b
 m + m1 =  and m + m2 =
3 3
 a = 5 and b = 1
11. Given equation of pair of lines is
3x2 – 48xy + 23y2 = 0
 a = 3, h = –24, b = 23
2 576  69
 tan  =
3  23

2 507 2  13 3
 tan  = = = 3
26 26

  = tan–1 ( 3 ) =
3

2
 Chapter 05: Vectors

05 Vectors

1. Since, a  b and b  c are collinear with c 4. aˆ  aˆ 1, bˆ  bˆ 1, cˆ  cˆ  1,

1
and a respectively aˆ  bˆ  bˆ  cˆ  cˆ  aˆ 
2
 a  b  tc …(i)
aˆ  aˆ aˆ  bˆ aˆ  cˆ
2
b  c  sa …(ii)    ˆ ˆ ˆ ˆ
 a b c   b  aˆ b  b b  cˆ
From (i) and (ii), we get cˆ  aˆ cˆ  bˆ cˆ  cˆ
a  c  tc  sa  a(1  s)  c(1  t) 1 1
1
But a and c are non-collinear 2 2
1 1 1
 1 + s = 0, 1 + t = 0  s = 1, t = 1  1 
2 2 2
Substituting value of t in (i) and value of s in 1 1
(ii), we get 1
2 2
a  b  c and b  c  a a b c  
1
   cubic units
Hence, a  b  c  0 . 2

a 1 1
2. Given, r  1 r 1   2 r 2   3 r 3
5. Since, 1 b 1  0
 2a  3b  4c  (1   2   3 )a 1 1 c

 ( 1   2   3 )b  (1   2   3 ) c Applying R2  R2  R1 and R3  R3  R1,

we get
 1   2   3  2, 1   2   3 = 3, a 1 1
1   2   3  4 1 a b 1 0  0
7 1 1 a 0 c 1
 1  , 2 = 1, 3 = 
2 2  a(b  1)(c  1)  (1  a)(c  1)  (1  a)(b  1)  0
 1 + 3 = 3 Dividing by (1  a)(1  b)(1  c), we get
a 1 1
3. Since, the given vectors are coplanar   0 ….(i)
1 a 1 b 1 c
a a c 1 1 1
Consider,  
 1 0 1 =0 1 a 1 b 1 c
c c b 1 a
=  ….[From (i)]
1 a 1 a
Applying C2  C2 – C1, =1
a 0 c
6. Volume of the parallelopiped formed by
1 1 1 = 0 vectors is
c 0 b 1 a 1
 a (–b) + c (c) = 0  c2 = ab i.e., V = 0 1 a = 1  a + a3
a 0 1
Hence, c is the geometric mean of a and b.
1
Std. XII : Triumph Maths 

dV d2V 9. Let c  2iˆ  3jˆ  4kˆ

 = 1 + 3a2, = 6a
da da 2  a c  c b
dV
For max. or min. of V, =0  a  c  b  c
da
1  
 a  b c  0
 a2 =
3   a  b  || c
1
 a= Let  a  b   c
3
2
d V 1  ab   c
2
= 6a > 0 for a =
da 3
 29  |  | . 29
1
 V is minimum for a =   = 1
3
 a  b = (2iˆ  3jˆ  4k)
ˆ
7. Given, a .b  b.c = c.a = 0   
Now, a  b . 7iˆ  2ˆj  3kˆ   (14  6  12)
The scalar triple product of three vectors is
=4
[ a b c ] = (a  b).c
10. Given,
 a .b  0  ab
l a  mb  nc l b  mc  na l c  ma  nb   0
 
 angle between a and b is  = 90
Similarly, [ a b c ] = | a | | b |n.c ˆ where n̂ is a  l a  mb  nc na  l b  mc ma  nb  l c   0
normal vector. l m n
 n̂ and c are parallel to each other  n l m a b c   0
 [ a b c ] = | a | | b | | nˆ |.| c |  | a | | b | | c | . m n l

8. Given, r  b = c  b l m n
 n l m 0 ....   a b c   0 
 
 r  c  b= 0
m n l
 

 r  c is parallel to b  l3 + m3 + n3  3lmn = 0
 r  c =  b for some scalar   (l + m + n) (l2 + m2 + n2  lm  mn  nl) = 0
 r = c  b ….(i) l+m+n=0

 
 r . a = c.a +  b  a 11.
P
A

 0 = c.a +   b  a  M

….  r  a  0(given)  H

O
a.c
= B C
a.b D
Substituting the value of  in (i), we get
a.c
r= c b
a.b
Let point O be the circumcentre of ABC.
a.c
 r.b = c.b  (b.b)
a.b Let a , b , c , p , d , h , m be the position
(4) vectors of the respective points.
 r.b 1  2=9
1 Since, h = a + b + c ….(Standard formula)

2
 Chapter 05: Vectors

ph pabc  mx  nx1 my2  ny1 mz 2  nz1 

 m= = D  2 , , 
2 2  mn mn mn 
pabc bc  3(9)  13(5) 3(6)  13(3) 3( 3)  13(2) 
 DM  m  d =   , , 
2 2  3  13 3  13 3  13 

pa  27  65 18  39 9  26 
=  , , 
2  16 16 16 
 38 57 17 
pa  , , 
 DM  PA =   ap    16 16 16 
 2 
 19 57 17 
1  , , 
=  a 2  p2   8 16 16 
2
17. A(x1, y1, z1)
=0
….[ O is circumcentre,  OA = OP i.e., a = p] (l, 0, 0) (0, 0, n)
 DM is perpendicular to PA.

15. Let position vector of Q be r B(x2, y2, z2) (0, m, 0) C(x3, y3, z3)
Since, p divides PQ in the ratio 3 : 4 x1 + x2 = 2l, x2 + x3 = 0, x3 + x1 = 0
3r  4(3p  q) On solving we get x1 = l, x2 = l, x3 = l
 = p
3 4 y1 + y2 = 0, y2 + y3 = 2m, y3 + y1 = 0
On solving we get y1 = m, y2 = m, y3 = m
 7 p = 3 r + 12 p + 4 q z1 + z2 = 0, z2 + z3 = 0, z3 + z1 = 2n
 – 5p – 4q = 3 r On solving we get z1 = n, z3 = n, z2 = n
1  A(l, m, n), B(l, m, n), C(l, m, n)
 r =
3

5p  4q  By distance formula,
AB2 = (l  l)2 + (m m)2 +(n + n)2 = 4m2 + 4n2
16. A(3, 2, 0) BC2 = (l + l)2 + (m  m)2 +(n n)2 = 4l2 + 4n2
 CA2 = (l + l)2 + (m m)2 +(n  n)2 = 4l2 + 4m2
3 13 AB2  BC2  CA 2

l 2  m2  n 2
4m 2  4n 2  4l 2  4n 2  4l 2  4m 2
=
B(5, 3, 2) D C (–9, 6, –3) l 2  m2  n 2

By distance formula, =8
l 2
 m2  n 2 
=8
l 2  m2  n 2
AB = (5  3)  (3  2)  (2  0)
2 2 2

18. A(1, 0, 3)
= 4 1 4
= 9=3

AC = (3  9) 2  (2  6) 2  (0  3) 2

= 144  16  9
B(4, 7, 1) D C(3, 5, 3)
= 169 = 13
Let D be the foot of perpendicular and let it
 Point D divides seg BC in the ratio of 3 : 13 divide BC in the ratio  : 1 internally
 By section formula,

3
Std. XII : Triumph Maths 

 3  4 5  7 3  1 
 D  , , 
  1  1  1 
AD = d  a
 3  4  ˆ  5  7  ˆ  3  1  ˆ ˆ ˆ
= i    j  k  i  3k
  1    1    1 
 2  3  ˆ  5  7  ˆ  2  ˆ
= i    j k
  1    1    1

BC = 3iˆ  5jˆ  3kˆ  4iˆ  7ˆj  kˆ 
= ˆi  2ˆj  2kˆ
Since, AD  BC .
AD . BC = 0
 2  3   5  7   2 
  ( 1)    ( 2)    (2)  0
  1    1    1
 2  3  10  14  4 = 0
  12  21 = 0
7
= 
4
  7  7  7 
 3  4   4 5   4   7 3  4   1 
 D    ,   ,   
  7 1 7
 1
7
  1 
 4 4 4
 
 21  16 35  28 21  4 
 , , 
 7  4 7  4 7  4 
 5 7 17 
 , , 
3 3 3 

4
 Chapter 06: Three Dimensional Geometry

06 Three Dimensional Geometry 

1.  =  = 2 Since, AD is equally inclined to the axes

 p5 r  12
  = ,  =  =1=
2 2 2
Since, cos2 + cos2 + cos2 = 1  p = 7, r = 14

 cos2 + cos2 + cos2 =1 4. The d.r.s of AB are 3  1, 2  4, 6  5
2
i.e. 2, –2, 1
1  cos 
 2cos2 + =1 Let a1, b1, c1 = 2, 2, 1
2
d.r.s. of BC are 1 – 3, 4 – 5, 5  3
 4 cos2 + cos  1 = 0
i.e., 2, 1, 2
1  1  16 1  17 Let a2, b2, c2 = 2, 1, 2
 cos  = 
2(4) 8
 a1a2 + b1b2 + c1c2 = 2(2) + (2)(1) + 1(2)
If  is acute, then cos  is positive. =4+2+2=0
17  1  AB and BC are perpendicular.
 cos  =
8  mABC = 90

2. cos = l1l2 + m1m2 + n1n2 5. The given equations are

2 2
sin = 1 – cos  6mn  2nl + 5lm = 0, and ….(i)
2
= 1.1  cos  3l + m + 5n = 0
=  l12  m12  n12   l22  m 22  n 22   m = –3l  5n ….(ii)
Substituting value of m in equation (i),
 (l1l2 + m1m2 + n1n2)2
we get
= l12 l22  l12 m 22  l12 n 22  l22 m12  m12 m 22  m12 n 22 6(3l  5n)n  2nl + 5l(– 3l  5n) = 0
+ l n  m n  n n l l  m m  n n
2 2
2 1
2 2
2 1
2
1
2
2
2 2
1 2
2
1
2
2
2
1
2
2  18ln  30n2  2nl  15l2  25nl = 0
2l1l2 m1m 2  2m1m 2 n1n 2  2n1n 2l1l2  15l2 + 45ln + 30n2 = 0
= l12 m 22  2l1l2 m1m 2  l22 m12  m12 n 22  2m1m 2 n1n 2  l2 + 3ln + 2n2 = 0
 (l + n)(l + 2n) = 0
 m 22 n12  l22 n12  2l1l2 n1n 2  l12 n 22
 l = n or l = 2n
= (l1m2  l2m1)2 + (m1n2  m2n1)2 + (n1l2  n2l1)2
If l = n, then m = 2n
3. Given, A(2, 3, 7), B(1, 3, 2), C(p, 5, r) l n m n
  and 
Let D be the midpoint of BC. 1 1 2 1
 1  p 3  5 2  r   p  1 r  2  l m n
 D  , ,   , 4,    
 2 2 2   2 2  1 2 1
p 1 r2  d.r.s. of the 1st line are 1, 2, 1.
 d.r.s. of AD are 2, 4 – 3, 7
2 2 If l = 2n, then m = n
p5 r  12 l n m n
i.e., , 1,   and 
2 2 2 1 1 1
1
Std. XII : Triumph Maths 
l m n 1
    cos  = 
2 1 1 3
 d.r.s. of the 2nd line are 2, 1, 1. 1
 l = m = n = cos  = 
1  (2)  2  1  (1)  1 3
 cos  =
1  22  (1) 2 ( 2) 2  12  12
2

2  2  1 1
= 
6 6 6
 1 
  = cos1  
 6 

6. Since, (l  m)2  0
 l2  2lm + m2  0
 l2 + m2  2lm ….(i)
2 2
Similarly, m + n  2mn ….(ii)
2 2
and n + l  2nl ….(iii)
Adding (i), (ii) and (iii), we get
2(l2 + m2 + n2)  2(lm + mn + nl)
 lm + mn + nl  1
 The maximum value of lm + mn + nl is 1.
7. Let A = (a, 2, 3), B  (3, b, 7) and
C  (3, 2, 5)
d.r.s of AB are 3  a, b  2, 4
d.r.s of BC are 6, 2b, 12
Since the points are collinear
3a b  2 4
  
6 2  b 12
 a = 2, b = 4
8. Let the d.r.s of the line perpendicular to both
the lines be a, b, c.
d.r.s of lines is 1, 1, 0 and 2, 1, 1
 ab=0 ….(i)
2a  b + c = 0 ….(ii)
On solving (i) and (ii), we get
a b c
 
1 1 1
 d.r.s of the line are 1, 1, 1
1 1 1
 the required d.c.s are , ,
3 3 3

9. Since cos2  + cos2  + cos2  = 1

 cos2  + cos2  + cos2  = 1 ….[  =  = ]
 3 cos2  = 1
2
Target Publications Pvt. Ltd. Chapter 07: Line

07 Line

x  1 y  12 z  7 63 63
1. Let   =r  k2 = = 81
1 5 2 49
 x =  r  1, y = 5r + 12, z = 2r + 7  k=9
 Co-ordinates of any point on the line are Since, the line makes obtuse angle with X-axis
(r  1, 5r + 12, 2r + 7). component along X-axis is negative.
This point lies on the curve 11x2 – 5y2 + z2 = 0  k = 9
 11( r  1)2  5(5r + 12)2 + (2r + 7)2 = 0  The components of the line vector are 3k, 2k, 6k
 11r2 + 22r + 11  125r2 – 600r  720 i.e., 27, 18, 54
+ 4r2 + 28r + 49 = 0 4. Let M be the foot of the perpendicular drawn
2
  110r – 550r – 660 = 0 from the point P(3, 1, 11) to the given line.
 r2 + 5r + 6 = 0 x y 2 z3
 (r + 2)(r + 3) = 0 Let   
2 3 4
 r = 2 or r = 3
 x = 2, y = 3 + 2, z = 4 + 3
If r = 2, then the point is (1, 2, 3)
 M  (2, 3 + 2, 4 + 3)
and if r = 3, then the point is (2, 3, 1)
d.r.s. of PM are 2  3, 3 + 3, 4  8
 option (A) is correct. Since, PM is perpendicular to the given line
2. The given equation of line is  (2  3)(2) + (3 + 3)(3) + (4  8)(4) = 0
x = 4y + 5, z = 3y  6.  4  6 + 9 + 9 + 16  32 = 0
It can be written as =1
x5 z6  M  (2, 5, 7)
y= = r, say
4 3  length of perpendicular (PM)
 co-ordinates of the any point on the line are = (3  2) 2  (1  5) 2  (11  7) 2
(4r + 5, r, 3r  6).
= 1  36  16
This point is at a distance of 3 26 from the
= 53
point (5, 0, 6)
 
2
 (4r + 5  5)2 + (r  0)2 + (3r  6 + 6)2 = 3 26 5. When square is folded co-ordinates will be
D(0, 0, a), C(a, 0, 0), A(– a, 0, 0), B(0, – a, 0).
 16r2 + r2 + 9r2 = 234 Y
 26r2 = 234
 r2 = 9 D
 r = 3
a
If r = 3, then the point is
(4  3 + 5, 3, 3  3  6)  (17, 3, 3) a a
X A C X
3. Let the components of the line vector be a, b, c. a
 a2 + b2 + c2 = (63)2 ….(i)
a b c B
Also,    k , say
3 2 6 Y
 a = 3k, b = 2k, c = 6k xa y z
Substituting value of a, b and c in equation (i), Equation AB is,  
we get a a 0
9k2 + 4k2 + 36k2 = 632 x y za
and equation of DC is  
 49k2 = 63  63 a 0 a
1
Std. XII : Triumph Maths 
 shortest distance Let S divide AB in the ratio  : 1
a 0 a  3  4 5  7 3  1 
 S  , ,  ….(i)
a a 0   1  1  1 
a 0 a Now, d.r.s. of PS are
=
(a 2  0) 2  (0  a 2 ) 2  (0  a 2 ) 2 3  4 5  7 3  1
1 , 0 , 3
 1  1  1
2  3 5  7 2
i.e., , ,
 1  1  1
a(a 2 )  a(a 2 ) 2a 3 2a i.e., 2 + 3, 5 + 7, 2
= = =
a a a
4 4 4
3a 4
3 Also, d.r.s. of AB are 1, 2, 2
6. Given equation of motion of a rocket is Since, PS  AB
x = 2t, y = 4t, z = 4t  (2 + 3)(1) + (5 + 7)(2) + (2)(2) = 0
x y z   2  3  10  14  4 = 0
i.e., the equation of the path is  
2 4 4 7
=
x y z 4
i.e.,  
1 2 2 Substituting the value of  in (i), we get
Thus, the path of the rocket represents a  5 7 17 
straight line passing through the origin. S=  , , 
3 3 3 
For t = 10 sec.
we have, x = 20, y = –40, z = 40 9. Equation of the line passing through the points
Let M(20, 40, 40) (5, 1, a) and (3, b, 1) is
 OM = x2  y 2  z2 x  3 y  b z 1
  ….(i)
5  3 1 b a 1
= 400  1600  1600 = 60 km
 17 13 
 Rocket will be at 60 km from the starting The line passes through the point  0, , 
 2 2 
point O(0, 0, 0) in 10 seconds.
17 13
7. d.r.s. of L1 are 3, 1, 2 and d.r.s. of L2 are 1, 2, 3 b 1
3
ˆi ˆj kˆ  = 2  2 ….[From (i)]
2 1 b a 1
 vector perpendicular to L1 and L2 = 3 1 2 15
1 2 3  a–1= 2 =5
3
= ˆi(3  4)  ˆj(9  2)  k(6
ˆ  1)
2
= ˆi  7ˆj  5kˆ  a=5+1=6
ˆi  7ˆj  5kˆ ˆi  7ˆj  5kˆ and 3 + 3b = 17 – 2b
 unit vector = =  5b = 20  b = 4
1  49  25 5 3
 a = 6, b = 4
8. Let S be the foot of perpendicular drawn from
P(1, 0, 3) to the join of points A(4, 7, 1) and
B(3, 5, 3)
P (1, 0, 3)

