936 Functions, Limits , Continuity and Differentiability

Limits

1 b 2 c 3 d 4 d 5 c
6 b 7 b 8 a 9 a 10 b
11 a 12 c 13 a 14 c 15 d
Functions 16 a 17 d 18 b 19 d 20 a
21 b 22 c 23 a 24 c 25 c
1 d 2 c 3 b 4 b 5 a 26 b 27 b 28 c 29 a 30 a
6 c 7 d 8 b 9 c 10 a 31 a 32 a 33 c 34 d 35 a
11 a 12 c 13 b 14 d 15 b 36 b 37 c 38 a 39 c 40 b

16 c 17 d 18 c 19 a 20 c 41 b 42 c 43 a 44 a 45 c
46 d 47 b 48 b 49 b 50 b
21 b 22 d 23 d 24 d 25 b
51 b 52 a 53 d 54 a 55 c
26 b 27 a 28 d 29 a 30 b
56 b 57 d 58 a 59 c 60 d
31 c 32 d 33 a 34 b 35 d
61 c 62 a 63 b 64 c 65 c
36 b 37 a 38 c 39 c 40 b
66 c 67 c 68 b 69 b 70 b
41 c 42 a 43 c 44 a 45 c 71 a 72 c 73 b 74 a 75 b
46 b 47 a 48 d 49 d 50 a 76 c 77 b 78 a 79 a 80 c

51 c 52 d 53 c 54 b 55 d 81 b 82 a 83 c 84 b 85 c

56 b 57 b 58 a 59 c 60 c 86 c 87 b 88 c 89 d 90 b
91 c 92 b 93 a 94 c 95 d
61 a 62 d 63 b 64 b 65 b
96 a 97 b 98 a 99 a 100 d
66 c 67 b 68 c 69 c 70 b
101 a 102 d 103 a 104 a 105 d
71 d 72 b 73 b 74 b 75 c
106 b 107 d 108 d 109 b 110 c
76 d 77 d 78 a 79 c 80 d
111 a 112 c 113 a 114 c 115 a
81 c 82 d 83 a 84 a 85 d 116 c 117 b 118 a 119 b 120 c
86 d 87 b 88 a 89 c 90 d 121 c 122 a 123 a 124 c 125 b

91 b 92 b 93 b 94 d 95 b 126 b 127 c 128 d 129 a 130 c

96 b 97 b 98 b 99 b 100 b 131 a 132 a 133 a 134 a 135 b

136 c 137 a 138 d 139 d 140 c
101 c 102 b 103 c 104 c 105 d
141 b 142 b 143 a 144 d 145 a
106 b 107 d 108 b 109 b 110 a
146 a 147 d 148 a 149 d 150 d
111 a 112 a 113 b 114 b 115 b
151 a 152 d 153 b 154 d 155 a
116 b 117 d 118 a 119 c 120 a
156 a 157 d 158 d 159 a 160 d
121 a 122 a 123 d 124 c 125 b 161 c 162 c 163 d 164 c 165 a
126 a 127 c 128 c 129 b 130 d 166 d 167 c 168 c 169 d 170 d

131 c 132 c 133 b 134 c 135 a 171 d 172 b 173 d 174 c 175 a

136 a 137 d 138 a 139 b 140 c 176 b 177 c 178 d 179 d 180 c
181 c 182 a 183 a 184 b 185 a
141 b 142 b 143 d 144 d 145 b
186 a 187 d 188 c 189 a 190 b
146 c 147 b 148 c 149 c 150 b
191 d 192 c 193 d 194 d 195 c
151 e 152 c 153 b 154 c 155 a
196 b 197 d 198 b 199 b 200 c
156 e 157 a,b,c 158 c 159 d 160 c
201 b 202 c 203 d 204 d 205 b
161 a 162 c 163 b 164 b 165 d 206 d 207 d 208 c 209 b 210 b
 Functions, Limits, Continuity and Differentiability 937
211 d 212 c 213 a 214 c 215 b
216 b 217 c 218 a 219 a 220 b
221 c 222 a 223 a 224 b 225 d
226 b 227 b

Continuity Functions

1 d 2 b 3 b 4 a 5 a 1. (d) Given f (x) = cos (log x) ⇒ f (y) = cos (log y)

6 b 7 d 8 b 9 c 10 c
1 x 
11 b 12 c 13 c 14 d 15 c Then f (x). f (y) −  f   + f (xy )
2   y  
16 b 17 d 18 c 19 c 20 a
1  x 
21 b 22 b 23 b 24 b 25 c = cos (log x) cos (log y) − cos log  + cos (log xy )
2   y 
26 b 27 c 28 d 29 b 30 b
1
31 b 32 c 33 a 34 c 35 a = cos (log x) cos (log y) − [ 2 cos (log x) cos (log y)] = 0.
2
36 a 37 d 38 b 39 b 40 b
 1 − cos 2θ 
41 b 42 d 43 d 44 d 45 d 2. (c) f [ f (cos 2θ )] = f  
 1 + cos 2θ 
46 d 47 c 48 a 49 a 50 c
1 − tan 2 θ
51 c 52 a 53 c 54 a 55 c = f (tan 2 θ ) = = cos 2θ .
1 + tan 2 θ
56 a 57 c 58 a 59 a 60 c
3. (b) f (xy ) = sin log xy = sin (log x + log y) .....(i)
61 a 62 d 63 c 64 b 65 b
f (x / y) = sin log( x / y) = sin (log x − log y) .....(ii)
66 d 67 c 68 c 69 a 70 a
∴ f (xy ) + f (x / y) = 2 sin log x cos log y
71 d 72 b 73 b 74 c 75 e
Hence required value of the function is
76 c 77 a 78 d 79 b 80 b,d
2 sin log x cos log y − 2 sin log x cos log y = 0 .
4. (b) f (x + 1) − f (x) = 8 x + 3
Differentiability
⇒ [b (x + 1) 2 + c (x + 1) + d] − (bx 2 + cx + d) = 8 x + 3
1 d 2 d 3 b 4 d 5 a,c,d ⇒ (2b) x + (b + c) = 8 x + 3
6 d 7 a 8 a 9 b 10 b ⇒ 2b = 8, b + c = 3 ⇒ b = 4, c = −1.
11 d 12 d 13 b 14 d 15 c
5. (a) We have f (x + y) + f (x − y)
16 c 17 a 18 d 19 c 20 c
21 b 22 d 23 b 24 b 25 d =
2
[
1 x+y
a + a− x −y + a x −y + a− x +y ]
26 c 27 a 28 d 29 c 30 b
31 b 32 b 33 b 34 c 35 d
=
2
[
1 x y
a (a + a − y ) + a − x (a y + a − y ) ]
36 d 37 d 38 d 39 c 40 c 1 x
= (a + a − x ) (a y + a − y ) = 2 f (x) f (y) .
41 d 42 c 43 d 44 c 45 a 2
46 b 47 c 48 c f (a) a /(a − 1) a2
6. (c) = = 2 = f (a 2 ) .
f (a + 1) (a + 1) / a a − 1
Critical Thinking Questions 7. (d) The given expression is