 1
A(4,7,1) S B(3,5,3)

2
Target Publications Pvt. Ltd. Chapter 08: Plane

08 Plane

1. Given planes are 3. Given euation of line and plane are

x  cy  bz = 0
cx  y + az = 0
….(i)
….(ii)
 
r = ˆi  ˆj   2iˆ  ˆj  4kˆ , and

bx + ay  z = 0 ….(iii) r.  ˆi  2ˆj  kˆ   3
Equation of a plane passing through the line of
intersection of planes (i) and (ii) is  b  2iˆ  ˆj  4kˆ and
x  cy  bz + k(cx  y + az) = 0 n  ˆi  2ˆj  kˆ
 (1 + ck)x  (c + k)y  (b  ak)z = 0 ….(iv)
Now, planes (iii) and (iv) are same for some
 
Consider b  n = 2iˆ  ˆj  4kˆ  ˆi  2ˆj  kˆ 
value of k, =2+2–4
1  ck c  k (b  ak) =0
 =  =
b a 1  the line lies in the plane.
1  ck ck
 =  4. The equation of the given line is
b a
1 1
 a + ack = bc – bk x = 2 + t, y = 1 + t, z =   t
2 2
 k(b + ac) = (a + bc)
1
 a  bc  z
k=  x  2 y 1 2
   
 b  ac  1 1 1

ck 2
Also,  = b  ak
a  The given line passes through the point
 a  bc   1 1
 c  b  ac   2,1,   and it’s d. r.s are 1, 1, 
 a  bc   2 2
    b  a 
 a   b  ac  The equation of the given plane is
  x + 2y + 6z = 10
 bc  ac 2  a  bc  d.r.s of the normal to the plane are 1, 2, 6
 = b2 + abc + a2 + abc
a ax1  by1  cz1  d
 p=
 1 – c2 = a2 + b2 + 2abc a 2  b2  c2
 a2 + b2 + c2 + 2abc = 1
 1
1(2)  2(1)  6     10
2. Let a, b, c be the intercepts form by the plane  2
=
on co-ordinate axes. 12  22  62
1 1 1 1
Since,   
a b c 2
2  2  3  10 9
2 2 2 = =
   1 1  4  36 41
a b c
 9
 The point (2, 2, 2) satisfies the equation of the  =
x y z  41
plane    1 .
a b c   = 9,  = 41
 the required point is (2, 2, 2).  5   = 5(9)  41 = 45 – 41 = 4
1
Std. XII : Triumph Maths 

5. Let a be the vector along the line of Let A  (2, 1, 3), AM be  to the given plane
intersection of the planes 3x  7y  5z = 1 and and let B  (x, y, z) be the image of A in the
8x – 11y + 2z = 0. the d.r.s of the normals to Plane.
the planes are 3, 7, 5 and 8, 11, 2. the d.r.s. of the normal to the plane are 3, 2, 1
ˆi ˆj kˆ  The equation of the line AM is
 a = 3 7 5 x  2 y 1 z  3
  = k, say
8 11 2 3 2 1
= ˆi(14  55)  ˆj(6  40)  k(
ˆ 33  56)  x = 3k + 2, y = 2k  1, z = k + 3
Let M  (3k + 2, 2k  1, k + 3)
= 69iˆ  46ˆj  23kˆ
 equation of plane becomes
Similarly, let b the vector along the line of
3(3k + 2)  2(2k  1)  (– k + 3) = 9
intersection of the planes 5x  13y + 3z + 2 = 0
and 8x – 11y + 2z = 0 2
 k=
the d.r.s of the normals to the planes are 7
5, 13, 3 and 8, 11, 2 6 4 2   20 11 19 
 M    2,   1,   3    ,  , 
ˆi ˆj kˆ 7 7 7   7 7 7
 b = 5 13 3 Since, M is the mid point of AB.
8 11 2 x1  2 20 y1  1 11 z  3 19
 = ,  , 1 
2 7 2 7 2 7
= ˆi(26  33)  ˆj(10  24)  k(
ˆ 55  104)
26 15 17
= 7iˆ  14ˆj  49kˆ  x1 = , y1 =  , z1 =
7 7 7
Consider,
 26 15 17 
  
a . b = 69iˆ  46ˆj  23kˆ . 7iˆ  14ˆj  49kˆ Image of A is B  ,  , 
 7 7 7 
= 69  7 + (46)  14 + 23  49
= 483  644 + 1127 8. Since, a and b are coplanar, a  b is a vector
=  1127 + 1127 perpendicular to the plane containing a and b .
=0
 a and b are perpendicular Similarly, c  d is a vector perpendicular to the
  = 90 plane containing c and d .
 sin = sin 90 = 1 The two planes will be parallel, if their normals
6. The equation of the given plane is a  b and c  d are parallel.
2x  (1 + )y + 3z = 0
 2x  y  y + 3z = 0  a  b   c  d   0
 (2x  y)  (y  3z ) = 0
1 9. Equation of the plane containing the given
 (2x  y)  (y  3z) = 0 lines is

 The plane passes through the point of x 1 y  2 z  3
intersection of the planes 2x  y = 0 and 2 3 4 =0
y  3z = 0 3 4 5
7. A(2, 1, 3)  (x  1) (15  16)  (y  2) (10  12)
+ (z  3) (8  9) = 0
 (x  1) (1)  (y  2) (2) + (z  3) (1) = 0
3x – 2y – z = 9
M  x + 1 + 2y  4  z + 3 = 0
 x + 2y  z = 0
B  x  2y + z = 0 ….(i)
2
 Chapter 08: Plane
Given equation of plane is The equation of the plane is
Ax  2y + z = d ….(ii) x y z
  =1
The planes given by equation (i) and (ii) are a b c
parallel. Since, this plane is at a distance of 1 unit from
 A=1 the origin,
distance between the planes (D) is
1
d d =1
D=  1 1 1
12   2   12
2
6 2
 2 2
a b c
d 1 1 1
 = 6  2
 2  2 =1
6 a b c
 |d| = 6 1 1 1
 2  2  2 = 1 ….[From (i)]
9x 9y 9z
10. P(2, 1, 2)
1 1 1
 2
 2  2 =9
x y z
k=9
Q 2x + y + z = 9
12. Let the equation of the plane OAB be
ax + by + cz = d
Since, direction cosines of PQ are equal and This plane passes through the points A(1, 2, 1)
positive and B(2, 1, 3)
1 1 1  a + 2b + c = 0, …(i)
 the d.r.s. of PQ are , ,
3 3 3 and 2a + b + 3c = 0 …(ii)
 The equation of the line PQ is  on solving (i) and (ii), we get
x  2 y 1 z  2 a b c
   
1 1 1 5 1 3
3 3 3 Similarly, let the equation of the plane ABC be
 x – 2 = y + 1 = z  2 = k, say a(x + 1) + b(y  1) + c(z  2) = 0
 Co-ordinate of the point Q are Substituting the co-ordinates of A and B, we get
(k + 2, k  1, k + 2) 2a + b  c = 0,
The point Q lies on the plane 2x + y + z = 9 and 3a + c = 0
 2(k + 2) + k  1 + k + 2 = 9 a  b c
  
 4k + 5 = 9 k=1 1 5 3
 Q  (3, 0, 3) If  is the angle between two planes, then it is
the angle between their normals.
3  2   0  1   3  2 
2 2 2
 PQ =
51 (1)  ( 5)  (3)  (3)
 cos  =
= 111 = 3 25  1  9 1  25  9
11. Let A  (a, 0, 0), B  (0, b, 0), C  (0, 0, c) 559
=
a b c 35 35
 G  (x, y, z)   , , 
 3 3 3 19
=
a b c 35
 = x, = y, = z
3 3 3  19 
  = cos1  
 a = 3x, b = 3y, c = 3z ….(i)  35 
3
Std. XII : Triumph Maths 
13. The equation of the given plane can be written
x y z
as   =1
20 15 12
Let the plane intersects the x, y and z axes in the
points A(20, 0, 0), B(0, 15, 0), C(0, 0, 12)
 ˆ b 15jˆ , and c   12kˆ
a  20i,
1
 Volume of tetrahedron = a b c 
6 
20 0 0
1
= 0 15 0 = 600 = 600
6
0 0 12

14. Given lines are coplanar.

1 2 4  3 5  4
 1 1 k = 0
k 2 1
  1(1 + 2k)  1(1 + k2) + 1(2  k) = 0
  1  2k  1  k2 + 2  k = 0
  k2  3k = 0  k(k + 3) = 0
 k = 0 or k = 3

4
 Chapter 09: Linear Programming

09 Linear Programming

1. Let no. of model M1 = x and no. of model M2 = y Y

 x ≥ 0, y ≥ 0
Constraints are 4x + 2y ≤ 80  2x + y ≤ 40, 2x + 5y ≤ 180 (0, 40)
Maximize z = 3x + 4y
The corners of feasible region are C(0, 36)
B(2.5, 35)
O(0, 0), A(20, 0), B(2.5, 35), C(0, 36)
2x + 5y = 180
 At A (20, 0), z = 3(20) + 0 = 60
(90,0)
At B (2.5,35), z = 3(2.5) + 4(35) = 147.5 X
A(20, 0) X
O
At C (0, 36), z = 0 + 3(36) = 108 2x+y=40
 z is maximum at B(2.5, 35). Y

3. Objective function P = 2x + 3y
The corner points of feasible region are Y
B(12, 12), C(3,3), D(20, 3), E(20, 10), F(18, 12) x = 20
(0,30)
At B = PB = 2 (12) + 3 (12) = 60
xy=0
At C = PC = 2 (3) + 3 (3) = 15
At D = PD = 2 (20) + 3 (3) = 49
B(12,12) F(18,12)
At E = PE = 2 (20) + 3 (10)= 70 y = 12
E(20,10)
At F = PF = 2 (18) + 3 (12) = 72
C(3,3) D(20,3) y=3
 P is maximum at F(18, 12). X X
O (30, 0) x + y = 30
4. For (1, 3), 3x + 2y = 3 + 6 > 0, Y
for (5, 0), 3  5 + 0 > 0,
and for (1, 2), 3 + 4 > 0
Similarly, other inequalities satisfies the given points.
 Option (D) is the correct answer.
5.
Y

(0,1500)

(0,1000)

B(800,600) x2 = 600
A(0,600) C(1000,500)

(2000,0)
X X
O D(1500,0)
x1 + 2x2 = 2000
Y x1 + x2 = 1500
1
Std. XII : Triumph Maths 
OABCD is the feasible region
 O(0, 0), A(0, 600), B(800, 600), C(1000, 500), D(1500, 0)
z = x1 + x2
At point C and D, z is maximum. Max z = 1500
 Infinite optimal solutions exist along CD.
6. Consider option (C)
3 + 2(4)  11
3(3) + 4(4) ≤ 30
2(3) + 5(4) ≤ 30
 All the above three in-equalities hold for point (3, 4).
 Option (C) is the correct answer.
7. Let the manufacturer produce x and y bottles of medicines A and B.
3x y
He must have +  66, x + y  45000, x  20000, y  40,000, x  0, y  0.
1000 1000
 the number of constraints is 6.
8. Let the company produce x telephones of A type and y telephones of B type.
 Constraints are 2x + 4y  800  x + 2y  400, x + y  300
Maximize z = 300x + 400y
Y

(0, 300)
(0, 200)
x + 2y = 400
(400, 0)
X X
O (300,0)

Y x + y = 300

 the feasible region of the LPP is bounded.

9. Given that 4x + 2y  8, 2x + 5y  10
 the feasible region lies on origin side of 4x + 2y = 8 and 2x + 5y = 10.
Also, x, y  0
 the feasible region lies in first quadrant.
 option (C) is correct. X2

10. Objective function z = x1 + x2

The corner points of feasible region are 2 7
C , 
2 7 3 3
O(0, 0), A(2, 0), B(2, 1), C  ,  and D(0, 1)
3 3
D(0,1) B(2, 1)
2 7
At B(2, 1) and C  ,  , z is maximum. Max z = 3
3 3 X1 X1
O A(2,0)
 Infinite number of solutions exists along BC.
2x1 + x2 = 1 x1 + x2 = 3
X 2 x1 = 2
2
 Chapter 09: Linear Programming
11. Objective function z = 3x + 2y
The corner points of feasible region are
1 5 1 5 5 7 Y
A  ,  , B  ,  , C(1, 0), D(3, 0), E(3, 3), F  ,  y  5x = 0
4 4 6 6 2 2 (0,6) x=3
1 5
At A = zA = 3    2   = 3.25
4 4
F(5/2, 7/2)
1 5
At B = zB = 3    2   = 2.167 E(3,3)
6 6
At C = zC = 3(1) + 2(0) = 3
A(1/4, 5/4)
At D = zD = 3(3) + 2(0) = 9 B(1/6,5/6)
At E = zE = 3(3) + 2(3) = 15 D(3,0) (6,0)
X O C(1,0) X
5 7 x  y = 1 x+y=6
At F = zF = 3    2   = 14.5 x+y=1
2 2
Y
 Maximum value of z at (3,3) is 15.

3
Target Publications Pvt. Ltd. Chapter 10: Continuity

10 Continuity 

1  cos(1  cos x) a 2  ax  x 2  a 2  ax  x 2
1. f(0) = lim = lim
x 0 x4 x 0 ax ax
 2  x 
 2sin  2   ax ax a 2  ax  x 2  a 2  ax  x 2
2sin 2   
 
 2  ax ax a 2  ax  x 2  a 2  ax  x 2
 
= lim
= lim 
x 0 x4  a 2  ax  x 2    a 2  ax  x 2    a  x  a  x 
 
 2
 a  x    a  x    a  ax  x  a  ax  x 
2 x 0 2 2 2
  x    x 
2sin 2 sin 2    sin 2   
  2    2 
= lim
x 0
  x 
2
= lim
2ax  ax ax 
 
x 4 sin 2    x 0
  2  2x a  ax  x  a  ax  x 2
2 2 2

 x
= 2lim
sin 4  
2  1  1
4 3 =
a  a a 
x 0
 x 4 2 8 a2  a2
  2
2
 f(0) =  a
2. f is continuous at x = 0.
 log(1  x 2 )  log(1  x 2 )  5 x .2 x  7 x  7 x.2 x  5 x
 f (0) = lim   4. lim
x 0
 sec x  cos x  x 0 x
2sin 2
  2
 2 
log(1  x )  log(1  x ) 
2
= lim  = lim
5x (2 x 1)  7 x (2 x 1)
x 0  1  cos 2 x   x 0 x
  cos x   2sin 2
    2
 cos x  log 1  x 2   log 1  x 2    1  2 x 1   5x 1 7 x 1  1
= lim    = lim     2
x 0  sin 2 x 
x 0 2  x  x x  sin x / 2 1

  x2 / 4 4
  log 1  x 2  log 1  x 2   
 cos x     5
 x2  x2 = 2(log 2)  log 
    7
= lim  
x 0
  sin 2 x  
It is discontinuous at x = 0 and it is removable.
 2 
 x 
sin 3 x log 1  3 x 
1  1  5. a = lim
= (cos 0)  2  = 2
 tan  e 
2
x0 1
 1  x 5 x
1 x

3. For f(x) to be continuous at x = 0, we must sin 3 x

 x   log(13x 3x)  3x
3

 x
3
have f (0)  lim f ( x )
x 0 = lim
 tan x   x
x 0 2
1
e5 x  1
 
2
a  ax  x  a  ax  x
2 2 2 2
 5 x  x
= lim
 x
2
5 x
x 0
ax ax
1
Std. XII : Triumph Maths 

 x  . (1)3x 10. Given, f(x) = [x]2  [x2]

3
(1)3
=  1 < x < 0, f(x) = (1)2  0 = 1
(1)  x  (1)5 x .x
2
2
x = 0, f(x) = 02  0 = 0
0 < x < 1, f(x) = 02  0 = 0
3
= x = 1, f(x) = 12  1 = 0
5 1  x  2, f ( x)  12  1  0
6. For f to be continuous at x = 2, x  2, f ( x)  12  2  1
1
f(2) = lim  x  1  2 x  2  x  3, f ( x)  12  2  1
x 0
1
x  3, f ( x)  12  3  2
= lim 1  ( x     x -2 = e1
x 0
3  x  2, f ( x)  12  3  2
7. Given function is continuous at (– , 6). x  2, f ( x)  22  4  0
 at x = 1 and x = 3, function is continuous.
2  x  5, f ( x)  22  4  0
If the function f(x) is continuous at x = 1, then
lim f ( x)  lim f ( x ) x  5, f ( x)  22  5  1
x 1 x 1
Hence, the given function is discontinuous at
 all integers except 1.
 1 + sin =a+b
2
 a+b=2 .....(i)
If the function is continuous at x = 3, then
lim f ( x )  lim f ( x )
x  3 x  3

3
 3a + b = 6 tan
12
 3a + b = 6 .....(ii)
From (i) and (ii), a = 2, b = 0
8. Since, x and | x | are continuous for all x.
 x + | x | is continuous for x  (– , ).
9. For f(x) to be continuous at x = 0, we must have
lim f(x) = f(0) = lim f(x)
x  0 x  0

lim f(x) = lim e tan 2 x / tan 3 x

x  0 x 0
 tan 2 x   tan 3 x 
 2 x  3 x 
 2x   3x 
= lim e
x  0
2
 e3
f(0) = lim f(x)
 x 0
2
 b  e3
lim f(x) = lim (1 | sin x |) a / | sin x |

x 0  x0
 a 
lim  | sin x |  
 e x 0  | sin x | 
 ea
f(0) = lim f(x)
 x 0
2
 b = e  e = ea
a 3

2
a=
3
2
 Chapter 11: Differentiation

11 Differentiation 

x  1  x  1  x  1  x  1
3 5 7
 3x   
1. y= + + + +… 3. y = f 
4 12 20 28  5x  4 
  x  1  x  1  x  1 
3 5 7
1
=  x  1     .... dy  3x    d  3x   
4  3 5 7   = f    
dx  5 x  4  dx  5 x  4 
x 2 x3 x 4
Now, log(1 + x) = x – + – + ….   5 x  4  3  5  3x    
2 3 4  3x   
= f    
 5x  4  5x  4
2 3 2
x x x 4 x5  
 log(1 – x) = – x  –   – ….
2 3 4 5
1 x   dy     12  5 
 log   = log(1 + x) – log(1 – x)    = f    
1 x   dx  x  0  4   16 
 x3 x5 
= 2  x    ....     12  5 
 3 5  = tan2    
 4   16 
 x  1  x  1  x  1
3 5 7

 x–1+ + + + ….  12  5 
3 5 7 = (1)2  
 16 
1  1 x 1 
= log   12  5
2 1   x  1  =
16
1  x 
= log  
2 2 x 4. y = |cos x| + |sin x|
1  x  d x
 y= log   Since, |x| =
8 2 x dx x

dy 1  2  x    2  x 1  x  1  dy cos x d sin x d

 =      =  (cos x) +  (sin x)
2  x
2
dx 8  x    dx cos x dx sin x dx

1 2 x 2  x  x 1 cos x sin x

=     = = (–sin x) + cos x
8 x    2  x  
2
4x  2  x cos x sin x

2. y = (cos x + i sin x) (cos 3x + i sin 3x) 2 2 1 1

When x = , cos x = cos = , |cos x| =
….(cos(2n – 1)x + i sin(2n – 1)x) 3 3 2 2
Since, cos  + i sin  = ei 2 3 3
 y = eix  ei3x  ei5x …. ei(2n – 1)x and sin x = sin = , |sin x| =
3 2 2
= eix[1 + 3 + 5 + …. + (2n – 1)]
= ein x
2
 dy   3  1 
   2  = –1   + 1  
dy 2  dx  x   2   2 
 = in 2 ein x 3
dx
d2 y 2 3 1
 2
= i 2 n 4 ein x = – n4y =
dx 2
1
Std. XII : Triumph Maths 

 1  2  n d   x  1  1
5. y = 1    1   ….  1   8.  a tan 1 x  b log   = 4
 x  x   x dx   x  1  x  1
dy  1   2  3  n  x 1 
 =    1    1   ….  1    a tan–1 x + b log 
dx  x 2   x  x  x 
 x 1
 1  2   3  n
+  1     2   1   ….  1   1
 x  x   x  x =
x 1
4

 1  2  3   n 1
+  1    1     2  ….  1   + ….
 x  x  x   x =
1 1
x 2
 1 x 2  1
When x = –1, 1 + = 1 + =1–1=0
x  1 1  1 1 
=   2  2
2  x 1 x 1
 dx
 Except 1st term all terms are 0.

 dy  1 1  x 1  1 –1
  = (–1) (–1) (–2) …. (1 – n) =  log   – tan x
 dx ( x 1) 2 2  x 1 2
= (–1)n (n – 1)! 1 1
 a=– ,b=
 x 2 4
 , x0
f (x) = 1  x
1 1 1 1
6.  a – 2b = – – 2   = – – = –1
 x , x0 2 4 2 2
1  x
x 9. f(x) = cos x cos 2x cos 4x cos 8x cos 16x
0
f ( x)  f (0) 1 16
 Lf (0) = lim = lim 1  x =1 =  (2 sin x cos x cos 2x cos 4x
x  0 x0 x 0 x 32 sin x
x cos 8x cos 16x)
0
Rf (0) = lim 1  x =1 =
1

16
(sin 2x cos 2x cos 4x cos 8x
x  0 x0 32 sin x
 f(x) is differentiable at x = 0 and f (0) = 1.
cos 16x)
7. f(x) = sin(log x) 1 8
1 =  (sin 4x cos 4x cos 8x
 f (x) = cos(log x) 32 sin x
x
cos 16x)
 2x  3 
y = f  1 4
 3  2x  =  (sin 8x cos 8x cos 16x)
32 sin x
dy  2x  3  d 2x  3 
 = f    1 2
dx  3  2 x  dx
 3  2x  =  sin 16x cos 16x
32 sin x
  2x  3  
= cos  log   sin 32 x
  3  2x   =
32sin x
  3  2 x  2    2  2 x  3   3  2 x  1  sin x  32cos32 x  sin 32 x cos x 
.     f (x) =  
3  2x    2 x  3 
2
 32  sin 2 x 
  2x  3    6  4x  4x  6   1 
= cos  log      32 1  0 
  1  2 
  3  2x    3  2x   f  = 2
1  4  32  1 
  
2x  3  2
12   2 x  3  1 1 2
= cos log   =   32  2 = = 2
9  4 x2   3  2 x  32 2 2
2
 Chapter 11: Differentiation
10. 1 + x4 + x8 = 1 + 2x4 + x8 – x4  dy 
= (1 + x4)2 – x4  1 dy   x dx  y 
= (1 + x4 + x2) (1 + x4 – x2)  x y = x y  2
2 2
2 
x y  dx   x y 
2 2

1  x 4  x8  
 = 1 – x2 + x4
1  x2  x4  tan 1
y

d  1  x 4  x8  d ….  ae x
 x2  y 2 
  4 
= (1 – x2 + x4)  
dx  1  x  x 
2
dx
dy dy
= 4x3 – 2x = ax3 + bx  x+y =x y ….(ii)
dx dx
 a = 4, b = –2
Diff. w.r.t.x, we get
1 1 2

11. 2x = y 5 + y 5 d2 y  dy  d 2 y d y dy
1+y +   = x + 
1 dx 2  dx  dx 2 dx dx
Let y 5 = a 2
d 2 y  dy  d2 y

1
1 1  1+y 2
+   =x 2
 y = ,
5
 a+ = 2x dx  dx  dx
a a 2   dy  2 
d y
 a2 – 2ax + 1 = 0  (y  x) 2 =   1    
dx   dx  
2x  4x  4 2  
 a= 
2 From (i), when x = 0, y  ae 2
1
 y 5 = x + x2  1 dy
From (ii), when x = 0, = 1
 
5 dx
 y = x  x2  1 
d2 y
 ae 2 2 = 2
 
dy 4  1  dx
 = 5 x  x2  1 1  2x 
dx d y
2
2  2
 2 x 1 
2
  2 =  e
 dx  x  0 a
dy
  x  
4
 x2  1 = 5 x  x2  1 x2  1
dx f g h
= 5y
2 13. f(x) = f  g h 
 dy 
 (x – 1)   = 25y2
2
f  g h 
 dx 
2 f  g h  f g h
2 2dy d 2 y  dy  dy
 (x  1)   2 +   (2x) = 25  2y  f (x) = f  g h  + f  g h 
dx dx  dx  dx
f  g h  f  g h 
dy
Dividing both sides by 2 , we get
dx f g h
d 2
y d y + f  g h 
(x2  1) 2 + x = 25y
dx dx f  g h 
 k = 25 =0+0+0
 y
tan 1   ….[ f, g, h are polynomials of 2nd degree,
12. x 2  y 2 = ae ….(i) x
f  = g = h = 0]
Diff. w.r.t.x, we get =0
1  dy 
 2x  2 y  y y1 y2 cos ax  a sin ax  a 2 cos ax
2 x2  y 2  dx 
14. y3 y4 y5 = a 3 sin ax a 4 cos ax  a 5 sin ax
 dy 
 y
tan 1  
x 1  x dx  y  y6 y7 y8  a 6 cos ax a 7 sin ax a 8 cos ax
= ae .  
y2 x2  =  a2  0 ….[ C1  C3]
1 2 
x   =0
3
Std. XII : Triumph Maths 

15. y = sin cos 1{sin(cos 1 x)} f(x) = x3 + ex/2

 f(0) = 1
      0 = f1(1)
= sin cos 1 sin   sin 1 x  
  2    g(1) = 0 ….[ g(x) = f1(x)(given)]
1 1
= sin[cos (cos(sin x)] From (i), we get
= sin(sin1 x) = x 1
dy g(1) =
 =1 
f (0)
dx
Now, f(x) = x3 + ex/2
 dy 
   1=1 1
 dx  x   f (x) = 3x2 + ex/2
2 2
1
1  f (0) =
16. 8 f(x) + 6f   = x + 5 ….(i) 2
x 1
1  g(1) = =2
Replacing x by , we get 1/ 2
x
1 1 19. y = f(x3)
8f   + 6f(x) =  5 dy
 x x  = f (x3).3x2 = 3x2 tan(x3)
dx
1 1
 6f(x) + 8f   =  5 ….(ii) z = g(x5)
 x x
(i)  8  (ii)  6 gives dz
 = g (x5).5x4 = 5x4 sec(x5)
6 dx
64 f(x)  36 f(x) = 8x + 40   30
x dy
dy 3x 2 tan x 3 3tan x 3

6
28 f(x) = 8x   10  = dx = =
x dz dz 5 x 4 sec x 5 5 x 2 sec x 5
x2  dx
6 
Given, y = x2f(x) =  8 x   10 
28  x 
20. 1  x 6  1  y 6 = a3(x3  y3)
1
 y= (8x3  6x + 10x2) Put x3 = sin  and y3 = sin 
28
dy 1  1  sin 2   1  sin 2  = a3(sin   sin )
 = (24x2  6 + 20x)
dx 28
 cos  + cos  = a3(sin   sin )
 dy  1 2 1
   = (24  6  20) =  =    
d x
  x 1 28 28 14  2 cos   cos  
 2   2 
17. f(x3) = x5
Diff. w.r.t. x, we get    
= a 3 .2sin   cos  
f (x3) . 3x2 = 5x4  2   2 
5    3
 f (x3) = x2  cot  =a
3  2 
5
 f (27) = f  (33) = (3)2 = 15     = 2 cot1 a3
3
 sin1 x3  sin1 y3 = constant
18. Since, g(x) is the inverse of f(x). Diff. w.r.t. x, we get
 f[g(x)] = x
3x 2 3y2 dy
 f   g( x)  g ( x) = 1   =0
1  x6 1  y6 dx
 f   g(1)  g (1) = 1
1 dy x 2 1  y6
 g(1) = ….(i)  =
f (g(1)) dx y 2 1  x6

4
 Chapter 11: Differentiation
21. Let f(x) = px2 + qx + r 1
sin x cos x
 f(1) = f(1)  p + q + r = p  q + r  q = 0 f ( x) tan x
 =2  x2
 f(x) = px2 + r x x
2 sin 2 x 5 x
 f (x) = 2px
 f (a) = 2ap, f (b) = 2bp and f (c) = 2cp 1 cos x cos x 1 sin x  sin x
Since, a, b, c are in A.P. + x sec x 2
x 3
+ x tan x 3x 2
 2ap, 2bp, 2cp are in A.P. 2 2cos 2 x 5x 2 sin 2 x 5

 f (a), f (b), f (c) are in A.P. 1 0 1 1 1 1 1 0 0

f ( x)
 lim = 2 1 0+ 0 1 0+ 0 0 0
dx
x 0 x
22. = sec  tan  + sin  2 0 0 2 2 0 2 0 5
d
=  2  2 + 0 = 4
dy
and  n sec n 1  .sec  tan   n cos n 1  .( sin )
d 24. Since, g is the inverse of f.
= n secn  tan   n cos n 1  sin   f[g(x)] = x
Diff. w.r.t.x, we get
dy
dy d n sec n  tan   n cos n 1  sin  f (g(x)) g(x) = 1
  
dx dx sec  tan   sin  1
 g(x) = = 1 + [g(x)]5
d f (g( x))
Dividing Nr and Dr by tan , we get
sin x sin x sin x
25. y=   ... 
dy n(sec n   cos n ) sin x sin 2 x sin 2 x sin 3 x sin nx sin(n  1) x

dx sec   cos  sin(2 x  x) sin(3x  2 x) sin((n  1) x  nx)
=   ... 
2 n 2
 dy  n (sec   cos )
n 2 sin x sin 2 x sin 2 x sin 3 x sin nx sin(n  1) x
   
 dx  (sec   cos ) 2 sin 2 x cos x cos 2 x sin x sin 3x cos 2 x
=  
sin x sin 2 x sin x sin 2 x sin 2 x sin 3 x
n 2 [(sec n   cos n ) 2  4sec n  cos n ]
 
cos3x sin 2 x
 .... 
sin(n  1) x cos nx cos(n  1) x sin nx

(sec   cos ) 2  4sec .cos  sin 2 x sin 3x sin nx sin (n  1) x sin nx sin (n  1) x
n 2 ( y 2  4) = cot x  cot 2x + cot 2x  cot 3x
= + …. + cot nx  cot(n+ 1) x
x2  4
2  y = cot x  cot(n + 1)x
 dy 
 (x + 4)   = n2 (y2 + 4)
2 dy
 dx   =  cosec2x  [ cosec2(n + 1)x] (n + 1)
dx
= (n + 1) cosec2(n + 1)x  cosec2x
x sin x cos x
23. f(x) = x 2
tan x  x3 a
26. If |r| < 1, a + ar + ar2 + …. +  =
2 x sin 2 x 5x 1 r
sin 2 x
1 sin x cos x  sin2x + sin4x + sin6x + …. =
1 sin 2 x
 f (x) = 2 x tan x  x3
sin 2 x
2 sin 2 x 5x = = tan2x
cos 2 x
x cos x cos x x sin x sin x 2x
 y = e tan
+ x 2 sec 2 x  x3 + x 2
tan x 3 x 2
dy 2 2
2 x 2cos 2 x 5x 2 x sin 2 x 5  = e tan x .2 tan x sec2x = 2e tan x tan x sec2x
dx
5
Std. XII : Triumph Maths 
1 1 f (1) n(1) n 1
27. y = tan1 + tan1 2 = = n = nC1
1 x  x 2
x  3x  3 1! 1
+ tan1 2
1
+ …. to n terms f (1) n(n  1)(1) n  2 n(n  1) n
= = = C2
x  5x  7 2! 2! 2!
1 1 f (1) n(n  1)(n  2)(1) n 3 n(n  1) (n  2)
= tan1 + tan1 = =
1  (1  x) x 1  ( x+ 2) ( x+1) 3! 3! 3!
n
1 = C3
+ tan1 + …. to n terms
1  ( x+ 3) ( x+ 2) f (1) f (1) f (1) f n (1)
 f(1)  +  + ….+(1)n
 ( x  1)  x   ( x  2)  ( x + 1)  1! 2! 3! n!
= tan1   + tan1   n n n n
= C0  C1 + C2  C3 + …. + (1) Cn nn
1  ( x  1) x  1  ( x  2) ( x + 1) 
=0
 ( x  3)  ( x  2) 
+ tan1   + …. to n terms 30. p = a2cos2 + b2sin2
1  ( x  3) ( x + 2) 
dp
= tan1(x + 1)  tan1 x + tan1(x + 2)  = a2.2 cos  ( sin ) + b2.2 sin  cos 
d
 tan1(x + 1) + tan1(x + 3)  tan1(x + 2) = (b2  a2) sin 2
+ …. + tan1(x + n)  tan1(x + (n  1))
d2p
 y = tan1(x + n)  tan1 x  2
= 2(b2  a2) cos 2
d
dy 1 1 = 2(b2  a2) (cos2  sin2 )
 = 
dx 1  ( x  n) 2
1  x2
d2p
 4p + = 4a2 cos2 + 4b2 sin2 
 dy  1 d2
   = 1
d x
  x 0 1  n2 + 2(b2  a2) (cos2   sin2 )
1  1  n2 = cos2 (4a2 + 2b2  2a2)
=
1  n2 + sin2 (4b2  2b2 + 2a2)
n2 = cos2 (2a2 + 2b2) + sin2 (2a2 + 2b2)
=
1  n2 = (2a2 + 2b2) (cos2  + sin2 )
= 2a2 + 2b2
28. y = a sin(bx + c)
= 2(a2 + b2)
 
 y1 = a cos(bx + c).b = ab sin   bx  c  = 2c2 ….[ a2 + b2 = c2 (given)]
2 
2
 y2 =  ab sin(bx + c).b = ab sin( + bx + c)
 3 
y3 =  ab2 cos(bx + c).b = ab3 sin   bx  c 
 2 
3 4
y4 =  ab ( sin(bx + c).b) = ab sin(2 + bx + c)
 4 
= ab4 sin   bx  c 
 2 
 n 
In general, yn = abn sin   bx  c 
 2 
29. f(x) = xn
f (x) = nxn1
f (x) = n(n  1) xn2
f ( x) = n(n  1) (n  2)xn3
 f(1) = 1n = 1 = nC0
6
 Chapter 12: Applications of Derivatives