1 a 2 d 3 c 4 d 5 d 1 x2 x2 
cos (log x 2 ) cos (log y 2 ) − cos log + cos log 2 
6 b 7 a 8 b 9 d 10 d 2  2 y 
11 d 12 d 13 d 14 a 15 d
=
1
2
[
cos (log x 2 + log y 2 ) + cos (log x 2 − log y 2 ) ]
16 a 17 a 18 c 19 c 20 b
21 b 22 b 23 c 24 c 25 a 1 x2 
− cos log + cos (log x 2 − log y 2 )
26 b 27 c 28 c 29 c 30 c 2  2 
31 c 32 c 33 a 34 d 35 d
1 2 2 x2 
36 b 37 c 38 a,b 39 a,c 40 a
= cos log x y − cos log .
2  2 
 938 Functions, Limits, Continuity and Differentiability
8. (b) As log x is defined for only positive values of x. But 16. (c) Given f (x + ay, x − ay) = axy …..(i)
2
log x defined for all real values of x, also log | x | is Let x + ay = u and x − ay = v
also defined ∀ real x. Hence log x 2 and 2 log | x | are u+v u−v
Then x = and y =
identical functions. 2 2a
1+ x  Substituting the value of x and y in (i), we obtain
9. (c) f (x) = log  
1− x  u2 − v2 x 2 − y2
f (u, v) = ⇒ f (x, y) = .
4 4
 2x 
 2x  1 +   x 2 + 1 + 2x  17. (d) f (x) = cos [π 2 ] x + cos [ −π 2 ] x
∴ f  = log  1 + x2  = log  2 
2
1+ x   1 − 2x   x + 1 − 2 x  f (x) = cos(9 x) + cos( −10 x) = cos(9 x) + cos(10 x)
 1+ x2 
 19 x  x
2
= 2 cos  cos 
1 + x  1 + x   2   2
= log   = 2 log  1 − x  = 2 f (x) .
1 − x    π   19π  π  π  −1 1
f   = 2 cos  cos  ; f   = 2 × × = −1 .
10. (a) φ (x) = ax ⇒ φ ( p) = ap  2  4  4  2 2 2
1 1
∴ [φ ( p)]3 = [ ap ]3 = a3 p = φ (3 p) 18. (c) f ( x) = +
f ( x) − 3 x + 2 2x − 4 x − 2 2x − 4
11. (a) f [ f (x)] =
f ( x) + 1 1 1
f (11) = +
 x−3 11 + 2 18 11 − 2 18
 −3
x +1  x − 3 − 3x − 3 3 + x
= = = 1 1 3− 2 3+ 2 6
 x−3 x − 3 + x +1 1− x = + = + = .
  +1 3+ 2 3− 2 7 7 7
 x +1 
10 + x  10 + x 
3+ x 19. (a) e f ( x ) = , x ∈ (−10,10) ⇒ f (x) = log 
Now f [ f ( f (x))] = f   10 − x  10 − x 
1− x 
 200 x 
3+ x  200 x   10 + 100 + x 2   10(10 + x) 
2
 −3 ⇒ f  = log  = log
1− x  3 + x − 3 + 3x 
 10 − 200 x
2
= = =x.  100 + x    10(10 − x) 
3+ x 3+ x +1− x  100 + x 2 
  + 1
1− x   10 + x 
= 2 log  = 2 f ( x)
12. (c) f (x) = cos (log x)  10 − x 
1 x  1  200 x  1
Now let y = f (x) . f (4) −  f   + f (4 x) ∴ f ( x) = f 2
 ⇒ k = = 0.5.
2 4  2  100 + x  2

  π2 3 1 1
⇒ y = cos (log x). cos (log 4) − 1 cos log  x  + cos (log 4 x) 20. (c) ( f + g) = + = 3+ .
2 4  3 2 4 4
⇒ y = cos (log x) cos (log 4) 21. (b) f (x) = f (− x) ⇒ f (0 + x) = f (0 − x) is symmetrical about
x =0.
1
− [cos (log x − log 4) + cos (log x + log 4)] ∴ f (2 + x) = f (2 − x) is symmetrical about x = 2 .
2
x 1 −1 y −1
1 22. (d) f (x) = = ⇒ = {Applying dividendo}
⇒ y = cos (log x) cos (log 4) − [2 cos (log x) cos (log 4)] x −1 y x −1 y
2
x −1 y
⇒ y =0. ⇒ = ⇒ − x + 1 = f (y).
−1 y −1
−1− | −1 | −1 − 1
13. (b) f (−1) = = = −2. ax + b ay + b
| −1 | 1 23. (d) y = ⇒ x(cy − a) = b + ay ⇒ x = = f (y) .
cx − a cy − a
1  4 3 3 
14. (d) x 3 f   = x 3  3 + 2 + + 4  x2 −1 x2 +1− 2 2
x x x x  24. (d) Let f ( x) = = = 1− 2
x2 +1 x2 +1 x +1
= 4 + 3 x + 3 x 2 + 4 x 3 = f ( x) . 2
 x2 +1 > 1 ; ∴ 2 ≤2
15. (b) f (2 x) = 2(2 x)+ | 2 x | = 4 x + 2| x | , x +1
f (− x) = −2 x + | − x | = −2 x + | x | , 2
So 1 − 2 ≥ 1 − 2 ; ∴ −1 ≤ f (x) < 1
x +1
f (x) = 2 x + | x | ⇒ f (2 x) + f (− x) − f (x)
Thus f (x) has the minimum value equal to –1.
= 4 x + 2 | x | + | x | −2 x − 2 x − | x | = 2 | x | .
25. (b) It is a fundamental concept.
 Functions, Limits, Continuity and Differentiability 939
26. (b) Obviously, it is an irrational number for maximum 41. (c) Function f : R → R is defined by f (x) = e x . Let
x = 2, y = 3 then 2 3 is an irrational number. x1 , x 2 ∈ R and f (x1 ) = f (x 2 ) or e x1 = e x 2 or x1 = x 2 .
3
3x + 1 3(−27) + 1 − 80 Therefore f is one-one. Let f (x) = e x = y . Taking log on
27. (a) = = =4.
2x 2 + 2 x = −3
2(9) + 2 20 both sides, we get x = log y . We know that negative
28. (d) Since the mapping is many-one into. real numbers have no pre-image or the function is not
29. (a) It is obvious. onto and zero is not the image of any real number.
30. (b) Width of both interval is same, which can mapped by Therefore function f is into.
these function y = 1 − x and y = 1 + x. 42. (a) | x | is not one-one; x 2 is not one-one;
31. (c) Y
x 2 + 1 is not one-one. But 2 x − 5 is one-one because
f (x) = f (y) ⇒ 2 x − 5 = 2y − 5 ⇒ x = y
}x=1/2 X
O Now f (x) = 2 x − 5 is onto. ∴ f (x) = 2 x − 5 is bijective.
x2 − 4 y2 − 4
Which is step function. 43. (c) Let f (x) = f (y) ⇒ 2
=
x +4 y2 + 4
32. (d)  f (0) = f (−1) = 0 hence f (x) is many one. But there is
no pre-image of −1 . Hence f (x) is into function. So x2 − 4 y2 − 4
⇒ −1 = − 1 ⇒ x2 + 4 = y2 + 4
function is many-one into. x2 + 4 y2 + 4
33. (a) Let x1 , x 2 ∈ R, then f (x1 ) = cos x1 , f (x 2 ) = cos x 2 , so ⇒ x = ± y , ∴ f (x) is many-one.
f ( x1 ) = f ( x 2 ) Now for each y ∈ (−1,1), there does not exist x ∈ X
⇒ cos x1 = cos x 2 ⇒ x1 = 2nπ ± x 2 such that f (x) = y . Hence f is into.
⇒ x1 ≠ x 2 , so it is not one-one. 44. (a) f ′(x) = 2 + cos x > 0 . So, f (x) is strictly monotonic
Again the value of f-image of x lies in between –1 to 1 increasing so, f (x) is one-to-one and onto.
⇒ f [ R] = {f (x) : −1 ≤ f (x) ≤ 1)} 45. (c) f:N→I
So other numbers of co-domain (besides –1 and 1) is f (1) = 0, f (2) = −1, f (3) = 1, f (4) = −2, f (5) = 2
not f-image. f [ R] ∈ R, so it is also not onto. So this
and f (6) = −3 so on. 0
mapping is neither one-one nor onto. 1
34. (b) We have f (x) = (x − 1)(x − 2)(x − 3) 2 –1
3 1
and f (1) = f (2) = f (3) = 0 ⇒ f (x) is not one-one. –2
4
For each y ∈ R , there exists x ∈ R such that f (x) = y . 5 2
Therefore f is onto. Hence f : R → R is onto but not 6 –3
one-one. In this type of function every element of set A has
35. (d) f (−1) = f (1) = 1 ; ∴ function is many-one function. unique image in set B and there is no element left in set
Obviously, f is not onto so f is neither one-one nor onto. B. Hence f is one-one and onto function.
36. (b) It is obvious. 1
46. (b) f ′(x) = > 0, ∀x ∈ [ 0, ∞) and range ∈ [ 0, 1)
37. (a) Let x, y ∈ N such that f (x) = f (y) (1 + x)2
Then f (x) = f (y) ⇒ x 2 + x + 1 = y 2 + y + 1 ⇒ function is one-one but not onto.
⇒ (x − y)(x + y + 1) = 0 ⇒ x = y or x = (−y − 1) ∉ N 47. (a) − 1 + (− 3 ) 2 ≤ (sin x − 3 cos x) ≤ 1 + (− 3 ) 2
∴ f is one-one.
− 2 ≤ (sin x − 3 cos x) ≤ 2
Again, since for each y ∈ N , there exist x ∈ N
∴ f is onto. − 2 + 1 ≤ (sin x − 3 cos x + 1) ≤ 2 + 1
38. (c) f (x1 ) = f (x 2 ) ⇒ x12 = x 22 ⇒ x1 = x 2 , [if X = R + ] − 1 ≤ (sin x − 3 cos x + 1) ≤ 3 i.e., range = [−1, 3]
⇒ f is one-one. Since Rf = R+ ⊆ R = Y ; ∴ f is not onto. ∴ For f to be onto S = [−1, 3] .
39. (c) The total number of injective functions from a set A 48. (d) Let f (x1 ) = f (x 2 ) ⇒ [ x1 ] = [ x 2 ] ⇒
/ x1 = x 2
containing 3 elements to a set B containing 4 elements
{For example, if x1=1.4, x2=1.5, then [1.4]=[1.5] =1}
is equal to the total number of arrangements of 4 by
∴ f is not one-one.
taking 3 at a time i.e., 4 P3 = 24 .
Also, f is not onto as its range I (set of integers) is a
40. (b) For any x, y ∈ R, we have proper subset of its co-domain R.
x −m y−m
f (x) = f (y) ⇒ = ⇒x=y 49. (d) We have f (x) = x + x 2 = x + | x |
x −n y−n
Clearly f is not one-one as f (−1) = f (−2) = 0 but −1 ≠ 2 .
∴ f is one-one.
x −m m − nα Also f is not onto as f (x) ≥ 0, ∀x ∈ R,
Let α∈R such that f (x) = α ⇒ =α ⇒ x =
x −n 1−α Also, range of f = (0, ∞) ⊂ R .
Clearly x ∉ R for α = 1 . So, f is not onto. 50. (a) It is obvious.
 940 Functions, Limits, Continuity and Differentiability
51. (c) Let f (x) be periodic with period T. 63. (b) The function sec −1 x is defined for all x ∈ R − (−1, 1)
Then, f (x + T ) = f (x) for all x ∈ R 1
and the function is defined for all x ∈ R − Z.
⇒ x + T − [ x + T ] = x − [ x] , for all x ∈ R x − [ x]
⇒ x + T − x = [ x + T ] − [ x] So the given function is defined for all
x ∈ R − { (−1, 1) ∪ (n | n ∈ Z)} .
⇒ [ x + T ] − [ x] = T for all x ∈ R ⇒ T = 1, 2, 3, 4,........
The smallest value of T satisfying 64. (b) x 2 − 6 x + 7 = (x − 3) 2 − 2
f (x + T ) = f (x) for all x ∈ R is 1. Obviously, minimum value is – 2 and maximum ∞ .
Hence range of function is [–2, ∞].
Hence f (x) = x − [ x] has period 1.
1
52. (d) It is a fundamental concept. 65. (b) f (x) = log ⇒ sin x ≠ 0 ⇒ x ≠ nπ + (−1)n 0
| sin x |
53. (c) We have f (x) = ax + b, g(x) = cx + d
⇒ x ≠ nπ . Domain of f (x) = R − { nπ , n ∈ I } .
and f (g(x)) = g( f (x))
66. (c) f (x) = log( x − 4 + 6 − x )
⇒ f (cx + d) = g(ax + b) ⇒ a[ cx + d] + b = c[ ax + b] + d
⇒ x − 4 ≥ 0 and 6 − x ≥ 0 ⇒ x ≥ 4 and x ≤ 6
⇒ ad + b = cb + d ⇒ f (d) = g(b) .
∴ Domain of f (x) = [ 4, 6] .
54. (b) Domain of f (x) = R − {3}, and range {1, –1}. 1/ 2
  5x − x 2 
55. (d) [ x] = I (Integers only). 67. (b) We have f (x) = log10   …..(i)
4 
−1 1   
56. (b) − 1 ≤ 5 x ≤ 1 ⇒ ≤ x ≤ . Hence domain is  −1 , 1  .
5 5  5 5 From (i), clearly f (x) is defined for those values of x for