12 Applications of Derivatives

1. Let f(x) = ax4 + bx3 + cx2 + dx 4

3
 f(0) = 0  y
 tan =
and f(3) = a.34 + b.33 + c.32 + d.3 4 1 4 3
= 81a + 27b + 9c + 3d y
= 3(27a + 9b + 3c + d) 4  3y
1=
=30 y  12
 f(0) = f(3) = 0  y = 2 or y = 8
f(x) is a polynomial function, it is continuous Putting y = 2 in the equation of the curve,
and differentiable. we get
Now, f (x) = 4ax3 + 3bx2 + 2cx + d 1
x=
By Rolle’s theorem, there exist at least 1 root 2
of the equation f (x) = 0 in between 0 and 3. 1 
 the point of contact is  ,  2  .
2 2 
2. The equation of the curve is y = x + bx + c.
dy 1
 = 2x + b ….(i) 4. f(x) = tan1x  log x
dx 2
Since, the curve touches the line y = x at (1,1). 1 1 ( x  1) 2
 f (x) =  = 
 [2x + b](1, 1) = 1 1  x2 2x 2 x(1  x 2 )
 2(1) + b = 1 Now, f (x) = 0  x = 1
 b = 1 1  3.14
Substituting the value of b in equation (i), f(1) = tan1 1  log 1 = = = 0.785
we get 2 4 4
Since, we are finding maxima on an interval
dy
= 2x  1  1 
dx  , 3  . We have to find the value of f(x) at
Since, gradient is negative.  3 
 1 

dy
dx
<0 
 3
 and 3 
 2x  1 < 0  1  1 1 1  1
f  = tan + log 3 =  log 3
 2x < 1  3 3 4 6 4
1
x< 3.14 1 1
2 =  log 3 = 0.52 +  1.0986
6 4 4
3. The equation of the parabola is y2 = 8x. = 0.52 + 0.2746 = 0.7946
 1

dy
2y = 8
dx
  1
f ( 3) = tan1 3  log 3 =  log 3
4 3 4
dy 8 4 3.14
 = = = m1 =  0.2746
dx 2y y 3
= 1.04  0.2746
Slope of given line, m2 = 3
= 0.7654
m1  m 2  1
Since, tan  =  the greatest value of f(x) is + log 3.
1  m1m 2 6 4
1
Std. XII : Triumph Maths 
 f(x) is a polynomial function, it is continuous
5. += and differentiable.
2
   There exists at least one value of x in (0, 1) at
 cos  = cos     = sin  which f (x) = ax2 + bx + c = 0
2 
 one root of the equation ax2 + bx + c = 0 has
1 value between 0 and 1.
Let y = cos  cos  = cos  sin  = sin2
2
8. f(x) = sin x(1 + cos x)
dy 1
 = cos 2.2 = cos 2 = sin x + sin x cos x
d 2
1
dy   f(x) = sin x + sin 2x
Now, = 0  cos 2 = 0  2 = 2
d 2
x 3x
  f (x) = cos x + cos 2x = 2 cos cos
= 2 2
4
x 3x
d2 y  f (x) = 0  cos = 0 or cos = 0
Also, =  2 sin 2 =  2 < 0 2 2
d 2 x  3x 
  = or =
 y is maximum when  = 2 2 2 2
4 
  x =  or x =
 it is maximum at  = 3
4 f (x) =  sin x  2 sin 2x < 0, only when
6. Let P(x1, y1) be the point on the curve at which 
x=
tangent is drawn. 3
The equation of the curve is xy = c2. 
dy  The maximum value of function is at
 x + y(1) = 0 3
dx
 3  1 3 3 3 3
 dy  y  f = 1   =  =
   =  1 3 2  2 2 2 4
 dx ( x1 , y1 ) x1
9. f(x) = sin4x + cos4x
 The equation of the tangent is
= (sin2x + cos2x)2  2 sin2x cos2x
y
y  y1 =  1 (x  x1) 1
x1  f(x) = 1  sin22x
2
 yx1  x1y1 =  xy1 + x1y1 1
 xy1 + yx1 = 2x1y1  f (x) =  (2 sin 2x cos 2x)  2
2
x y
 + =1 f (x) =  2 sin 2x cos 2x
2 x1 2 y1
Now, f (x) = 0
 The tangent meets the X-axis in the point  sin 2x = 0 or cos 2x = 0
A(2x1, 0) and the Y-axis in the point B(0, 2y1) 
 P is the mid point of AB  x = 0 or x =
4
 The ratio is 1 : 1
Since, f (x) = 2 sin 2x cos 2x
ax3 bx 2  f (x) =  sin 4x
7. Let f(x) =   cx  d  f(x) =  4 cos 4x
3 2
 f (x) = ax2 + bx + c For x = 0, f(x) =  4 < 0
a b 2a  3b  6c  6d 
Now, f(1) = + + c + d = For x = f(x) = 4 > 0
3 2 6 4
0  6d 
 f(1) = = d ….[ 2a + 3b + 6c = 0]  At x = , f(x) is minimum
6 4
Also, f(0) = d 1 1 1
 Minimum value of f(x) = 1  (1) = 1  =
 f(0) = f(1) 2 2 2
2
 Chapter 12: Applications of Derivatives

10.
2 3
2( x 3)  27 is minimum when  x2 3
3
is b
 27  f(x) is minimum when tan2x =
minimum. a
Since, (x2  3)3 + 27  Minimum value of f(x) = a (1 + tan2 x)
2

= x6  9x4 + 27x2 + b2(1 + cot2 x)

= x2(x4  9x2 + 27)  b  a
= a2 1   + b2 1  
 9  27 
2
 a  b
= x2  x 2      0, for all x
 2 4  ab 2ab
= a2  b  
 Minimum value of (x2  3)3 + 27 is 0.  a   b 
 Minimum value of 2( x
2  3)3  27
= 20 = 1 = a(a + b) + b(a + b) = (a + b)2

11. f(x) = 3 cos|x|  6ax + b ax  b ax  b

13. y= = 2
= 3 cos x  6ax + b ( x  4) ( x  1) x  5x  4
….[ cos ( x) = cos x] dy ( x 2  5 x  4)a  (ax  b) (2 x  5)
 =
 f (x) =  3 sin x  6a dx ( x 2  5 x  4) 2
The function f(x) is increasing for all x  R. For extreme (i.e., maximum or minimum)
 f (x) > 0
dy
  3 sin x  6a > 0 =0
dx
 6a <  3 sin x
1  a(x2  5x + 4)  (ax + b) (2x  5) = 0
 a <  sin x Since, y has an extreme at P(2, 1)
2
1  x = 2 satisfies above equation
a< 
2 a(4  10 + 4)  (2a + b) (1) = 0
  2a + 2a + b = 0
12. Let f(x) = a2sec2x + b2cosec2x
 f (x) = a2.2 sec x sec x tan x b=0
+ b2.2 cosec x ( cosec x cot x) x = 2, y = 1 satisfies the equation of the curve
= 2a2 sec2 x tan x  2b2 cosec2x cot x a(2)  b
 1=
Now, f (x) = 0 4  10  4
 2a2 sec2 x tan x  2b2 cosec2 x cot x = 0 2a  0
 1= = a
1 sin x 1 cos x 2
2a2. 2
 = 2b2 2 
cos x cos x sin x sin x  a=1
4 2
sin x b  a = 1, b = 0
 =
cos 4 x a 2
14. Let f(x) = x tan x
b2
 tan4x = 2  f (x) = x sec2 x + tan x
a
 
b a  f (x) > 0 for x   0, 
 tan2x = and cot2x =  2
a b
Also,  
 f(x) is increasing in the interval  0, 
f (x) = 2a 2 sec2 x.sec2 x  tan x.2sec x sec x tan x   2
cosec 2 x (cosec 2 x)  
2b 2  Since, 0 <  <  <
 2
  cot x.2cosec x ( cosec x cot x) 
 f() < f()
= 2a 2 sec4 x+ 2sec2 x tan 2 x 
  tan  <  tan 
 2b 2 cosec4 x  2cosec2 x cot 2 x   tan 
 <
 f (x) > 0 for all x.  tan 

3
Std. XII : Triumph Maths 
15. The point of intersection of the given curves is 18. f(x) = x3  12ax2 + 36a2x  4
(0, 1). Diff. w.r.t. x, we get
Now, y = 3x f (x) = 3x2  12a(2x) + 36a2(1)
dy = 3x2  24ax + 36a2
  3x log 3
dx Now, f (x) = 0  3x2  24ax + 36a2 = 0
 dy   x2  8ax + 12a2 = 0
   = log 3 = m1 (say)
 dx  (0,1)  (x  2a) (x  6a) = 0
Also, y = 5x  x = 2a or x = 6a
dy Also, f (x) = 6x  24a
  5x log 5 [f (x)]x=2a = 12a  24a =  12a < 0
dx
[f (x)]x=6a = 36a  24a = 12a > 0
 dy 
    log 5  m 2 (say)  Maxima at p = 2a and minima at q = 6a
 dx (0,1) 3p = q2 ….(given)
m1  m 2 log3  log 5  3  2a = (6a)2
 tan  = =
1  m1m 2 1  log 3log5 1
 6a = 36a2  a=
2 6
16. Let f(x) = ax + bx + c
 f (x) = 2ax + b 19. The functions ex , sin x, cos x are continuous
since,  and  are roots of the equation and differentiable in their respective domains.
ax2 + bx + c = 0  f(x) is continuous and differentiable
 f() = f() = 0   5 
Also f   = 0 = f  
 f(x) being a polynomial function in x, 4  4 
it is continuous and differentiable. Now,
 There exists k in (, ) such that f (k) = 0 f (x) =  ex (sin x  cos x) + ex (cos x + sin x)
b = ex ( sin x + cos x + cos x + sin x)
 2ak + b = 0,  k=
2a = 2ex cos x
But k  [, ] Also, f (x) = 0  cos x = 0
 <k<    5 
b  x=   , 
 < < 2 4 4 
2a
20. ay2 = x3 ….(i)
17. f(x) = tan1 (sin x + cos x)
Diff. w.r.t.x, we get
1
 f (x) =  (cos x  sin x) dy
1  (sin x  cos x)2 2ay = 3x2
dx
  dy 3 x 2
2 cos  x    =
 4 dx 2ay
=
1  (sin x  cos x ) 2
2ay
For f(x) to be increasing, f (x) > 0  slope of the normal = 
3x 2
  Since, the normal to the given curve makes
 2 cos  x   > 0
 4 equal intercepts with the axis.
  2ay
 cos  x   > 0   2 = 1
 4 3x
   3x 2
 <x+ < y=
2 4 2 2a
3  3x 2
  <x< Substituting y = in (i) and solving, we get
4 4 2a
    4a 8a 
 f(x) is an increasing function in   ,  . the point  ,  .
 2 4  9 27 

4
 Chapter 13: Integration

13 Integration

5 x10 x
1. Let I =  dx 3. Let I =  sin 1 dx
x16 ax
Put x = a tan2t
5 x 1
10
 dx = 2a tan t sec2t dt
=  . dx
x10 x11 a tan 2 t
 I =  sin 1  2a tan t sec2 t dt
5 1 a  a tan 2 t
=  x10
 1. 11 dx
x =  sin 1 (sin t)  2a tan t sec 2 t dt
5 = 2a  t tan t sec 2 t dt
Put 1 t
x10
 d  
 5(10)x11 dx = dt = 2a  t  tan t sec2 t dt    (t)  tan t sec2 t dt  dt 
  dt  
1 1
 dx =  dt  tan 2 t tan 2 t 
x11
50 = 2a  t.   1. dt 
 2 2 
1
 1 
 I= t 2
   dt
 50 
= a  t tan 2 t   (sec2 t  1)dt 

x
1 t 3/2 = a  t tan 2 t  tan t  t   c , where t = tan1
=  . +c a
50 3 / 2
x x x x
3/2 = a  tan 1   tan 1 +c
1  5 
=   1 10  +c a a a a
75  x 
4. Let I =  cosec x 1 dx
2. Multiplying Nr and Dr by sin 3x, we get 1
cos5 x  cos 4 x =  sin x
 1dx
 1  2cos3x dx
1  sin x
=
sin 3 x cos5 x  sin 3 x cos 4 x
 sin 3x  2sin 3x cos3x dx
=  sin x
dx

sin 3 x  cos5 x  cos 4 x  1  sin x 1  sin x

=  dx =  
sin x 1 sin x
dx
sin 3 x  sin 6 x
cos x
 3x 3 x  9x x =  dx
 2sin cos  2cos cos  sin x  sin x
2
2 2  2 2
=  dx
9x 3x Put sin x = t
2cos sin
2 2  cos x dx = dt
3x x
=   2cos cos dx  I= 
1
dt
2 2
t2  t
=   cos 2 x  cos x  dx 1
1 
=  1 
dt
=   sin 2 x  sin x  + c t t 
2

2  4 4
1
Std. XII : Triumph Maths 
1 t2 1
=  2 2
dt I2 =  t 4  1 dt
 1 1
t     1
 2 2 1 2
=  t dt
1 1
= log t   t 2  t  c , where t = sin x t  2
2

2 t
1  1
1 =  1  2  dt
= log sin x +  sin 2 x  sin x + c  1
2
 t 
2 t   2
 t
1 1
5. Let I =  tan x dx =  2 dm , where t + = m
m 2 t
Put tan x = t2 1
t  2
1 m 2 t 1
 sec2x dx = 2tdt = log = log
2 2 m 2 2 2 1
t  2
2t t
 dx = dt
1 t4 1 t 2  2t  1
= log
2t 2 2 t 2  2t  1
 I =  t2 .
1  t4
dt
 From (i),
t 2 1  t2 1 1 t 2  2t  1
= 2 dt I= tan1   + log +c
1  t4 2  2t  2 2 t 2  2t  1
t2  1  t2 1 1  tan x 1 
=  t 4  1 dt =
2
tan1 
 2 tan x 
 
t2  1 t2 1 1 tan x  2 tan x  1
=  t 4  1 dt +  t 4  1 dt +
2 2
log
tan x  2 tan x  1
+c

= I1 + I2 (say) ….(i)
 x 1 x  dx
1/ 2
13/ 2 5/ 2
6. Let I =
t 12
I1 = t 4
1
dt
=  x .(1  x ) .x dx
5 5/2 1/ 2 3/ 2

1
1 Put 1  x5/2  t
=  t 2 dt
1 5 3/2
t2  2  x dx  dt,
t 2
1  1 2
 x 3/ 2 dx = dt
=  1
2 1  2  dt
 t 
5
t   2 2
 t  I =  (t  1)2 .t1/ 2 . dt
5
1 1
= y 2
2
dy , where t  = y
t
2
=   t 5/2  2t 3/2  t1/2  dt
5
 1 2 2 4 2 
1  y  1 tt  =  t 7/ 2  t 5/ 2  t 3/ 2  + c, where t = 1  x 5/ 2
5 7 5 3
= tan  =1
tan 1
  

2  2 2  2  2 2
1 x5/ 2   54 1 x5/ 2 
7/ 2 5/ 2
  =
5  7
1  t2 1 2 3/ 2 
=
2
tan 1 
 2t 
 
3
1  x5/ 2   + c


2
 Chapter 13: Integration
tan x 2  2 tan x 1 
7. Let I =  1  tan x  tan 2
x
dx =
3
tan1 
 3 
 + c1

tan x  From (i),

=  sec 2
x  tan x
dx
2  2 tan x 1 
I=x tan1  + c
sin x 3  3 
cos x
=  dx  A 3
1 sin x
  A=3
cos 2 x cos x
sin x cos x  x 1  1
=  1 sin x cos x dx 8. Let I =  log   2
 x 1 x 1
dx

1
sin 2 x  x 1 
Put log  =t
=  2 dx  x 1
1
1  sin 2 x  1 1 
2     dx = dt
 x 1 x 1
sin 2 x
=  dx 1 1
2  sin 2 x  dx = dt
x 1
2
2
2  sin 2 x  2
=  2  sin 2 x
dx

1 1 1   x  1 
I =  t. dt  t 2  c =  log 
2

 +c
2 4 4   x  1  
 2 
=  1   dx 1
 2  sin 2 x   A=
4
= x  I1 (say) ….(i)
1
2 9. Let I =  dx
I1 =  x  2x  2
2
dx 2
2  sin 2 x
1
2
Put tan x = t  sec x dx = dt  dx =
1
dt
=  ( x  1) 2
 1
2
dx
1  t2 
2 tan x 2t Put x + 1 = tan 
sin 2x = =
1  tan x 1  t 2
2  dx = sec2  d
2 1 sec 2 
 I1 =   dt  I= d
2t 1  t 2 (tan 2   1) 2
2
1  t2 sec 2  x2  2x  2
1
=  sec4  d x+1
=  2 dt
t  t 1 =  cos 2  d 
1
=  dt 1
=  (1  cos 2)d
1
1 3
t t 
2
2
4 4 1  sin 2 
=   c
= 
1
2
dt 2 2 
 1 
2
 3  1
t     = ( + sin  cos ) + c
 2   2  2
1  1 x 1 1 
 1 =  tan ( x  1)  2  c
2 t  2  x  2 x  2 x2  2 x  2 
= tan1  2+c
1
3  3  1  1 x 1 
  =  tan ( x  1)  2 +c
 2  2 x  2 x  2 
3
Std. XII : Triumph Maths 
1 = b2 ( cos2 x)  a2(sin2 x)
10. Let I =  dx
cos x  sin 6 x
6 1
  =  b2 cos2x  a2sin2x
Since, a3 + b3 = (a + b)3  3ab(a + b) f ( x)
 cos6x + sin6x = 1  3 sin2x cos2x 1
 f(x) =
….[ a + b = cos2x + sin2x = 1] a sin x  b 2 cos 2 x
2 2

1
 I =  1  3sin 2 2
dx 12. Let I =  esin   log (sin )  cosec 2   cos  d
x cos x
Put sin  = t
1
=  3
dx  cos  d = dt
1  sin 2 2 x  1
4  I =  e t log t  2  dt
 t 
4
=  dx  1 1 1
4  3sin 2 2 x =  e t log t    2  dt
 t t t 
4cosec2 2 x
=  4cosec2 2 x  3 dx  1
= e t  log t    c
 t
4cosec 2 2 x
=  4(1 cot 2 2 x)  3 dx  d 1 1 1 
….   log t     2 
1  dt  t t t 
=  ( 4 cosec2 2x) dx  1 
4cot 2 2 x 1 = esin  log (sin )   c
Put 2 cot 2x = t  sin  
  4 cosec2 2x dx = dt = esin   log(sin )  cosec   c
1
 I =  2 dt
t 1  cos   sin    1  tan  
13. log    log  
=  tan1(t) + c  cos   sin    1  tan  
=  tan1 (2 cot 2x) + c  
 log tan    
 2cos 2 x  2sin 2 x  4 
=  tan1   +c
 2sin x cos x  1  
1 Since,  sec 2d  log tan    
=  tan (cot x  tan x) + c 2 4 
= tan1(tan x  cot x) + c d  
 log tan      2sec 2 ….(i)
1 d 4 
11.  f ( x)sin x cos x dx = 2(b2  a 2 ) log[f ( x)] + c Integrating the given expression by parts, we
d  1  get
  log  f ( x)   c  = f(x) sin x cos x
dx  2(b  a )
2 2
   1 sin 2
I  log tan      sin 2    2sec 2 d
1 1 4  2 2
  f (x) = f(x) sin x cos x
2(b  a ) f ( x)
2 2 ….[From (i)]
f ( x ) 1  
 = 2(b2  a2) sin x cos x  sin 2 log tan       tan 2 d
[f ( x )] 2 2 4 
Integrating on both sides, we get 1   1
= sin 2 log tan     – log (sec 2)  c
f ( x) 2 4  2
 f ( x)2 dx = (b  a )  2sin x cos x dx
2 2

x  4
1 14. Let =t
  = (b2  a2)  2sin x cos x dx 3x  4
f ( x)
(3 x  4)  (3 x  4) t  1
 =
= b 2  2sin x cos x dx  a 2  2sin x cos x dx (3 x  4)  (3 x  4) t  1

4
 Chapter 13: Integration
6x t  1 t2
 =
8 t  1  I1 = 1 t2
.2t dt

4  t 1  t2
x=   = 2 dt
3  t 1 1 t
4t  4 1  t2 1
x+2= +2 =  2 dt
3t  3 1 t
4t  4  6t  6 2t  10  1 
= = =  2  1  t   dt
3t  3 3t  3  1  t
 x  4   1 
Given, f  = 2   1  t  dt
= x + 2 1 t 
 3x  4 
2t  10 1 t2 
 f(t) = = 2  log(1  t)  t   + c1
3t  3  1 2
2 t 5  1 
=   =  2  log(1  x )  x  x  + c1
3  t 1   2 
 From (i),
2  t 1  4 
=   I = x log (1  x )
3  t 1 
1  1 
2 4  2 8  .2  log (1  x )  x  x  + c
= 1  =  2  2 
3  t  1  3 3(t  1)


2
f(x) = 
8  
= (x  1) log 1 x  x  x  c
1
2
3 3( x  1) 3
1
x
16. P(x) =  3 dx, Q(x) =  3 dx
2 8  x x 2
x  x2
  f ( x)dx =   3  3( x  1)  dx x3 1
 P(x) + Q(x) =  3 dx
2 8 x  x2
= x  log|x  1| + c
3 3 x3  x 2  x 2 1
=  dx
x3  x 2
15. 
Let I =  log 1  x dx  
=  1  3
x2  1 
dx
2 


=  log 1  x (1)dx   x x 
=x+I ….(i)
d
= log(1  x )  1dx    log 1  x
 dx
  
1dx  dx x 1
I=  2
2
dx
x ( x 1)
1  1  x 2 1
= log(1  x ) .x  1   x dx
x 2 x Put =
A B C
+ + 2
x ( x  1)
2
x 1 x x
1 x  x2 + 1 = Ax 2  Bx( x 1)  C( x  1) ….(ii)
= x log(1  x ) + 
2 1 x
dx
Putting x = 0 in (ii), C = 1
1 Putting x = 1 in (ii), A = 2
= x log(1  x ) + I1 ….(i)
2 Putting x = 1 in (ii), B = 1
x  2 1 1 
 I =    2  dx
Now, I1 = 1 x
dx
 x 1 x x 
Put x = t2, 1
= 2 log|x  1|  log|x| +
 dx = 2t dt x
5
Std. XII : Triumph Maths 
 From (i), 1 1
1
=  1 2
  dt
t
P(x) + Q(x) = x + 2 log|x  1|  log|x| + + c  1 4
x t 2
t
 (P + Q) (2) = P(2) + Q(2) 1
1
= 2 + 2 log 1  log 2 + + c
=  1 2t  5t 2
dt
2
1 1

5 5
=  log 2 + c
 5
….  (P  Q) (2)   = 
5
 2 1
dt
2 2  2 t  t
2

 c = log 2 5 5
 P(x) + Q(x) = x + 2 log|x  1|  log|x| 1 1
1
=–
5
 2 1 4
dt
+ + log 2 t  t 
2

x 5 25 25
1 1 1
 P(3) + Q(3) = 3 + 2 log 2  log 3 + + log 2
3
=
5
 2 2
dt
 1    2
10 8 t   
= + log  5  5
3 3
1 1 2t 1
= log t   t2   + c
2a sin x  bsin 2 x 5 5 5 5
17. Let I =  (b  a cos x)3
dx
1 1 1 1 2 1
(a  b cos x) = log     c
= 2 sin x dx 5 x 1 5 ( x  1) 2
5( x  1) 5
(b  a cos x)3
Put b + a cos x = t 1 1 1 x2  4
=  log   +c
  a sin x dx = dt 5 x 1 5 5( x  1) 2
1
 sin x dx =  dt 1  x cos x
19. Let I =  dx
 
a
x 1   xesin x 
2

tb
a  b 
 a   1
 I = 2    dt esin x (1  x cos x)
=  dx
 
t3  a
xesin x 1   xesin x 
2

2 a  bt  b
2 2

a 
= dt
at 3 Put xesin = t
x

2  [xesin x cos x + esin x(1)] dx = dt

=  2  (a 2  b 2 ) t 3  bt 2  dt
a
 esin x (1 + x cos x) dx = dt
2  a 2  b2 b 
= 2   +c 1 1
a  2t 2 t   I=  t(1  t 2
)
dt =  t(1  t) (1  t) dt
1 a 2  b2 2b
= 2 + 2 +c 1 A B C
a t 2
a t Put = + +
t(1  t) (1  t) t 1 t 1 t
1
18. Let I =  ( x  1) x2  4
dx  1 = A(1  t) (1 + t) + Bt(1 + t) + Ct(1  t)
….(i)
1 1 Putting t = 0 in (i), we get
Put x  1 = ,  dx =  2 dt
t t A=1
1  1
 I=    2  dt
Putting t = 1 in (i), we get
1 1 
2
 t  1
  1  4 B=
t t  2
6
 Chapter 13: Integration
Putting t = 1 in (i), we get  x2  x + 1 = t2  2tx + x2
1 t2 1
C=  x=
2 2t  1
1 1/ 2 1/ 2  dx (2t  1).2t  (t 2  1).2
 I =     =
 dt dt (2t  1)2
 t 1 t 1 t 
2t 2  2t  2
1 1  dx = dt
= log|t|  log|1  t|  log|1 + t| + c (2t  1)2
2 2
1 2t 2  2t  2
1 t2 1 x 2 e2sin x  I =   dt
= log  c  log +c t (2t  1)2
2 1  t2 2 1  x 2 e 2sin x
t2  t 1
= 2 dt
t(2t  1)2
 tan(sin
1
20. Let I = x)dx
t2  t 1 A B C
  x  Put = + +
=  tan  tan 1    dx t(2t  1) 2
t 2t  1 (2t  1) 2
  1  x  
2
 t2  t + 1 = A(2t  1)2 + Bt(2t  1) + Ct
x ….(i)
=  dx
1  x2 Putting t = 0 in (i), we get
Put 1  x2 = t,   2x dx = dt, A=1
1 1
 x dx =  dt Putting t = in (i), we get
2 2
1  1 3
 I =     dt C=
2
t  2
Putting t = 1 in (i), we get
1
=  dt =  t + c =  1 x 2 + c 1=A+B+C
2 t 3
B=1–1–
21. Let I =  sec 25/13 x cos ec 27/13 x dx 2
3
B=
=  cos 25/13 x sin 27/13 x dx 2
25 27 52 1 3 3 1 
Now   =  = 4  I = 2    . 2 
dt
13 13 13  t 2(2t  1) 2 (2t  1) 
Multiplying and dividing by cos4x, we get 3 3 1
= 2 log t  log(2t  1)  . +c
I =  cos 4 x cos 25/13 x sin 27/13 x sec 4 x dx 2 2 2t  1
=  tan 27/13 x (1+ tan 2 x)sec2 x dx 3 1 3 
= 2 log t  log(2t  1)   + c,
2 2  2t  1 
Put tan x = t,  sec2x dx = dt
 I =  t 27/13 (1+ t 2 )dt where t = x + x2  x  1
and 2t  1 = 2x  1 + 2 x 2  x  1
t + t 1/13  dt
27/13
=
3 1
13 14/13 13 12/13  P = 2, Q =  , R = 
= t  t +c 2 2
14 12
1
13 13
=  (tan x)14/13  (tan x)12/13 + c 23. Let I =  (1  x 2
) 1  x2
dx
14 12
1 1
1 Put x = ,  dx =  dt
22. Let I =  x x2  x  1
dx t t2
1  1
Put x + x2  x  1 = t
 I =  1
  2  dt
1  t 
1  2  1  2
 x2  x  1 = t  x  t  t

7
Std. XII : Triumph Maths 
t dt log x
=  25.  ( x  1) dx =  log x .( x  1) 2 dx
t 2
 1 t  1 2 2

 x 1
1
1  x  1
1
Put t2  1 = m2
= log x.  . dx
 2t dt = 2m dm, 1 x 1
 t dt = m dm log x  1 
mdm = +   dx
 I =  x 1  x( x 1) 
 m  2 m2
2
log x 1 1 
= +    dx
1 x 1  x x 1 
=  dm
 
2
m2  2 log x
= + log |x|  log |x + 1| + c
1  m  x 1
= tan 1  c
2  2 26. In =  sin n x dx
1  t2 1 
= tan1  c =  sin n 1 x .sin x dx
2  2 

= sinn1x  sin x dx
 1 
 1  d 
1     sin n 1 x   sin x dx  dx
2
= tan1  x c
2  2   dx 
  n1
= sin x( cos x)
 
 1  x2    (n  1)sin n  2 x cos x ( cos x)dx
1
=  tan1  +c
2  2 x  =  sinn1x cos x + (n  1)  sin n  2 x cos 2 x dx

1   1  x2  =  sinn1x cos x + (n  1)  sin n  2 x 1  sin 2 x  dx
=   cot 1   + c
2  2 
 2x 
 =  sinn1x cos x + (n  1)   sin n  2 x  sin n x  dx

1  1
 2x  =  sinn1x cos x + (n  1)  sin n  2 x dx
=    tan    + c
2  2  1 x
2
  (n  1)  sin n x dx
1  2x    In =  sinn1 x cos x + (n  1) In2  (n  1) In
= tan1    +c
2  1 x
2
 2 2  In + (n  1)In  (n  1)In 2 =  sinn 1 x cos x
 nIn  (n  1) In 2 =  sinn1x cos x
x2  1
24. Let I =  dx 27. u =  f () sin  + f () cos 
x3 2 x 4  2 x 2  1 du
Dividing Nr and Dr by x5, we get  =  f () sin   f () cos 
d
 1 1  + f () cos   f () sin 
 3 5
x x 
I=   dx =  f () sin   f () sin 
2 1 v = f () cos  + f () sin 
2 2  4
x x dv
 =  f () sin  + f () cos 
2 1  4 4 d
Put 2  2  4  t   3  5  dx  dt
x x x x  + f () cos  + f () sin 
1 dt 1 = f () cos  + f () cos 
 I=   t c 2 2
4 t 2  du   dv  2
   +   = [ f () sin   f () sin ]
1 2 1 
   
d d
= 2 2  4 c
2 x x + [f () cos  + f () cos ]2

8
 Chapter 13: Integration
= [f ()]2 sin2 + 2 f () f () sin2  sin x  cos x
+ [f ()]2 sin2  + [f ()]2 cos2 
= 2  1  1 2sin x cos x 
dx

+ 2f () f () cos2  + [f ()]2 cos2  sin x  cos x

= [f ()]2 + 2f () f () + [f ()]2 = 2  1   sin 2 x  2sin x cos x+ cos 2 x 
dx

….[ sin2  + cos2  = 1]

1
= [f () + f ()]2 = 2  (sin x + cos x) dx
1   sin x  cos x 
2
1/ 2
 du  2  dv  2 
   d    d   d =  f ()  f () d 
Put sin x  cos x = t
(cos x + sin x)dx = dt
 
= f () + f() + c 1
 I = 2  dt
x 1 1  t2
28. Let I =  ( x  1) x3  x 2  x
dx = 2 sin 1 (t)  c
= 2 sin1(sin x  cos x) + c
x2  1
=  ( x  1) 2
x3  x 2  x
dx

x2  1
=  (x 2
 2 x  1) x3  x 2  x
dx

1
1
=  x2 dx
 1  1
 x   2  x  1
 x  x
1
Put x +  1 = t2
x
 1 
  1  2  dx = 2t dt
 x 
2t
 I =  dt
(t  1) t 2
2

1
= 2 dt
t 1 2

= 2 tan1 t + c
 1 
= 2 tan1  x   1  + c
 x 

29. Let I =   tan x  cot x dx 

 sin x cos x 
=     dx
 cos x sin x 
sin x+ cos x
=  dx
sin x cos x
2 (sin x  cos x)
=  2sin x cos x
dx

sin x  cos x
= 2  1  1 2sin x cos x
dx

9
Target Publications Pvt. Ltd. Chapter 14: Definite Integrals

14 Definite Integrals 

1 3
log(1  x) 1
1. Let I = 0 1  x 2 dx 3. Let I = 1 2
0
f (x)
dx …. (i)