sin −1 (3 − x)  5x − x 2 
57. (b) f (x) = which log10  ≥0
log[| x | −2]  4 
 5x − x 2   2 
Let g(x) = sin −1 (3 − x) ⇒ −1 ≤ 3 − x ≤ 1 ⇒   ≥ 10 0 ⇒  5 x − x  ≥1
4   4 
Domain of g(x) is [2, 4]    
and let h(x) = log[| x | −2] ⇒ | x | −2 > 0 ⇒ x 2 − 5 x + 4 ≤ 0 ⇒ (x − 1)(x − 4) ≤ 0
⇒ | x | > 2 ⇒ x < −2 or x > 2 ⇒ (−∞, − 2) ∪ (2, ∞) Hence domain of the function is [1, 4].
1  1
we know that
 2 (− x − 1), x < −1 − 2 , x < −1
f ( x)  
( f / g)(x) = ∀x ∈ D1 ∩ D2 − {x ∈ R : g(x) = 0} 68. (c) f (x) = tan −1 x, − 1 ≤ x ≤ 1 ; f ′(x) =  1 , − 1 < x < 1
g(x) 2
1 1 + x
∴ Domain of f (x) = (2, 4] − { 3} = (2, 3) ∪ (3, 4] .  (x + 1), x > 1 1
 2  , x >1
2
  x  x
58. (a) y = sin −1 log 3   ⇒ − 1 ≤ log 3   ≤ 1 1 1 1
 3
  3 f ′(−1 − 0) = − ; f ′(−1 + 0) = =
2 1 + (−1 + 0)2 2
1 x
⇒ ≤ ≤ 3 ⇒ 1 ≤ x ≤ 9 ⇒ x ∈ [1, 9] . 1 1 1
3 3 f ′(1 − 0) = = ; f ′(1 + 0) =
1 + (1 − 0)2 2 2
59. (c) For x = −3, 3, | x 2 − 9 | = 0
∴ f ′(−1) does not exist; ∴ domain of f ′(x) = R − { −1} .
Therefore log| x 2 − 9 | does not exist at x = − 3, 3. 69. (c) f (x) is to be defined when x 2 − 1 > 0
Hence domain of function is R − {− 3, 3}. ⇒ x 2 > 1, ⇒ x < −1 or x > 1 and 3 + x > 0
60. (c) f (x) = log | log x | , f (x) is defined if | log x | > 0 and ∴ x > −3 and x ≠ −2
x > 0 i.e., if x > 0 and x ≠ 1 (| log x | > 0 if x ≠ 1) ∴ D f = (−3, − 2) ∪ (−2, − 1) ∪ (1, ∞) .
⇒ x ∈ (0,1) ∪ (1, ∞). 70. (b) According to question, as sin 2 x can’t be negative.
61. −1
(a) f (x) = sin [log 2 (x/2)] , Domain of sin −1
x is x ∈ [−1,1] So the option (b) is correct
Domain of function sin 2 x is [nπ , nπ + π /2] .
1 x
⇒ −1 ≤ log 2 (x / 2) ≤ 1 ⇒ ≤ ≤2 ⇒ 1≤ x ≤ 4 3
2 2 71. (d) f (x) = 2
+ log10 (x 3 − x) . So, 4 − x ≠ 0 ⇒ x ≠ ± 4
4 − x2
∴ x ∈ [1, 4] .
and x 3 − x > 0 ⇒ x(x 2 − 1) > 0 ⇒ x > 0, x > 1
62. (d) Here x + 3 > 0 and x 2 + 3 x + 2 ≠ 0
∴ x > −3 and (x + 1)(x + 2) ≠ 0, i.e. x ≠ −1, − 2 – + – +

∴ Domain = (−3, ∞) − { −1, − 2} . –1 0 1

∴ D = (−1, 0) ∪ (1, ∞) − { 4 } i.e., D = (−1, 0) ∪ (1, 2) ∪ (2, ∞) .
 Functions, Limits, Continuity and Differentiability 941
72. (b) The quantity under root is positive, when 83. (a) Here | x | > 1, therefore x ∈ (− ∞, − 1) ∪ (1, ∞).
− 1 − 3 ≤ x ≤ −1 + 3 . 84. (a) For it must | x | −x > 0
73. (b) Obviously, here | x | > 2 and x ≠ 1 | x | > x but | x | = x for x positive and | x | > x for x
i.e., x ∈ (− ∞, − 2) ∪ (2, ∞) . negative. So, domain will be (− ∞, 0) .

 5 x − x 2  5x − x 2 85. (d) f (x) = x 2 − 1 + x 2 + 1 ⇒ f ( x ) = y1 + y 2

74. (b) log  ≥ 0 ⇒ ≥ 1 or x 2 − 5 x + 6 ≤ 0
 6  6
Domain of y1 = x2 −1⇒ x2 −1 ≥ 0 ⇒ x2 ≥ 1
or (x − 2) (x − 3) ≤ 0 . Hence 2 ≤ x ≤ 3.
x ∈ (− ∞, ∞) − (−1, 1) and Domain of y 2 is real number,
2
75. (c) (i) x ≤ 2 (ii) 9−x > 0 ⇒ | x | < 3 or −3 < x < 3. ∴ Domain of f (x) = (−∞, ∞) − (−1, 1) .
Hence domain is (− 3, 2]. 86. (d) f (x) = e 5 x − 3− 2 x 2
⇒ 5 x − 3 − 2 x 2 ≥ 0 or (x − 1) x − 3  ≥ 0
+ve  2
76. (d) 1 + x ≥ 0 ⇒ x ≥ −1 ; 1 − x ≥ 0 ⇒ x ≤ 1, x ≠ 0 -ve
Hence domain is [ −1, 1] − { 0} . –1 0 1 3/2
∴ D ∈[1, 3 / 2] .
77. (d) f (x) = x − x 2 + 4 + x + 4 − x
87. (b) To define f (x) , 9 − x 2 > 3 ⇒ −3 < x < 3 .....(i)
Clearly f (x) is defined, if 4 + x ≥ 0 ⇒ x ≥ −4
−1 ≤ (x − 3) ≤ 1 ⇒ 2 ≤ x ≤ 4 .....(ii)
4−x≥0 ⇒ x ≤ 4 From (i) and (ii), 2 ≤ x < 3 i.e., [2, 3).
x(1 − x) ≥ 0 ⇒ x ≥ 0 and x ≤ 1 π 
88. (a) f (x) = sec cos 2 x 
∴ Domain of f = (−∞, 4] ∩ [ −4, ∞) ∩ [ 0,1] = [ 0,1] . 4 
1 We know that, 0 ≤ cos 2 x ≤ 1 at cos x = 0, f (x) = 1
78. (a) Clearly − 1 ≤ ≤1
1 + ex and at cos x = 1, f (x) = 2 ; ∴ 1 ≤ x ≤ 2 ⇒ x ∈ [1, 2 ] .
1 1 1
x x
But 2 < e < 3 ⇒ 3 < (e + 1) < 4 ⇒ < < 1
4 1 + ex 3 89. (c) f ( x) = 1 + 2
⇒ Range = (1, 7 / 3] .
 1 3
1 1 x +  +
∴Domain of f (x) =  ,  .  2  4
4 3 90. (d) f (x) = a cos(bx + c) + d …..(i)
79. (c) The function f (x) = log( x 2 − 6 x + 6) is defined when For minimum cos(bx + c) = −1
log( x 2 − 6 x + 6) ≥ 0 from (i), f (x) = −a + d = (d − a)
For maximum cos(bx + c) = 1
⇒ x 2 − 6 x + 6 ≥ 1 ⇒ (x − 5)(x − 1) ≥ 0
from (i), f (x) = a + d = (d + a)
This inequality holds if x ≤ 1 or x ≥ 5 . Hence, the
domain of the function will be (−∞,1] ∪ [ 5, ∞) . ∴Range of f (x) = [ d − a, d + a] .
91. (b) As shown in graph
1
80. (d) 1 − > 0 ⇒ x > 1 . Also, x ≠ 0 . ⇒ Range is (–1, 0]. O (1,0) (2,0)
x
∴ Required interval = (−∞, 0) ∪ (1, ∞) .
y=–1
81. (c) − 1 ≤ 1 + 3 x + 2 x 2 ≤ 1
92. (b) f (x) = cos( x/3)
Case I : 2 x 2 + 3 x + 1 ≥ −1 ; 2 x 2 + 3 x + 2 ≥ 0
We know that −1 ≤ cos( x/3) ≤ 1 .
− 3 ± 9 − 16 −3±i 7
x= = (imaginary). x+2
6 6 93. (b) f (x) =
| x + 2|
Case II : 2 x 2 + 3 x + 1 ≤ 1
−1, x < −2
 3 f ( x) = 
⇒ 2x 2 + 3 x ≤ 0 ⇒ 2x  x +  ≤ 0  1, x > −2
 2
∴ Range of f (x) is { −1,1} .
−3  3  94. (d) Since maximum and minimum values of cos x − sin x
⇒ ≤ x ≤ 0 ⇒ x ∈ − , 0
2  2  are 2 and − 2 respectively, therefore range of
In case I, we get imaginary value hence, rejected
f (x) is [− 2 , 2 ].
 −3 
∴ Domain of function =  , 0 . 95. +
(b) R {as y is always positive ∀ x ∈ R} .
 2 
 π 
(x − 2) (x − 1) 96. (b) f (x) = 2 sin − 2 x 
82. (d) f (x) =  4 
(x − 2) (x + 3) 
Hence domain is {x : x ∈ R, x ≠ 2, x ≠ −3} . ∴ − 2 ≤ f ( x) ≤ 2 and [ −1, 1] ⊂ [ − 2 , 2 ] .
 942 Functions, Limits, Continuity and Differentiability
1 1+ x 
97. (b) f (x) = , sin 3 x ∈ [ −1, 1] 107. (d) Here, f (x) = log 
2 − sin 3 x 1− x 
1  −1
Hence f (x) lies in  , 1 . 1− x  1+ x 
3  and f (− x) = log  = log 
 1 + x  1− x 
98. (b) f (x) = sin 2 (x 4 ) + cos 2 (x 4 ) = 1 . Hence range = {1}.
1+ x 
99. (b) y = f (x) = 9 − 7 sin x. Range = [ 2, 16]. = − log  = − f (x) ⇒ f (x) is an odd function.
1− x 
x 2 + 34 x − 71
100. (b) Let =y 108. (b) f (x) = sin  log (x + 1 + x 2 )
x 2 + 2x − 7  
⇒ x 2 (1 − y) + 2 (17 − y) x + (7y − 71) = 0
⇒ f (− x) = sin [log (− x + 1 + x 2 )]
2
For real value of x, B − 4 AC ≥ 0
 ( 1 + x 2 + x) 
⇒ y 2 − 14 y + 45 ≥ 0 ⇒ y ≥ 9, y ≤ 5 . ⇒ f (− x) = sin log ( 1 + x 2 − x)
 