Put x = tan t  dx = sec2 t dt 3

1
When x = 0, t = 0 and when x = 1, t =
 = 1 2
0
f (3  x )
dx
4

 a a

log(1  tan t)
4
….  f ( x)dx   f (a  x)dx 
 I=   sec 2 t dt  0 
0
1  tan t
2 0

 3
1
4
=  log(1  tan t) dt
= 1 2
0
f ( x)
dx
0
 ….[ f(x) + f(3  x) = 0 (given)]
4
   
=  log 1  tan   t   dt
  4  3
2f ( x )
0 2f ( x )  1 dx
0

 I= …. (ii)
 1  tan t 
4
=  log  1   dt
0  1  tan t  Adding (i) and (ii), we get
 3 3
1 2f ( x )
0 1  2f ( x ) dx + 0 2f ( x )  1 dx
4
 2  2I =
=  log   dt
0  1  tan t 
3 3

1  2f ( x )
4
=  [log 2  log(1  tan t)]dt
= 0 1  2f ( x ) dx = 0 1dx = 3
0
 3
4  I=
2
 I =  (log 2)dt  I
0

 x log x
 2I = log2  t 0
/ 4
 log 2
4
4. Let I =  (1  x
0
2 2
)
dx

 1
x log x x log x

 I= log 2
8 = 0 (1  x 2 )2 dx  1 (1  x 2 )2 dx
5 1 2 3 = I1 + I2 (Say) ….(i)
2.  f ( x ) dx =  f ( x ) dx +  f ( x ) dx +  f ( x ) dx 
x log x
0 0 1
4
2
5
I2 =  (1  x
0
2 2
)
dx
+  f ( x ) dx +  f ( x ) dx
1
3 4 Put x =
2 3 4 5 y
= 0 +  12 dx   22 dx   32 dx   42 dx
1
1 2 3 4  dx =  dy
= 1(2  1) + 4(3  2) + 9(4  3) + 16(5  4) y2
= 1 + 4 + 9 + 16 = 30 When x = 1, y = 1 and when x  , y  0
1
Std. XII : Triumph Maths 

1 1 1
0
y
log  
 y    1  dy
7. Let I =  ( x   )(  x)
dx
 I2 =  2  2 


1  1   y  Put x =  sin2 t +  cos2 t

1  2 
 y   dx = (.2 sin t cos t + .2 cos t ( sin t))dt
0
y log y  1  = 2 (  ) sin t cos t dt
=  dy ….  log     log y  When x = ,  =  sin2 t + (1  sin2 t)
1
(1  y )
2 2
  y 
1  =  + (  ) sin2 t
y log y
=  dy 
0
(1  y 2 ) 2  sint = 1,  t=
1
2
x log x When x = ,
=  dx
0
(1  x 2 ) 2  = (1  cos2 t) +  cos2 t
 I2 =  I1 =  + (  )cos2 t
 From (i), I = I1 + I2 = 0  cos t = 1, t = 0
n
(x  ) (  x) = ( sin2 t +  cos2 t  )
5.  [ x]dx
0
(   sin2 t –  cos2 t)
1 2 3 n = [ cos2t  (1 sin2)]
=  [ x]dx +  [ x]dx +  [ x]dx +… +  [ x]dx [(1 – cos2 t)   sin2 t]
0 1 2 n 1
1 2 3 n = (  ) cos2t (  ) sin2t
=  0dx +  1dx +  2dx +… +  (n  1)dx Since,  > 
0 1 2 n 1
 ( x  )(  x) = (  ) sin t cost
= 0 +  x 1  2  x 2 + …. + (n  1)  x n 1
2 3 n
0
(  )sin t cos t
= (2 – 1) + 2(3 – 2) + …. + (n  1)(n  n + 1)  I = 2 dt
= 1 + 2(1) + …. + (n  1)(1)  (   )sin t cos t
2
= 1 + 2 + …. + (n  1) 0
(n  1)n n(n  1) = 2 (1)dt
= 
2 2 
2
/ 2
sin 2 x  
6. I1 =  sin 2
x
dx 
2
2
0
= 2 1dt
/ 2 / 2
sin 2 2 x (2sin x cos x) 2 0
I2 = 
0
sin 2 x
dx = 
0
sin 2 x
dx
= 2  t 0
/ 2

/ 2

=  4 cos x dx = 4  =  
2

4 = 2   0
/ 2
0
 2 
sin 2 3x
I3 = 
0 sin 2 x
dx =
/ 2
(3sin x  4sin 3 x) 2  ex  1 
= 
0
sin 2 x
dx 8. Let h(x) = x3f(x) = x3  x
 e 1 

/ 2
 e x  1  3 1 e 
x
 (9  24sin x 16sin 4 x)dx
2
=  h(x) = (x)3   x = x
  x 
0  e 1  1 e 
9  3.1  3
=  24. + 16. . =  ex  1 
2 4 4.2 2 2 = x3  x 
 3  e 1 
 I1+I3 =   2  2I 2
2 2  h(x) = h(x)
 I1, I2, I3 are in A.P.  h(x) is an even function.
2
 Chapter 14: Definite Integrals
1 1 1 b
  t f (t)dt =  h(t)dt = 2 h(t)dt
3
11. Let I =  f ( x)dx
1 1 0 a
1
Put x = (b  a)t + a
= 2 t 3f (t)dt
0  dx = (b  a)dt
1
When x = a, t = 0 and when x = b, t = 1
= 2 x 3f ( x)dx 1
0
 I =  f  (b  a)t  a  (b  a)dt
 1 3  0
= 2 ….   x f ( x)dx    1
 0  = (b  a)  f  (b  a)t  a  dt
1 0
9. f(m, n) =  (log x) m x n 1dx 1
0 = (b  a)  f  (b  a) x  a  dx
1
1 d  0
= (log x)m  x n 1dx     (log x) m  x n 1dx  dx
0  dx 0   =ba
1
 xn 
1
1 xn 12. If 0  x < 1, then 0  x2 < 1,  [x2] = 0
=  (log x) m     m(log x) m 1   dx
 n 0 0 x n If 1  x < 2 , then 1  x2 < 2,  [x2] = 1
1
m If 2  x  1.5, then 2  x2  2.25,  [x2] = 2
n 0
=00 (log x) m 1  x n 1dx ….[ log 1 = 0]
1.5 1 2 1.5
   x  dx =   x  dx +   x 2  dx +   x  dx
2 2 2
m
=  f(m  1, n) 0 0 1 2
n
1 2 1.5
x
=  0dx +  1dx +  2dx
10. (x) =  (4sin t  3cos t)dt
7
0 1 2

= 0 +  x 1   2 x 
6 2 1.5

  (x) = 4 sin x + 3 cos x 2

 7  4  = 2  1 + 2 (1.5 – 2 )
If x   , ,
 6 3  = 2 – 1 + 3 2 2
then x is in the third quadrant.
=2 2
 sin x and cos x are both negative.
 (x) = 4 sin x + 3cos x < 0 1
13. f   + x2f(x) = 0 ….(given)
 7  4  x
 (x) is decreasing on the interval  , 
6 3 1 1
 7  4   f(x) =  2 f  
 Minimum (least) value of (x) on  , x x

 6 3 sec  sec 
1 1
 4 
4  /3
is    =  (4sin t  3cos t)dt
Let I = 
cos 
f ( x)dx = 
cos 
 f   dx
x2  x 
 3  7  /6
1 1
=  4cos t  3sin t 7  /6
4  /3
Put  t ,   2 dx = dt
x x
 4 7   4 7  When x = cos , t = sec 
=  4  cos  cos   3  sin  sin 
 3 6   3 6  and when x = sec , t = cos 
cos  sec  sec 
 1 3  3 1
= 4      3      I=  f (t)dt    f (t)dt =   f ( x)dx = I
 2 2   2 2 sec  cos  cos 

 I +I=0
=

7 1 3   2I = 0
2  I=0
3
Std. XII : Triumph Maths 

1 n n f ( x)
14. lim 1   16. f ( x)  f ( x)  1
n  n n 1 n2 f ( x)

Integrating on both sides, we get
n n 
  ...   log f ( x)  x  log c  f ( x)  ce x
n3 n  3(n  1) 
 f(0) = c  c = 1
   f(x) = ex
 
= lim 
1 1

1

1
 ... 
1
 Now, f(x) + g(x) = x2
n  n  0 1 2 3(n  1)   g(x) = x2  ex
1 
 1 n 1 1
n 
 n n 1 1
  f ( x) g( x) dx = e ( x 2  e x ) dx
x
1 3(n 1) 1
= lim  0 0
n  n r 1 1
r 0
1
 x e  e
2 x 2x
n = dx
0 0
3
1 1
= dx  e2 x 
1 x =  x 2 e x  2 xe x  2e x  
0
 2 0
3
=  2 1  x  1 2 3
0 =e e 
= 2  1 3  1 0  100 
2 2

= 2(2  1) = 2(1) = 2 17. Let I = 

0
(| sin 3 x |  | cos3 x |) dx

4n 
n 200
15. lim  2

   (| sin 3 x |  | cos3 x |) dx
2
n 
r 1 r 3 r 4 n =
0
4n 
1 1 1
= lim    2

  = 200 (| sin 3 x |  | cos3 x |)dx

2
r 1 n r 3 r 4 n
n 

n 0
n  | sin 3 x |  | cos3 x |is a periodic 
....  
4n
1 1 1
= lim    2  function with period
 
r 1 n r  r
n 
  2 
n 3 n  4 
  2
4  I = 200 (sin 3 x  cos3 x) dx
1
=  dx 0

 
2
0 x 3 x 4  2 
2

 
= 200   sin x dx +  cos x dx 
3 3
Put 3 x + 4 = t
 0 0

1
 3. dx = dt = 200[I1 + I2] (Say) ….(i)
2 x 
1 2 2
 dx = dt Where I1 =  sin 3 x dx
x 3 0
When x = 0, t = 4 and when x = 4, t = 10 
2
10
=   (1  cos 2 x)( sin x) dx
10
1 2 2 1 
 I= 4 t 2  3 dt =  3  t  4 0

Put cos x = t,  sin x dx = dt

2 1 1 2 25 
=     =    When x = 0, t = cos 0 = 1 and when x = ,
3  10 4  3  20  2
2 3  1 
=     t = cos =0
3  20  10 2
4
 Chapter 14: Definite Integrals
0
 I2  0
 I1 =   (1  t 2 )dt 
1 2

 sin
3
1 I3 = x dx = 0
=  (1  t 2 )dt 
0 2
1
 t3  1 2 ….[ sin3x is an odd function]
=  t   = 1– =
 3 0 3 3 1
1 x 
  I4 =  log   dx
2
 
2
0  x 
I2 =  cos3 x dx =  cos3   x  dx
 2  1
 11 x 
=  log 
0 0
  dx
2 0  1 x 
2
=  sin 3 x dx = I1 =
3  a a

0 ….   f ( x) dx   f (a  x) dx 
 From (i),  0 0 
 2 2  800 1
 x 
I = 200    = =  log 
3 3 3  dx
0 1 x 

1
sin x  cos x
2
1 x 
18. I1 =  dx =   log   dx
0
1  sin x cos x 0  x 

 I4 =  I4  2I4 = 0
cos x  sin x
2
=  dx  I4 = 0
0
1  sin x cos x
 I1 = I3 = I4 = 0, but I2  0
 2 


2
  
....   f ( x) dx =  f   x  dx  2
x sin 2n x
 0 0  2  

19. Let I = 0 sin 2n x  cos2n x dx

2
sin x  cos x
2 (2  x)sin 2n (2  x)
=  dx =  I1 = 0 sin 2n (2  x)  cos2n (2  x) dx
0
1  sin x cos x
2
 2I1 = 0  I1= 0 sin 2n x
2 = 2  dx  I
sin x  cos 2n x
2n
  cos
6
I2 = x dx 0


0
sin 2n x

= 2  cos 6 x dx
 I=2 0 sin 2n x  cos2n x dx
0
 sin 2n x 
6
….[ cos x is a periodic f with period ] n
 is a periodic f n with period  
 sin x  cos x
2n 2n


2 
= 2  (cos 6 x + cos6 (  x)) dx 2
sin 2n x
= 4  dx
0
0 sin
2n
x  cos 2n x
 2a a

….   f ( x)dx   [f ( x)  f (2a  x)]dx   2a a

 0 0   
 f ( x )d x  2  f ( x)dx, 

.…  0 0 
2  if f (2a  x)  f ( x) 
= 2.2  cos 6 x dx
0 
= 4 
 6  1 6  3 6  5  5 4
= 4. . =
6  6  2  6  4  2 8 = 2
5
Std. XII : Triumph Maths 
 x
4
 bsin x  22. f(x) =  sin 6 tdt
20.  a | sin x |  1  cos x  c  dx = 0 0
x+  x+
4
 f(x + ) =  sin 6 tdt =  sin 6 tdt +  sin
6
i.e., I1 + I2 + I3 = 0 tdt
 0 0 
4 x
I1 = a  | sin x | dx = f() +  sin 6 (u  )du , where t = u + 
 0
4
x
0 
4
 = f() +  sin 6 u du
 
= a   | sin x | dx   | sin x | dx  0

 4 0
 x
= f() +  sin 6 tdt = f() + f(x)
0 
4
 0
 
= a   ( sin x) dx   sin x dx   f( + x) = f() + f(x)
 4 0

23. Let f(x) = ax2 + bx + c
= a  cos x  /4   cos x 0 
0 / 4
   f (x) = 2ax + b

= a 1 

1
2

1
2

 1 = a 2  2

  f (x) = 2a
f(0) = c = 3

4
f (0) = b = –7
bsin x
I2 =  1  cos x dx

f (0) = 2a = 8
4  a=4
=  b log |1  cos x | / 4
 /4  f(x) = 4x2 – 7x + 3
2 2
    
 f ( x)dx =  (4 x  7 x  3)dx
2
=0 …. cos    cos 
  4  4 1 1

 2
 4 x3 7 x 2 
    c
4
=    3x 
I3 = c  1dx = c      3 2 1
 4 4 2
4 32 4 7 
=  14  6     3 
 I1 + I2 + I3 = 0 becomes 3 3 2 
c

a 2 2 + = 0
2
 ….[ I2 = 0]
=
32  42  18  8  21  18 
 
 The given equation is a relation between a and c. 3  6 
8 5 16  5 11

log   =  = 
   2xe
2
2 x2
3 6 6 6
21. Let I = 0
cos e x dx
1
2A
x2
Put e = t 2 xe dx = dt x2 24.  f ( x ) dx 
0

When x = 0, t = e0 = 1
1
    x   2A
When x = log , t = elog  /2 
2 2
  A sin  2   B dx 
0


1
2  2A  x   2A
 I =  cos tdt = sin t 1
 /2
  cos    Bx  
1
   2  0 
 2A 2A
= sin  sin 1 = 1 – sin1  B B=0
2  
6
 Chapter 14: Definite Integrals

 x   from the given condition,

Now, f(x) = A sin    B
 2  2a 2 a 2
+ a  20 
 x   3 3
 f (x) = A cos   .
 2  2 3a 2
  a  20  0
3
 1  A 1
 f   = .  a2 + a  20  0
2 2 2
 (a + 5) (a  4)  0
A 4
 2 A=  5  a  4
2 2 
The positive integer values of a satisfying the
sin x  sin 2 x  sin 3 x sin 2 x sin 3x above inequality are 1, 2, 3, 4.
25. f(x) = 3  4sin x 3 4sin x  There are 4 such values.
1  sin x sin x 1 27. Since, 1  sin x  1  2  2sin x  2
sin x sin 2 x sin 3x 3
2
5
6 

= 0 3 4sin x (C1  C1  C2  C3)  [2sin x]dx   [2sin x]dx   [2sin x]dx

  5
0 sin x 1 2 2 6

7 3
= sinx (3  4 sin2x) 6 2
  [2sin x]dx   [2sin x]dx
= 3sinx  4 sin3x  7
= sin 3x 6

5 7 3
 
6  6 2
2 2
  f ( x)dx =  sin 3x dx
=  (1)dx   (0)dx   (1)dx   (2)dx
 5  7
0 0 2 6 6

1  5   7   3 7 
cos3x 0
 /2
=  =     0      2  
3  6 2  6   2 6 
1 3  2  4 
=   cos  cos 0  =   
3 2  6 6 6 2

1 28. Applying R 1  R 1  sec x R 3 , we get

=   0  1
3
0 0 sec2 x  cot x cosec x  cos x
1 f ( x)  cos x cos x
2 2
cosec2 x
= 2
3 1 cos x cos2 x

 = (sec2 x + cot x cosec x  cos x) (cos4 x  cos2 x)

2
  cos3t 3   = (sec2 x + cot x cosec x  cos x) ( cos2x sin2 x)
 a  cos t   a sin t  20cos t  dt
2
26. 
0  4 4   =  sin2 x  cos3 x + cos3 x sin2 x
/ 2 =  sin2 x  cos3 x (1  sin2 x)
 1 3  
=  a 2  sin 3t  sin t   a cos t  20sin t  =  sin 2 x  cos5 x
  12 4  0
 

12 3  2 2
= a  (1)  (1)  0   a(0  1)  20(1  0)   f ( x)dx    (sin x  cos5 x) dx
2