 ( 1 + x 2 + x) 
π
101. (c) sin −1 x + cos −1 x = holds x lying in [0,1].
2  1 
−1
⇒ f (− x) = sin log  
102. (b) Let y = sin x ⇒ x = sin y  (x + 1 + x 2 ) 
⇒ x = sin 2 y, ∴ 0 ≤ x ≤ 1
⇒ f (− x) = sinlog( x + 1 + x 2 )−1 
2 4  
103. (c) y = f (x) = cos x + sin x
⇒ y = f (x) = cos 2 x + sin 2 x(1 − cos 2 x) ⇒ f (− x) = sin− log(x + 1 + x 2 )
 
⇒ y = cos 2 x + sin 2 x − sin 2 x cos 2 x
1 ⇒ f (− x) = − sinlog(x + 1 + x 2 ) ⇒ f (− x) = − f (x)
2 2  
⇒ y = 1 − sin x cos x ⇒ y = 1 − . sin 2 2 x
4
∴ f (x) is odd function.
3
∴ ≤ f (x) ≤ 1 , ( 0 ≤ sin 2 2 x ≤ 1)
4 109. (b) f (x) = log(x + x 2 + 1)
⇒ f (R) ∈ [ 3 / 4, 1] .
and f (− x) = − log(x + x 2 + 1) = − f (x)
2
x + 14 x + 9 f (x) is odd function.
104. (c) = y ⇒ x 2 + 14 x + 9 = x 2 y + 2 xy + 3y
x 2 + 2x + 3
110. (a) A function is invertible if it is one-one and onto.
⇒ x 2 (y − 1) + 2 x(y − 7) + (3y − 9) = 0 x+2 3 y+2
111. (a) y = ⇒ x= +1 = = f (y) .
Since x is real, ∴ 4(y − 7) 2 − 4(3y − 9)(y − 1) > 0 x −1 y −1 y −1
⇒ 4(y 2 + 49 − 14 y) − 4(3y 2 + 9 − 12 y) > 0 1− x
1−
2 2 1− x  1 + x = x, ∀x
⇒ 4 y + 196 − 56 y − 12 y − 36 + 48 y > 0 112. (a) Since fof (x) = f ( f (x) ) = f  =
1+ x  1+ 1− x
⇒ 8 y 2 + 8 y − 160 < 0 ⇒ y 2 + y − 20 < 0 1+ x
⇒ (y + 5)(y − 4) < 0 ; ∴ y lies between – 5 and 4. ∴ fof = I ⇒ f is the inverse of itself.
2
105. (d) −1 ≤ cosθ ≤ 1 ⇒ cos θ ≤ 1 e x − e−x e 2x − 1
π 113. (b) y = +2⇒ y= +2
and sec 2 θ ≥ 1 for θ > , sec θ ≥ 2
x
e +e −x
e 2x + 1
3
1− y y −1 1  y −1
⇒ sec 2 θ ≥ 4 . ∴ Required interval = [ 2, ∞) . ⇒ e 2x = = ⇒ x = log e  
y−3 3−y 2 3−y
a− x + 1 1 + ax ax + 1 1/ 2 1/ 2
106. (b) In (a), f (− x) = = =− = − f ( x)  y −1  x −1
a −1 1− a −x x
ax − 1 ⇒ f −1 (y) = log e   ⇒ f −1 (x) = log e   .
3−y 3−x
So, it is an odd function.
a− x − 1 1 − ax ax − 1 114. (b) Given f (x) = 2 x( x −1) ⇒ x (x − 1) = log 2 f (x)
In (b), f (− x) = (− x) = −x =x = f ( x)
a− x + 1 1 + ax ax + 1 1 ± 1 + 4 log 2 f (x)
So, it is an even function. ⇒ x 2 − x − log 2 f (x) = 0 ⇒ x =
2
a− x − a x
In (c), f (− x) = = − f ( x) 1 + 1 + 4 log 2 f (x)
a− x + a x Only x = lies in the domain
So, it is an odd function. 2
In (d), f (− x) = sin(− x) = − sin x = − f (x) 1
∴ f −1 (x) = [1 + 1 + 4 log 2 x ] .
So, it is an odd function. 2
 Functions, Limits, Continuity and Differentiability 943

115. (b) Let f (x) = y ⇒ x = f −1 (y). 2x − 1

125. (b) Let f (x) = y ⇒ x = f −1 (y) . Now, y = , (x ≠ −5)
x+5
Hence f (x) = y = 3 x − 5 ⇒ x = y + 5 ⇒ f −1 (y) = x = y + 5
3 3 xy + 5y = 2 x − 1 ⇒ 5y + 1 = 2 x − xy .
x+5 5y + 1 5y + 1
∴ f −1 (x) = ⇒ x(2 − y) = 5y + 1 ⇒ x = ⇒ f −1 (y) =
3 2−y 2−y
Also f is one-one and onto, so f −1 exists and is given by 5x + 1
∴ f −1 (x) = , x ≠ 2.
x+5 2− x
f −1 (x) = .
3  −5   −5 
126. (a) Given (gof )   − ( fog)  
116. (b) Given f (x) = 3 x − 4 . Now let y = f −1 (x) ⇒ f (y) = x  3   3 
⇒ 3y − 4 = x ⇒ 3y = x + 4   − 5    − 5  5
= g f  − f  g  = g(−2) − f   = 2 − 1 = 1 .
x+4 x+4   3    3   3
⇒y= ⇒ f −1 (x) = .
3 3 127. (c) ( fog)x = 2[ g(x)] and (g + g) x = 2 [ g(x)].
x
117. (d) f (x) = . Let y = f (x) ⇒ x = f −1 (y) 128. (c) (gof ) (x) = 3 (x 2 − 1) + 1 = 3 x 2 − 2.
1+ x
x y 129. (b) ( fog) (x) = f {g(x)} = f (log 1) = e o = 1 = log e e.
∴ y= ⇒ y + yx = x ⇒ x =
1+ x 1− y
130. (d) fog (x) = f { g(x)} = e2 log x
= x.
−1 y x
⇒ f (y) = ⇒ f −1 (x) = . 131. (c) gof (x) = g{ f (x)} = [| cos x |].
1− y 1− x
118. (a) For finding inverse of any function, function should be 132. (c) fof (x) = f { f (x)} = (x 2 + 1)2 + 1 = x 4 + 2 x 2 + 2.
1
bijective. Since f (x) = is bijective function.  x 
x −1 133. (b) ( fofof ) (x) = ( fof ) ( f (x)) = ( fof )  
 2 
119. (c) Here, f (θ ) = sinθ (sinθ + sin 3θ )  1+ x 
= sinθ (sinθ + 3 sinθ − 4 sin3 θ ) = 4 sin2 θ (1 − sin2 θ )    
  x  
2 2 2
= 4 sin θ cos θ = (sin 2θ )   2    
 1+ x  x 1+ x2
∴ f (θ ) ≥ 0 for all real θ . = f  = f 
   
x2 2
 1 + x 1 + 2x
2

10 x − 10 − x 1 1+ y   1+ 
120. (a) y = ⇒ x = log10    1+ x2 
x −x
10 + 10 2 1− y 
x
Let y = f (x) ⇒ x = f −1 (y)
 x  1 + 2x 2 x
1 1+ y  1 1+ x  = f = = .
−1
⇒ f (y) = log10   ⇒ f −1 (x) = log10  .  1 + 2x 2   2
1 + 3x 2
2 1 2   x
 − y  1− x  1 + 2
 1 + 2 x 
y+3
121. (a) y = 2 x − 3 ⇒ x =
2 134. (c) φ {ψ (x) } = φ (3 x ) = (3 x ) 2 + 1 = 3 2 x + 1
y+3 x+3
and ψ {φ (x) } = ψ (x 2 + 1) = 3 x
2
⇒ f −1 (y) = ⇒ f −1 (x) = . +1
.
2 2
2x + 1 1 1
122. (a) Let y = f (x) ⇒ y = ⇒ y − 3 xy = 2 x + 1 135. (a) (gof ) (x) = 2 x 2 − 5 x + 2 or g [ f (x)] = 2 x 2 − 5 x + 2
1 − 3x 2 2
y −1 y −1 x −1 ∴ [{ f (x)} 2 + { f (x)} − 2] = 2 [ 2 x 2 − 5 x + 2]
⇒ x= ⇒ f −1 (y) = ⇒ f −1 (x) = .
3y + 2 3y + 2 3x + 2
⇒ f (x) 2 + f (x) − (4 x 2 − 10 x + 6) = 0
2
123. (d) Let y = x + 1 ⇒ x = ± y − 1
− 1 ± 1 + 4 (4 x 2 − 10 x + 6)
−1 −1
⇒ f (y) = ± y − 1 ⇒ f (x) = ± x − 1 ∴ f ( x) =
2
⇒ f −1 (17) = ± 17 − 1 = ±4 2
− 1 ± (16 x − 40 x + 25 − 1 + (4 x − 5)
−1 = = = 2x − 3 .
and f (−3) = ± − 3 − 1 = ± − 4 , which is not possible. 2 2
124. (c) By definition of composition of function, 136. (a) F [ f (x)] = F (log a x) = a loga x = x
2
g( f (x)) = (sin x + cos x) − 1 ⇒ g( f (x)) = sin 2 x
f [ F (x)] = f (a x ) = log a a x = x log a a = x .
 π π
We know sin x is bijective only, when x ∈ − ,  x
 2 2 x  x
137. (d) ( fog)(x) = f (g(x)) = f  = 1− x = = x.
π π −π π 1− x  x x +1− x
Thus g(x) is bijective if − ≤ 2x ≤ ⇒ ≤x≤ . +1
2 2 4 4 1− x
 944 Functions, Limits, Continuity and Differentiability
138. (a) f (x) = (1 + x)2 and g(x) = x 2 + 1 149. (c) The set B satisfied the above definition of function f so
option (c) is correct.
and fog(−3) = f [ g(−3)] = f [ 9 + 1] = f [10]
150. (b) f (x) = 2 x 6 + 3 x 4 + 4 x 2
⇒ fog(−3) = f [10] = [11] 2 = 121 .
f (− x) = 2(− x)6 + 3(− x)4 + 4(− x)2 = f (x)
139. (b) g(x) = 1 + x and f (g(x)) = 3 + 2 x + x .....(i)
⇒ f (x) is an even function and derivative of an even
⇒ f (1 + x ) = 3 + 2 x + x function is always odd.
Put 1 + x = y ⇒ x = (y − 1)2  αx 
α 
αx  αx  x +1
= 
2 2
then, f (y) = 3 + 2(y − 1) + (y − 1) = 2 + y 151. (e) f (x) = ; f ( f (x)) = f 
x +1  x +1 αx
therefore, f (x) = 2 + x 2 . +1
x +1
140. (c) f : R → R, f (x) = sin x and g : R → R, g(x) = x 2
α 2x
⇒ fog(x) = f (g(x)) = f (x 2 ) = sin x 2 . But f ( f (x)) = x , ∴ =x
αx + x + 1
141. (b) ( fog)(x) = f (g(x)) = a(cx + d) + b
(−1)2 x x
and (gof )(x) = g( f (x)) = c(ax + b) + d Put α = −1 , = = x ; ∴ α = −1 .
(−1)x + x + 1 − x + x + 1
Given that, ( fog)(x) = (gof )(x) and at a = 1, b = 2
152. (c) It is well known fact that fractional function always a
⇒ cx + d + 2 = cx + 2c + d ⇒ c = 1 and d is arbitrary. periodic function whose period is 1.
142. (b) Here g(x) = 1 + n − n = 1, x = n ∈ Z Y
1 + n + k − n = 1 + k , x = n + k (where n ∈ Z, 0 < k < 1 )
− 1, g(x) < 0
 X
Now f (g(x)) =  0, g(x) = 0
O
 1, g(x) > 0
 −3 ≤ x < −2,−2 ≤ x < −1,−1 ≤ x < 0
Clearly, g(x) > 0 for all x. So, f (g(x)) = 1 for all x.
y = f (x), 0, ≤ x + 3 < 1, 0 ≤ x + 2 < 1 ,
 αx 
α  0 ≤ x +1 < 1
α f (x) x +1 α 2 .x
143. (d) f ( f (x)) = =  = 0 ≤ x < 1, 1 ≤ x < 2
f (x) + 1  αx  αx + x + 1
 + 1 0 ≤ x < 1, 0 ≤ x − 1 ≤ 1 .
 x +1 
α 2 .x 153. (b) g(x) = 1 + { x} ; f { g(x)} = f { 1 + { x} } = f (k) = 1
∴x= or x((α + 1)x + 1 − α 2 ) = 0
(α + 1)x + 1 where, k = 1 + { x} ,1 ≤ k < 2
2 2
or (α + 1)x + (1 − α )x = 0 . This should hold for all x. x2 + 1
154. (c) g(x) = x 3 + tan x +
⇒ α + 1 = 0, 1 − α 2 = 0 , ∴ α = −1 . P

5 (− x)2 + 1
144. (d) Here f (2) = g(− x) = (− x)3 + tan( − x) +
4 P
5 x2 + 1
2× +1 g(− x) = − x 3 − tan x +
5 4 P
Hence ( fof )(2) = f ( f (2)) = f   = = 2.
4 5 g(x) + g(− x) = 0 because g(x) is a odd function
3× − 2
4
 x2 + 1  3 x2 + 1
145. (b) (gof )(x) =| sin x | and f (x) = sin 2 x ∴  x 3 + tan x +  + − x − tan x + =0
 P   P 
⇒ g(sin 2 x) =| sin x | ; ∴ g(x) = x.
2(x 2 + 1) x2 + 1
146. (c) f [ f (x)] = [ a − {f (x)} ]
n 1/ n
= [ a − (a − x n )] 1 / n = x . ⇒ =0 ⇒ 0≤ < 1 because x ∈ [−2, 2]
P P
2x 5
147. (b) For – 1< x< 1, tan −1 = 2 tan −1 x ⇒ 0≤ <1⇒ P > 5.
1 − x2 P
 π π 155. (a) The domain of function loge {x − [ x]} is R, because [x]
Range of f (x) =  − ,  .
 2 2 is a greatest integer whose value is equal to or less than
 π π zero.
∴Co-domain of function = B =  − ,  .
 2 2 1
156. (e) −1 ≤ log3 x ≤ 1 ; 3−1 ≤ x ≤ 3 ⇒ ≤x≤3
148. (c) f (a − (x − a)) = f (a) f (x − a) − f (0) f (x) .....(i) 3
2 2 1 
Put x = 0, y = 0 ; f (0) = ( f (0)) − [ f (a)] ⇒ f (a) = 0 ∴ Domain of function =  , 3 .
3 
[ f (0) = 1]. From (i), f (2a − x) = − f (x) .
 Functions, Limits, Continuity and Differentiability 945
157. (a,b,c) When x1 = −1 and x 2 = 1 , then
Limits
 −1 − 1 
f (−1) − f (1) = f   = f (−1) ⇒ f (1) = 0 1. (b) Here f (0) = 0
 1 + 1(1) 
1 1
Since − 1 ≤ sin ≤ 1 ⇒ − | x | ≤ x sin ≤ | x |
1− x  x x
Which is satisfied when f (x) = tan −1  
1+ x  We know that lim | x | = 0 and lim −| x | = 0
x →0 x →0
When x1 = x 2 = 0 , then In this way lim f (x) = 0.
x →0
0 − 0 3
x cot x  x 3 cot x 1 + cos x 
f (0) − f (0) = f   = f (0) ⇒ f (0) = 0 2. (c) lim = lim  × 
1− 0  x →0 1 − cos x x →0  1 − cos x 1 + cos x 
When x1 = −1 and x 2 = 0 then 3
 x 
= lim   × lim cos x × lim (1 + cos x) = 2
 −1 − 0  x →0  sin x  x →0 x →0
f (−1) − f (0) = f   = f (−1) ⇒ f (0) = 0
 1− 0  x (e x − 1) 2 x (e x − 1)
3. (d) lim = lim
1− x  x →0 1 − cos x x →0 4. sin2 x
Which is satisfied when f (x) = log  and 2
1+ x 
 