 12 4  0 0

3 1   1  4.2 
= a2    + a  20     
 4 12   2 2 5.3.1 
2a 2  8
= + a  20 = 
3 4 15
7
Std. XII : Triumph Maths 
a
1 3 1 
29. 
a 2 1
x 1  dx < 4
x
a
 
1  3 x 3/ 2 
  . 3 +x  2 x  4
a 2 
 2 1
1 
 a a  1  a  1  2 a  2   4
a
 a+ a2<4
 a+ a 6<0
  a 3  a 2 <0
 3 < a < 2
But a cannot be negative and according to
the problem, a  0
 0< a <2
 0<a<4
4 3 4 3
3esin x 3x 2 esin x
30. Let I = 1 x dx = 1 x3 dx
Put x 3  t  3 x 2 dx  dt
64
esin t
 I = 1 t dt
64
esin x
= 1 x dx
 d esin x 
= [f ( x)]164 
….  dx [f ( x )]  
 x 
= f(64)  f(1)
 k = 64

8
 Chapter 15: Applications of Definite Integral

15 Applications of Definite Integral

3. Y
1. Y
x=– x=
y=x+1 y = 5  x2
(1, 2)
X X (2, 1)
O

Y X X
(1, 0) O (1, 0) (2, 0)

Required area =  | sin x | dx

0  Y
=  | sin x | dx   | sin x | dx 2 1

0
0

Required area = 
1
5  x 2 dx –  (1  x) dx
1
=    sin x  dx   sin x dx 2
 0
  ( x  1) dx
=  cos x    cos x 0
0 
1
2 1
= cos0  cos()  (cos  cos0) x 5 x   x2 
=  5  x 2  sin 1    x  
= 1  (1)  ( 1  1) 2 2 5  1  2  1
= 1 + 1  (2) = 4 sq. units 2
 x2 
2. Y Y    x
2 1

y=
2
y = sin1x
5 1  2   1 5  1 
O =1+ sin       2  sin 1   
X X 2  5  2 2  5 
  1  1   1 
y=
2  1    1      2  2   1
x=0  2  2   2 
Y 5  2  5  1  1
1
= 1 + sin1   + 1 + sin1   2 –
y = sin x 2  5 2  5 2
 x = sin y 5  1  2   1  1
 = sin    sin 1   
2 2  5  5  2
Required area = 2 x dy
0 5  2   2  1
= sin 1    cos 1   
 2  5  5  2
2
= 2 sin y dy
….  sin 1 x  cos 1 1  x 2 
0  
= 2  cos y 0
/ 2
5  1  5 1 
=       sq.unit
= 2(0  1) = 2 sq.units 2 2 2  4 2
1
Std. XII : Triumph Maths 
4. Y  x2 
5/ 2

=     2 x 2
5/2

 2 2
5/ 2
B(0,1)  x  3  1 1  x  3  
y=x–1 y=x–1 +   1  ( x  3)  sin 
2

 2  2  1  2
X X  1
(–1, 0) O A(1,0) 
1  25  5   2 1
=   4   2  2  +   1
y=x+1 y = – x +1 2 4  2   2  4
(0, –1)
 
1  1  1 1 
Y + sin–1      0  sin (1) 
2  2  2 
Required area = 4 (area of  OAB)
9 1 3 1  1 
= 1        
1 8 4 2 2 6  2 2 
= 4  y dx
1 3  
0
=   
8 8 4 12
1
= 4   x  1dx  3 1
=    sq.unit
0
6 8 
1
  x2 
=4   x  = 2 sq. units 6. The point of intersection of the curve y = 2x  x2
 2 0 and the line y = x are (0, 0) and (3, 3).

5. The given equation can be written as Y

x=3
x2  6x + 9 + y2 – 4y + 4 = 9 + 4 – 12
 (x  3)2 + (y  2)2 = 1
(2, 0)
This is a circle with centre at (3, 2) and radius X X
O (3, 0)
1.

Y
x = 5/2 y = x
x=2 (3, –3)
(x3)2 + (y2)2=1 y = 2x – x 2

Y
(2, 2)  Required area
(3, 2)
3
=  [(2 x  x 2 )  ( x)]dx
0

X O X 3
=  (3 x  x 2 ) dx
Y 0

3
 Required area  3x 2 x3 
=   
5 5  2 3 0
2 2
=  x dx   2  1  ( x  3) 2 dx 
 =
9
sq. unit
2 2 2
2
 Chapter 15: Applications of Definite Integral
7. Y 1 1  1 1  1 1 
=  (0 + 1)       
2 2  2 2  2 2 

(0, 1)  1 1 
+ (1) + 0     
 2 2
y = cos x
= 2 1+ 2 + 2 –1 + 2
X X
   O  
  ,0
 2 
 ,0
2  
= 4 2  2 sq.unit 
x = 1 x=1
9.
Y Y
2
|x| = 1  x = 1 or x = 1 y=x +2
1
Required Area =  cos x dx
1
y=x

1
= 2 cos x dx (0, 2)
0

….[ cos x is an even fn] X X

O
= 2 sin x 0 = 2(sin 1  0)
1
x=3
= 2 sin 1 Y
3
8. Y
Required area =  ( x 2  2  x)dx
0
y = cos x
y = sin x
5 3  x3 x2 
3

4 2 =   2x  
X
O 
X 3 2 0

4 9
=9+6 0
2
Y 21
= sq. unit
2
Required area
3 10. Y
2
2
x=2–y–y
=  | cos x  sin x | dx
0
1
 5
4 4
X X
=  (cos x  sin x)dx +  (sin x  cos x)dx O
0 
4

3
2
+  (cos x  sin x)dx –2
5
4
Y
= sin x  cos x 0   cos x  sin x  / 4
/ 4 5  /4

Putting x = 0 in the given equation, we get

+ sin x  cos x 5  /4
3 / 2
y = 1 or y = 2
3
Std. XII : Triumph Maths 
1 1 Required area
  x dy =  (2  y  y )dy
2
Required Area = 0 1
2 2 = 2  x  1 dx + 2   x  1 dx
1 1 0
 y 2 y3 
= 2 y    1
2    x  1 
3/ 2
2 3  2 2 3/ 2 0
 = 2.  x  1  + 2.  
3  1 3  1 
1 1  8 0
=2    4  2   4 4 8
2 3  3 =  = sq. units
3 3 3
1 1 8
=2  +6– 13.
2 3 3 x = 1
Y
1 9
= 8 – 3 – = sq.units
2 2 y = x2 + x + 1
D(1, 3)
11. Y x=1 C
y2 = 4a2(x – 1) B
y = 4a
(0, 4a)
X X
A y=0 O

Y
O X
(1, 0) (5, 0) dy
y = x2 + x + 1  = 2x + 1
dx
 dy 
   = 2(1) + 1 = 2 + 1 = 3
 dx (1,3)
Y
4a
 y2   The equation of the tangent at the point (1, 3)
Required area = 0  4a 2  1  1 dy is y  3 = 3(x  1) i.e., y = 3x.
 It passes through origin.
4a
4a
y 2
1 y  3
 Required area
= 2
dy =   = area of the region OABCO + area of the
0
4a 4a 2  3  0
region OCDO
1 1 0 1
=   (64a3 – 0)
2
4a 3 =  y dx   ( y
1 0
1  y 2 ) dx
16a 0 1
= sq. units
 ( x  x  1) dx   ( x  x  1  3x)dx
2 2
3 =
1 0

12. Y 0 1

 (x  x  1) dx   ( x 2  2 x  1) dx
2
=
1 0
y2 =  x + 1 (0, 1) y2 = x + 1 0 1
x x  3
 x3 2

=    x     x2  x 
3 2  1  3 0
 1 1  1
= 0      1   1  1  0
X X  3 2  3
(–1, 0) O (1, 0)
4 1 1
=  
3 2 3
83 2
=
(0, –1) 6
7
= sq. unit
Y 6
4
 Chapter 15: Applications of Definite Integral
14. Y  3 1  3  8 3 3 
2
4y = 3x =2  sin 1    0      0 
 2 4  2   3  3  8 
 3  3  8 3 3 2 2 3 3
x2 + 4y2 = 4 A 1,  = 2    =  
 2   4 3 9 8 3 4 3
X O X 2 2 3
= 
3 12
B  2 1 
=   sq. unit
 3  3 2 3
1,  
 2  Y
15.
y = x2 + x
Y
The equation x2 + 4y2 = 4 is of ellipse with
centre at origin and the equation 4y2 = 3x is of
a parabola with vertex at origin. x=1
Solving the equations, we get x2 + 3x  4 = 0
 (x + 4)(x  1) = 0 X X
O 1
But x =  4 is not possible, since both points
of intersection lie on the right hand side of
Y-axis.
3 Y
 x = 1 and y = ±
2 dy
Slope of tangent = = 2x + 1
 3 dx
 The points of intersection are A  1,  and  y =  (2 x  1)dx = x2 + x + c
 2 
The curve passes through the point (1, 2).
 3
B  1,  .  2 = 12 + 1 + c
 2   c=0
 Required area  The equation of the curve is y = x2 + x, which
3 is a parabola as shown in the figure.
2 1
1
 x3 x 2 
=   x2  x1  dy  Required area =  ( x  x)dx =   
2

 3 0 3 2 0
2
1 1
3 =  0
2
 4 y2  3 2
  
2
= 4 4 y  dy 5
 3
3  = sq. unit
2 6
3
16. Draw AP  to X-axis.
 2
4 y2 
= 2   4  4 y2   dy 1 a3
3  A1 = A(OAP) =  a  a2 =
0  2 2
….[ the function is even] A2 = Area bounded the curve OA and the lines
3 3
OP and AP
a
2
8 2 a a
 x3  a3
= 4 1  y dy   y 2 dy
2
=  y dx =  x dx =   
2

0
30 0 0  3 0 3
3 3  Required area = A1  A2
 2 8y  2
3
y 1 a3 a3 a3
= 4  1  y 2  sin 1 ( y )     =   sq. unit
2 2 0 3  3 0 2 3 6
5
Std. XII : Triumph Maths 
17. Y
3

y = sin1x
 required area = x
0
1  x2  dy
3
=  ( y  2) 2  1  (2 y  4)  dy
0
   1  3
 0,  C B , 
 y  6 y  9  dy
2
 4 =
 2 4
0
X O X  y3 
3
 1  =   3y2  9 y
A ,0 
 2  3 0
= 9 – 27 + 27 – 0
Y = 9 sq. units
Required area 
= area of the rectangle OABC  1  sin x4
1  sin x 
 area of the region OBCO 19. Required area =     dx
0
cos x cos x 


 1 4   1  sin x 1  sin x 
   cos y 0
/ 4
=    sin y dy = ….    0
4 2 0 4 2  cos x cos x 
   1    2 2 
=   1  sq. units 
4
  cos x  sin x  
 cos
x
 sin
x
 
 4 2  2  =    
2 2
 
2 2
x x x x  dx
0 cos 2  sin 2 cos 2  sin 2 
18. Y  2 2 2 2 
 
(y –2)2 = x – 1 
 x x x x 
(2, 3)  cos  sin
4 cos  sin 
=  2 2  2 2  dx
 x x x x 
 cos 2  sin 2 cos  sin
0
2 2 
 
x –2y + 4 = 0

 x x
 1  tan
4 1  tan 
X X =  2  2  dx
Q O R
 x x
(–4, 0) (5, 0) 0
 1  tan 2 1  tan 
 2
Y  x x  x
4 1  tan  1  tan 4 2 tan
The equation of the parabola is =  2 2 dx =  2 dx
(y  2)2 = x  1 2 x 2 x
0
1  tan 0
1  tan
Diff. w.r.t. x, we get 2 2
dy x 1 x
2(y  2) =1 Put tan = t  sec2 dx = dt
dx 2 2 2

dy 1 tan
 = 8
4t
dx 2( y  2)  required area =  dt
 dy  1 1
0 1  t  2
1  t2
     2 1
 dx (2,3) 2(3  2) 2 4t
=  (1  t 2 ) 1  t 2
dt
1 0
 Equation of tangent is y  3 = (x  2)
2   
….  tan  2  1
 2y  6 = x  2  8 
 x  2y + 4 = 0
It cuts the X-axis at the point Q (4, 0) and the
parabola cuts the X-axis at the point R(5, 0).
6
 Chapter 16: Differential Equations

16 Differential Equations 

1. The given equation is Both of these equations reduce to

1  x  1  y  a (x  y )
4 4 2 2
1 1
dx  dy  0
2 2
Put x = sin , y = sin  x y
The equation becomes Integrating both sides, we get
cos  + cos  = a (sin   sin ) log x + log y = log c
    
2 cos   cos    log (xy) = log c
 2   2 
 xy = c, which is the equation of rectangular
    
= 2a sin   cos   hyperbola.
 2   2 


  
cot 
 2 
=a
4. 
1  x2  1  y 2  A x 1  y 2  y 1  x2 
    = 2 cot1 a Put x = tan , y = tan 
 sin1 x2  sin1 y2 = 2 cot1 a The equation becomes
Differentiating w.r.t. x, we get sec  + sec  = A (tan  sec   tan  sec )
1 1 dy
 2x   2y  0 1 1  sin  1 sin  1 
1 x 4
1 y 4 dx    A    
cos  cos   cos  cos  cos  cos 
dy x 1  y 4
  cos   cos   sin   sin  
dx y 1  x 4   A 
cos  cos   cos  cos  
 Degree and order are both 1.
 cos  + cos  = A (sin   sin )
2. Since, the given differential equation cannot
be expressed as a polynomial in differential    
 2 cos   cos  
coefficients, so its degree is not defined.  2   2 

3. The equation of tangent at any point P(x, y) is    

= A.2 sin   cos  
dy  2   2 
Yy= (X – x)
dx   
 cot  =A
 dx   2 
This meets the X-axis at A  x  y ,0  .
 dy      = 2 cot1 A
Similarly, it meets the Y-axis at
 tan1 x  tan1 y = 2 cot1 A
 dy 
B  0, y  x  . Differentiating w.r.t. x, we get
 dx 
According to the given condition, 1 1 dy
  0
P is the mid-point of AB. 1  x 1  y 2 dx
2

dx dy dy 1  y 2
 2x = x  y and 2y = y  x  
dy dx dx 1  x 2
dx dy  Degree and order of the differential equation
 x+y = 0 and y + x = 0
dy dx are both 1.
1
Std. XII : Triumph Maths 

 y 2 1
f   y.e  x =  6
dy y x 1
   
x
5.
dx x  y 1  6 x  6 6x  5
f   = =
x x 1 x 1
Put y = vx  6 x  5  x2
 y = f(x) =  e
dy dv  x 1 
 =v+x
dx dx dy x 2  y 2  1
 The given equation becomes, 8. 
dx 2 xy
dv f (v)
v+x =v+  2xydy = (x2 + 1) dx + y2 dx
dx f (v)
 2xydy  y2dx = (x2 + 1) dx
1 f (v) Dividing by x2, we get
 dx = dv
x f (v) 2 xydy  y 2dx  x 2  1 
=  2  dx
Integrating on both sides, we get x2  x 
log x = log f(v) + log K
 y2   1 
 x = f(v)K  d   =  1  2  dx
 y  x   x 
 x = Kf   Integrating both sides, we get
x
y2 1
 y 1 1 = x c
 f   =  x = cx, where c = x x
x K K
 2 2
y = x  1 + cx
dy When x = 1, y = 0,  c = 0
6. The given equation is  f ( x) y = f ( x)f ( x)  the required solution is y2 = x2  1
dx
i.e., x2  y2 = 1, which is the equation of a
I.F. = e 
f  ( x ) dx
  ef ( x ) hyperbola.
 the required solution is x dy  y dx
9. x dx + y dy + 0
y.ef (x) =  ef ( x ) .f ( x)f ( x)dx x2  y2
=  e t .t dt , where f(x) = t 1 1  xdy  ydx 
 (2xdx + 2ydy) + = 0
2 y 2  x2 
= t.e t   e t .dt 1 2
x
= tet  et + c 1  y
 d ( x 2  y 2 )  d  tan 1  = 0
 y.ef (x) = f(x) ef (x)  ef (x) + c 2  x
 y = f(x)  1 + cef (x) Integrating both sides, we get
7. The given equation is 1 2  y c
( x  y 2 )  tan 1   
2 2 x 2
ex
(x + 1) f (x)  2(x2 + x) f(x) =  y
x 1  x2 + y2 + 2 tan1   = c
If y = f(x), the equation is x
2  y
dy ex  2 tan1   = c  x2  y2
 2 xy = , which is a linear equation x
dx ( x  1) 2
 y cx  y
2 2