1+ x   (x / 2) 2   e x − 1 
f (x) = log .
1− x  = 2 lim     = 2.

x →0
 sin 2 x   x 
158. (c) It is a fundamental concept.  2 
159. (d) It is a fundamental concept. 1 1
4. (d) lim = lim =∞
160. (c) It is direct consequence of the definition. x →1− | 1− x| h→ 0 1 − (1 − h)

 x, x ∈ Q 1 1
161. (a) ( f − g)(x) =  and lim = lim =∞
x →1 + | 1 − x | h→ 0 1 + h − 1
 − x, x ∉ Q
1
x2 Hence lim = ∞.
x →1 | 1− x|
162. (c) Let y = 2
x +1
n (2n + 1) 2 4n 3 + 4n 2 + n
⇒ (y − 1)x + 0 x + y = 1, y ≠ 1 for real values of x,
2 5. (c) lim = lim
n→ ∞ (n + 2) (n 2 + 3n − 1) n→ ∞ n 3 + 5n 2 + 5n − 2
we have D ≥ 0 ⇒ −4 y(y − 1) ≥ 0 ⇒ y(y − 1) ≤ 0 ⇒ y ∈ [ 0, 1)  4 1 
n3  4 + + 2 
x 2  n n 
0≤ < 1. = lim =4
x2 + 1 n→ ∞ 3  5 5 2 
n 1 + + 2 − 3 
 n n n 
 x + 59 
163. (b) 3 f (x) + 2 f   = 10 x + 30 1 1
 x −1  6. (b) lim = .
n→ ∞ 1 2
For x = 7 , 3 f (7) + 2 f (11) = 70 + 30 = 100 1+ 1+
n
For x = 11 , 3 f (11) + 2 f (7) = 140
3x − a − x + a
7. (b) lim
f (7) f (11) −1 x →a x−a
= = ⇒ f (7) = 4 .
− 20 − 220 9 − 4 3x − a − x + a 3x − a + x + a
= lim ×
x
164. (b)  e = y + 1 + y 2 (x − a)
x →a 3x − a + x + a
2 1
= =
∴ ex − y = 1 + y 2 2 2a 2a
Aliter : Apply L-Hospital’s rule
Squaring both the sides, (e x − y)2 = (1 + y 2 )
3x − a − x + a 3 1
e2 x + y 2 − 2ye x = 1 + y 2 ⇒ e2 x − 1 = 2ye x lim = lim −
x →a x−a x →a 2 3 x − a 2 x+a
e2 x − 1 3 1 1
⇒ 2y = ⇒ 2y = e x − e− x = − = .
ex 2 2a 2 2a 2a
8. (a) Hence lim f (x) = 1 Y
e x − e− x
Hence, y = . x →1
2 Aliter : lim f (x) = lim (1 − h) = 1 y=f(x)
x →1− h→0
165. (d) Given f : (2, 3) → (0, 1) and f (x) = x − [ x]
and lim f (x) = lim 2 − (1 + h) = 1
x →1+ h→0
∴ f (x) = y = x − 2 ⇒ x = y + 2 = f −1(y) ⇒ f −1(x) = x + 2 . X
Hence limit of function is 1. O 1 2
 946 Functions, Limits, Continuity and Differentiability
log [(x − 1) + 1] 2x − 1 2 x log 2
9. (a) lim = 1. 21. (b) lim = lim
x →1 x −1 x →0 (1 + x)1 / 2 − 1 x →0 1 (1 + x) −1 / 2
2
1
log x  f ( x) f ′(x) 
Aliter : Apply L-Hospital’s rule, lim = lim x = 1  xlim
→a g( x)
= lim
x →a g ′( x)

x →1 x − 1 x →1 1  
x n − 2n = 2 log 2 = log 4.
10. (b) lim = n . 2 n−1 ⇒ n. 2 n−1 = 80 ⇒ n = 5 .
x →2 x−2 1 − cos mx  2 sin2 mx 
2
2 2 22. (c) lim = lim  
x . 2 sin x  sin x  x → 0 1 − cos nx x → 0  2 sin 2 nx 
11. (a) lim = 2. lim  ⋅ lim x = 0 .  2
x →0 x2 x → 0 x  x →0
 
2  
  1  
x/2
2
12. (c)  lim  1 +   = e2 .  sin mx  m2 x 2 1 4 
x →∞ x / 2   = lim  mx 2  . . 2 2
 x →0   4 2
n x 
 2  sin nx2 
  nx  
(2 x − 3) ( x − 1) × ( x + 1) −1 −1  
13. (a) lim = = .   2  
x →1 (x − 1) (2 x + 3) × ( x + 1) 5 . 2 10
m2 m2
14. (c) lim kx cosec x = lim x cosec kx = ×1 =
.
x →0 x →0
n2 n2
x 1 kx 1 Aliter : Apply L-Hospital’s rule,
⇒ k . lim = lim ⇒ k = ⇒ k = ±1 .
x →0 sin x k x →0 sin kx k 1 − cos mx m sin mx m 2 cos mx m 2
lim = lim = lim 2 = 2 .
 e1 / x − 1  x →0 1 − cos nx x →0 n sin nx x →0 n cos nx n
15. (d) f (x) =  1 / x  , then

 e +1 e sin x − 1 e sin x − 1 sin x
23. (a) lim = lim ×
x →0 x x →0 sin x x
 1 
e1 / h  1 − 1 / h  e sin x − 1 sin x
 e1 / h − 1   e  =1 = lim × lim = 1×1 = 1.
lim f (x) = lim 1 / h  = lim
sin x
x → 0+ h→ 0 e
 + 1  h→ 0 1 / h  1  x →0 x →0 x
e 1 + 1 / h  Aliter : Apply L-Hospital’s rule,
 e 
Similarly lim f (x) = −1 . Hence limit does not exist. e sin x − 1 cos x e sin x
x → 0−
lim = lim = 1. e 0 = 1.
x →0 x x →0 1
 x
log 1 − 2 sin 2  ( x + 5 + x)
log cos x  2  24. (c) lim x ( x + 5 − x ) ×
16. (a) lim = lim x →∞ ( x + 5 + x)
x →0 x x →0 x
 2  x (5) 5
 x = lim = .
  2 sin 2   x →∞  5  2
x 2  + ......  x  1 + + 1
−  2 sin 2 +   
 2  2    x 
   
    x −1 1
= lim =0 25. (c) lim =− .
x →0 x x →1 (x − 1) (2 x − 5) 3
Aliter : Apply L-Hospital’s rule, Aliter : Apply L-Hospital’s rule.
log cos x − tan x sin x 1 1
lim = lim = 0. 26. (b) lim , let x = or y = , so that x → ∞ ⇒ y → 0
x →0 x x → 0 1 x →∞ x y x
2 sin 2 x  sin x   1 1
17. (d) lim = 2. ∴ lim   = lim  y. sin  = lim y × lim sin = 0 × ... = 0
x →0 2x x →∞  x  y →0 y  y →0 y →0 y