 I.F. = e 
 2 x dx
 e x
2  tan1   =
x 2
 the required solution is
y  c  x2  y 2 
2 1 1  = tan  
y.e  x =  dx + c =  c x  2 
( x  1) 2
x 1
When x = 0, y = 5
 c  x2  y 2 
 y = x tan  
 c=6  2 
2
 Chapter 16: Differential Equations
dy 1 dy 1
10. 2x2y = tan (x2y2)  2xy2    tan x =  sec x
dx y 2 dx y
dy 1 1 dy dt
 x2.2y + y2.2x = tan (x2y2) Put = t,   2  
dx y y dx dx
d 2 2
 (x y ) = tan (x2y2)  The equation becomes,
dx
dt
dz  (tan x) t = sec x, which is linear equation.
 = tan z, where z = x2y2 dx
dx
I.F. = e 
tan x dx
 dx = cot zdz 
Integrating both sides, we get = e log (sec x ) = sec x
x = log (sin z) + c  x = log(sin x2y2) + c  the required solution is
  t sec x =  sec2 x dx  c
When x = 1, y = , z=
2 2
1
 1 = log 1 + c,  c = 1  sec x = tan x + c
y
 the required solution is
x = log (sin x2y2) + 1  sec x = y (c + tan x)
 log (sin x2y2) = x  1 dy
 sin (x2y2) = ex 1 13. (xy  x2) = y2
dx
dy dx
11. x3 + 4x2 tan y = ex sec y  y2 = xy  x2
dx dy
Dividing by x3 sec y, we get Dividing by x2y2, we get
1 dy 4 tan y ex 1 dx 1 1 1
  = 3 =  
sec y dx x sec y x x 2 dy x y y 2
dy 1 ex 1 dx 1 1 1
 cos y + (4 sin y). = 3     
dx x x x 2 dy x y y 2
dy dt
Put sin y = t,  cos y = 1
dx dx Put t
x
 the equation becomes,
1 dx dt
dt  4  ex   2 =
   t  3 , which is a linear equation x dy d y
dx  x  x
4
 The equation becomes
 x dx log x 4 dt 1 1
 I.F. = e e e
4 log x
x 4
  t  2 , which is a linear equation
 the required solution is dy y y
tx4 =  e x . x dx  dy
1

 I.F. = e y = elog y = y
 tx4 = xex  ex + c  the required solution is
 sin y.x4 = xex  ex + c
1
When x = 1, y = 0 ty =  dy + c
y
 c=0
 sin y.x4 = xex  ex  ty = log y + c
 sin y = (x  1) ex.x4 y
 = log y + c
x
dy
12. We have, = y tan x  y2 sec x  y = x (log y + c)
dx
The curve passes through the point (1, 1).
1 dy 1
  = tan x  sec x  1 = 1(0 + c),  c = 1
y 2 dx y  the required solution is y = x (log y  1).
3
Std. XII : Triumph Maths 
dy  the equation becomes,
14. = exy (1  ey)
dx dv vx v
v+x = =
dy dx x  vx 1  v
 ey = ex (1  ey) dv v
dx  x = v
dx 1 v
dy
 ey = ex  ex.ey v  v  v2
dx =
1 v
dy
 ey + e x e y = ex 1 v 1
dx  2
dv = dx
v x
dy dt Integrating both sides, we get
Put e y = t,  e y 
dx dx  2 1  1
 The given equation becomes   v  v  dv   x dx + c
dt 1
 e x .t  e x , which is a linear equation.    log v = log x + c
dx v
x dx x  y
I.F. = e 
x

e
 ee    log   = log x + c
y x
 the required solution is
x
x x    log y + log x = log x + c
t.ee =  ee .e x dx y
x
=  e z .dz , where ex = z    log y = c
y
= ez + c This curve passes through (1, 1).
 t.ee
x x
= ee  c  c = 1
x

x
e y .e e  e e  c
x    log y = 1
y
x
15. The equation of the tangent to the curve  + log y = 1
y = f(x) at P (x, y) is y

dy x
Yy= (X  x)  log y = 1 
dx y
x x
1 
 dx   y= e y
= e.e y
This meets the X-axis at  x  y ,0  .
 dy  x

 yey = e
According to the given condition,
y dy 1 y 2
x =y 16. =
dy dx y
dx
y
 xy=
y   1 y 2
dy = 1dx
dy
dx   1  y2 = x + c
dy y  (x + c)2 = 1  y2
 = , which is a homogeneous d.E.
dx x  y  (x + c)2 + y2 = 1
Put y = vx  Radius is fixed, which is 1 and the centre is
dy dv (c, 0) which is a variable centre on the
 =v+x
dx dx X-axis.
4
 Chapter 16: Differential Equations

tx  ty  x y   the required solution is

17. (A) f(tx, ty) = = t1  2 2  tan y
t x t y x y 
2 2 2 2
x cot y =  cot y dy  c
sin 2 y
= t1 f(x, y)
 Homogeneous of degree 1. 1 sin y 1
=   
tan y cos y 2sin y cos y
dy  c
1

2
 tx 
(B) f(tx, ty) = (tx) 3 (ty ) 3 tan 1   1
 ty  = 2 tan y
sec2 y dy + c

1
 x   13
1

2
= (t) x y tan    t f ( x, y )
3 3 3 1
 x cot y = tan y  c
 y
 
1 The curve passes through  1,  .
 Homogeneous of degree  .  4
3
 1 = 1 + c,  c = 0
 the equation of the curve is
(C) f(tx, ty) = tx log t x  t y  log t y 
2 2 2 2
  x = tan y
tx

+ ty e ty 19. The equation of hyperbola is xy = 2

2
  y=
t 2 x2  t 2 y 2 
x

= t  x log   ty e y x
 ty  dy 2
 m1 = =  2 (slope of tangent to the
  dx x
 
x

= t  x log x 2  y 2  log y  y e y  hyperbola)

  dy
m2 = = slope of tangent to the required
= t f(x, y) dx
 Homogeneous of degree 1. family of curves.
The curves are intersecting orthogonally,
(D) f(tx, ty) m1m2 = 1
  2t 2 x 2  t 2 y 2   dy  2 
= tx log    log (tx  ty )       = 1
  tx   dx  x 2 
 tx  2ty  dy x2
+ t 2 y 2 tan    =
 3tx  ty  dx 2
x3
 2 x2 + y 2  2 2  x  2y  Integrating both sides, we get y = + c,
= tx log   t y tan   6
 x( x  y )   3x  y  which is the equation of required family of
 Non-Homogeneous. curves.

dy sin 2 y 20. The given equation is

18.  dy ( y  y3 ) ( y  y 3 )
dx x  tan y  
dx 1  x  xy 2 1  x(1  y 2 )
dx x  tan y
  dx 1  x(1  y 2 )
dy sin 2 y  
dy y (1  y 2 )
dx x tan y
  = , 1 x
dy sin 2 y sin 2 y =  2

y (1+ y ) y
which is a linear equation
1 dx 1 1
1     x=  , which is a linear
I.F. = e 
 cosec 2 y dy  log(tan y )
 e 2
e log (tan y ) 2
dy y y (1  y 2 )
= elog cot y equation
1
= cot y  y dy
 I.F. = e = elog y = y
5
Std. XII : Triumph Maths 
 the required solution is Integrating both sides, we get
2
1 x
xy =   dy + c  
1  y2 y
   = (1  log x)  x 2dx
 xy = tan1 y + c 2
 d  c
The curve passes through (0, 1)  c =    (1  log x)  x 2 dx  dx 
4  dx  2
 the required equation of the curve is x2 x3 1 x3 c

 
2y 2
= (1 + log x) .
3
   dx 
x 3 2
xy + tan1 y =
4 x2 x3 x3 c
  = (1 + log x).  
dy 2 y2 3 9 2
21. Slope of tangent =
dx x2 2 x3 2 x3
  = (1  log x )  c
1 y2 3 9
 slope of normal = 
dy 2 x3  1
dx
= 1  log x    c
3  3
The equation of the normal is x 2 2 x3  2 
  =   log x   c
1 y 2
3 3 
Yy=  (X – x)
dy
dx 23. The equation of the tangent at P(x, y) is
dy dy
 (Y – y) + (X – x) = 0 Yy= (X  x)
dx dx
This meets the Y-axis at  0, y  x  .
dy
The normal passes through the point (3, 0).
 dx 
dy
 (0  y) + (3  x) = 0 According to the given condition,
dx dy
dy y  x = x3
 y =3x dx
dx dy  1 
     y = x2
 ydy = (3  x) dx dx  x 
y2 x2 1
  x dx 1
Integrating both sides, we get = 3x  + c  I.F. = e = elog x =
2 2 x
7  solution of the given equation is
The curve passes through (3 , 4),  c =
2 1 1
y  =   x2   c
y2 x2 7 x x
 the equation of the curve is  3x   2
2 2 2 y x x3
 =   c  y =   cx
 x2 + y2  6x  7 = 0 x 2 2
x3
22. The given equation can be written as  f(x) =   cx ....(i)
2
xdy  ydx = xy3 (1 + log x) dx 1
 f(1) =   c
 ydx  xdy  2
   = xy (1 + log x) dx
 y2  1 3
 1 =  c c =
x 2 2
 d   = xy (1 + log x) dx x 3
3
 y  f(x) =   x ....[From (i)]
2 2
x x 27 9
  d   = x2 (1 + log x) dx  f(3) =  =9
y  y 2 2
6
 Chapter 16: Differential Equations
dV
24. = k(T  t)
dt
 dV = k(T  t)dt
Integrating on both sides, we get
 dV = k  (T  t) dt  c
k(T  t) 2
 V(t) = c ....(i)
2
Initially i.e., when t = 0, V(t) = I
kT 2 kT 2
 I= c c = I 
2 2
k(T  t) 2
kT 2
 V(t) = I  ....[From (i)]
2 2
When t = T,
kT 2
V(T) = I 
2
dy y y
25. =  cos2 ….(i)
dx x x
Put y = vx ….(ii)
dy dv
 =v+x ….(iii)
dx dx
Substituting (ii) and (iii) in (i), we get
dv dv
v+x = v  cos2v  x =  cos2 v
dx dx
Integrating on both sides, we get
dx
 sec v dv =   x  c
2

 tan v =  log x + c
y
 tan =  log x + c ….(iv)
x
 
Since, the required curve passes through 1,  .
 4

 tan =  log 1 + c  c = 1
4
y
 tan =  log x + 1 ….[From (iv)]
x
y
 tan =  log x + log e
x
  e 
 y = x tan1 log   
  x 

7
Target Publications Pvt. Ltd. Chapter 17: Probability Distribution

17 Probability Distribution 

1. Given, 1  3p 1 p
P(X= 3) = 2P(X= 1) and P(X= 2) = 0.3 ….(i) i.e., 0   1, 0   1,
4 4
Now, mean = 1.3
1  2p 1  4p
 0  P(X = 0) + 1  P(X = 1) + 2  P(X = 2) 0  1 and 0  1
4 4
+ 3  P(X = 3) = 1.3
1 1
 7P(X = 1) = 0.7 ….[From (i)]   p 
3 4
 P(X = 1) = 0.1
Also, P(X = 0) + P(X = 1) + P(X = 2) 1  3p 1 p 1  2p
Mean(X) = 1  +0 +1 
+ P(X = 3) = 1 4 4 4
 P(X = 0) + 3P(X = 1) = 0.7 1  4p
+2
….[From (i)] 4
 P(X = 0) + 0.3 = 0.7 2  9p
=
 P(X = 0) = 0.4 4
8
1 1
2.  P(X  x) 1
x 0
Now,   p 
3 4
 a + 3a + 5a + 7a + 9a + 11a + 13a 9
 3 ≥ 9p  
+ 15a + 17a = 1 4
 81a = 1 1
   2  9p  5
1 4
a=
81
1 2  9p 5
   
3. P(E) = P(X = 2 or X = 3 or X = 5 or X = 7) 16 4 4
= P(X = 2) +P(X = 3) +P(X = 5)+P(X= 7)
2
= 0.23 + 0.12 + 0.20 + 0.07 = 0.62 x  x2 
2
5. P(X > 1.5) =  dx =   = 0.4375
P(F) = P(X < 4) 1.5
2  4 1.5
= P(X = 1) + P(X = 2) + P(X = 3)
2
= 0.15 + 0.23 + 0.12 = 0.50 x  x2 
2
and P(X > 1) =  dx =   = 0.75
P(E  F) = P(X is a prime number less than 4) 1
2  4 1
= P(X = 2) + P(X = 3)
 X  1.5  P(X  1.5) 0.4375 7
= 0.23 + 0.12 = 0.35  P   
 X  1  P(X  1) 0.75 12
 P(E  F) = P(E) + P(F)  P(E  F)
= 0.62 + 0.50  0.35 = 0.77 6. P(X = xi) = ki, where 1  i  10

4. Here,
1  3p 1  p 1  2p
, , and
1  4p
are
  P(X  x ) 1 i

4 4 4 4
probabilities when X takes values 1, 0, 1 and  (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)k = 1
2 respectively. Therefore, each is greater than 1
k=
or equal to 0 and less than or equal to 1. 55
1
Std. XII : Triumph Maths 

7. We have,  P(X  x) 1
x 0

 x
1
 k  ( x+ 1)   1
x 0 5
 1 1
2
1
3

 k 1  2    3    4    .... = 1
 5 5 5 
 1 
 1 
 1 5 =1
k 
 1  1 2 
1  5  1   
  5 
 a  (a  d) r  (a  2d) r 2  ....
….  a dr


 
 1  r (1  r) 2 
5 5 
 k   = 1
 4 16 
25k
 =1
16
16
k=
25

2
Target Publications Pvt. Ltd. Chapter 18: Binomial Distribution

18 Binomial Distribution

10 1 3 1
1. We have, p = = 4. Here p = =
100 10 6 2
1 9 1 1
 q=1– = and q = 1  p = 1  =
10 10 2 2
According to the given condition, n = 100
50
P(X  1)  1 1
100  variance = npq = 100   = 25
2 2
1
 1 – P(X = 0) 
2 5. Here n = 8,
1 2 1
 P(X = 0)  p = Probability of getting 1 or 3 = =
2 6 3
n
9 1 1 2
   ,  q=1 =
 10  2 3 3
which is possible if n is at least 7.
1 2 16 4
 n=7  S.D. = npq = 8  = =
3 3 9 3
2. P(X = 1) = 8 . P(X = 3), if n = 5
 5 C1q 4 p1 = 8. 5 C3q 2 p3 6. Let X denote the number of failures in 5 trials.
Then, P(X = r) = 5Cr (1  p)r p5r ; r = 0,1,2,...., 5
5q 2 q2
 2 = 8(10)  2 = 16 31
p p  P(X  1) 
32
 q = 4p
 1  p = 4p 31
 1  P(X = 0) 
 5p = 1 32
1 31
 p=  1  p5 
5 32
1
16   p5
3. P(X = 0) = 32
81
16 1  1
 4C0 p0 q4 = p p 0, 2 
81 2  
4
2
 q4 =   7. Required probability
3 3 3 3 3
1 1 1 1
2 = C0    3C0   + 3C1    3C1  
3
q= 2 2 2 2
3
3 3 3 3
1 1 1 1 1
p= + 3C2    3C2   + 3C3    3C3  
3 2 2 2 2
4
1 1 1 1 5
 P(X = 4) = 4C4p4q0 = p4 =   = = (1 + 9 + 9 + 1)   =
3 81 8 8 16
1
Std. XII : Triumph Maths 
9 Since, P(X= 4), P(X= 5) and P(X= 6) are in A.P.
8. P(at least one success) 
10  2P(X = 5) = P(X = 4) + P(X = 6)
n n n
9 1 1 1
 P(X  1 )   2 nC5   = nC4   + nC6  
10 2 2 2
9  2 nC5 = nC4 + nC6
 1  P(X = 0) 
10 n! n! n!
2 = +
1 5!(n  5)! 4!(n  4)! 6!(n  6)!
 P(X = 0) 
10 2 1 1
0 n  = +
1 3 1 5(n  5) (n  4) (n  5) 6  5
 nC0     
 4   4  10  n2  21n + 98 = 0
3
n
1  (n  7) (n  14) = 0
    n = 7 or 14
 4  10
n 11. Let the probability of success and failure be p
4
    10 and q respectively.
3  p = 2q
4 Since, p + q = 1
 nlog10    log10 10
3 1
 3q = 1  q =
n(log10 4 – log10 3)  1 3
1 1 2
n  p=1 =
log10 4  log10 3 3 3
 required probability
9. According to the given condition, 4 2 5
6  2 1 6  2 1
np + npq = 15 and (np)2 + (npq)2 = 117 = C4     + C5    
n 2 p 2 (1  q 2 ) 117  3  3  3  3
 = 6 0
(np  npq) 2 152  2 1
+ 6C6    
1  q2 117  3  3
 = 240 192 64
(1  q) 2 225 =  
729 729 729
2
 6q2  13q + 6 = 0  q = 496
3 =
729
2 1
 p=1 = 12. Mean = np and variance = npq
3 3
Since, np + npq = 15  np = 20 and npq = 16
1 2  20q = 16
 n  + n  = 15 4
3 9 q=
 n = 27 5
1 4 1
 mean = np = 27  = 9  p=1 =
3 5 5
Since, np = 20
1 1
10. P(getting head) = p =  n  = 20
2 5
1 1  n = 100
 q=1 =
2 2
r n r
1 1
Here, P(X = r) = nCrprqnr = nCr    
2 2
n
n 1
= Cr  
2
2