18. (b) Applying L-Hospital’s rule,
1 f ′(9) 1 + sin x − 1 − sin x
⋅ f ′(x) 4 27. (b) Apply L-Hospital‘s rule, lim
x
x →0
2 f ( x) f (9)
lim = = 3 =4 cos x cos x 1 1
x →9 1 1 1 = lim + = + = 1.
x →0 2 1 + sin x 2 1 − sin x 2 2
2 x 9 3
| x| | x| 2 × 9 sin 2 3 x
19. (d) Since lim = −1 and lim = 1, hence limit 28. (c) lim = 18
x →0 − x x →0 + x x →0 (3 x) 2
does not exist. sin α − cosα
29. (a) lim
x+h− x ( x + h)2 − ( x )2 1 α →π / 4 α −π /4
20. (a) lim = lim = .
h→0 h h→0 h( x + h + x ) 2 x   1 1 
 2  sin α . − cos α . 
Aliter : Apply L-Hospital rule,   2 2  
= lim  
x+h− x 1 1 α →π / 4
  π 
lim = lim = . α − 
h→0 h h→0 2 x + h 2 x   4 
 
 Functions, Limits, Continuity and Differentiability 947

 π cosec 2θ − 4
sin α −  41. (b) lim = lim cosecθ + 2 = 4.
 4 x →π / 6 cosec θ − 2 x →π / 6
= 2 lim = 2 ×1 = 2.
α →π / 4  π
α −  x [ 5 C1 + 5 C2 x + 5 C3 x 2 + 5 C4 x 3 + 5 C5 x 4 ] 5
 4 42. (c) lim = .
x →0 x [ 3 C1 + 3 C2 x + 3 C3 x 2 ] 3
Aliter : Apply L-Hospital’s rule,
Aliter : Apply L-Hospital’s rule.
sin α − cos α cos α + sin α 1 1
lim = lim = + = 2. x 9 + a9 2a 9
α →π / 4 α − (π / 4) 1
α →π / 4 2 2 43. (a) lim =9⇒ = 9 ⇒ a8 = 9 ⇒ a = 91 / 8
x →a x + a 2a
log sin x
30. (a) lim tan x log sin x = lim x
x→
π
x→
π cot x 44. (a) lim = 0 as e −1 / x → 0 when x → 0 +
2 2 x →0 + 1 + e −1 / x
1 45. (c) lim [ x] = lim [1 − h] = lim 0 = 0
cos x x →1− h→0 h→0
= lim sin x 2 = 0 (Applying L-Hospital’s rule)
x → − cosec x
π and lim [ x] = lim [1 + h] = lim 1 = 1
2 x →1+ h→0 h→0

31. (a) lim (x − [n]) = lim x − lim[n] = n − n = 0 . Hence limit does not exist.
x →n + 0 x →n + 0 x →n + 0
2 sin 4 x cos 2 x  sin 4 x   x  cos 2 x
2 46. (d) lim = lim 4   =4.
 θ θ x →0 2 sin x cos 4 x x → 0  4 x   sin x  cos 4 x
 cos − sin 
32. (a) lim 1 − sinθ = lim  2 2
= 0. 2 sin 2 x 6 sin 6 x
θ →π / 2 cosθ θ →π / 2  θ θ  θ θ +
 cos − sin   cos + sin  2x 6x 2+6
 2 2  2 2 Aliter : lim = = 4.
x →0 5 sin 5 x 3 sin 3 x 5−3
 2 tan 2 x  −
− 1 5x 3x
tan 2 x − x  2 x  1
33. (c) lim = lim  = . θ θ
x →0 3 x − sin x sin x sin sin
x →0
 3−  2 4 = lim 1 . 4 = 1.
 x  47. (b) lim
θ →0 θ θ →0 4 (θ / 4) 4
Aliter : Apply L-Hospital‘s rule
 b 4 
tan 2 x − x 2 sec 2 2 x − 1 2 − 1 1 1 + +
x x2 
lim = lim = = . 48. (b) lim   = 1.
x →0 3 x − sin x x →0 3 − cos x 3 −1 2 x →∞  a 5
0−h −1 1+ + 2
34. (d) lim f (x) = lim = lim = −1  x x 
x →0 − h→0 h + h 2 h→0 1 + h
d
h 1 49. (b) f (r ) = 2πr .
and lim f (x) = lim = lim =1 dr
x →0 + h→0 h + h 2 h→0 1 + h
50. (b) lim x log sin x = lim log (sin x) x = log [ lim (sin x) x ]
Hence limit does not exist. x →0 x →0 x →0
sin ax a sin ax bx a  x (sin x −1) 
35. (a) lim = lim = .
x →0 sin bx x →0 b ax sin bx b = log  lim (1 + sin x − 1) sin x −1 
 x →0 
πx
sin
sin x o 180 π  πx  lim x (sin x −1)
36. (b) lim = lim =  x ° = radian . = log e [ e x →0 ] = log e 1.
x →0 x x → 0 x 180  180 
x x  a −1
x  x 
x 2 − a2 51. (b) lim
a −b
= lim   − lim  b − 1 
37. (c) lim = lim (x + a) = 2a. x   
x →a x−a x →a
x →0 x →0
 x  x → 0
 x 
f ( x) f ′(x) = log a − log b = log (a / b) .
38. (a) Apply the L-Hospital‘s rule, lim = lim .
x →a g(x) x →a g′(x) 52. (a) Expand sin x and then solve.
2 Aliter : Apply L-Hospital’s rule
39. (c) lim f (x) = 5 − 3 = 2, lim f (x) = = 1.
x →3+ x →3− 5−3 x3 3x 2
sin x − x + cos x − 1 +
cos ax − cos bx lim 6 = lim 6
40. (b) lim
x →0 x2
x →0 x5 x →0 5x4
6x
a+b  b − a − sin x +
2 sin   x . sin  x 6 = lim − cos x + 1 = lim sin x
 2   2  b 2 − a2 = lim
= lim = x →0 20 x 3 x →0 60 x 2 x → 0 120 x
x →0  a + b  2 2  b − a 2
 x . . . x cos x 1
 2  a+b b−a  2  = lim = .
x →0 120 120
Aliter : Apply L-Hospital’s rule,
cos ax − cos bx − a sin ax + b sin bx  a1 / x − 1 
lim = lim 53. (d) lim x (a1 / x − 1) = lim  
2 x →∞ x →∞  1 / x 
x →0 x x → 0 2x  
1/ x
− a 2 cos ax + b 2 cos bx b 2 − a 2 [ e loge a − 1] 1
= lim = . = lim = log e a = − log e .
x →0 2 2 x →∞ 1/ x a
 948 Functions, Limits, Continuity and Differentiability
54. (a) Expand log (1 + x) and then solve. x cos x − sin x − sin x
63. (b) lim = lim
 x − log (1 + x) 
x →0 x 2 sin x 2 sin x + x cos x
x →0
Aliter : Apply L-Hospital’s rule, lim   (By L-Hospital’s rule)
x →0  x2  − cos x 1
= lim = − , (Again by L-Hospital’s rule)
1 x →0 3 cos x − x sin x 3
1− 2
= lim 1 + x = lim 1  1  = 1 . (x − 1) (2 x + 3) 2x 2 + x − 3
x →0 2x x →0 2  1 + x  2 64. (c) lim 2
= lim = 2.
x →∞ x x →∞ x2
2
 1 1  1
1 +   2 +  1 + 
 n (n + 1) (2n + 1)   n n 1  n 1
55. (c) lim  3  = nlim = . 65. (c) lim = .
n→ ∞  6n  → ∞ 6 3 n→ ∞ 4 4
Note : Students should remember that ax + bx 2 + cx 3 a + bx + cx 2
66. (c) lim = lim = a.
x →0 x x →0 1
∑n 1 ∑ n2 1 ∑ n3 1
lim = , lim 3 = and lim 4 = . (1 + x)1 / 2 + (1 − x)1 / 2
n→ ∞ n2 2 n → ∞ n 3 n → ∞ n 4 67. (c) Multiply function by and solve.
f (a)[ g(x) − g(a)] − g(a)[ f (x) − f (a)] (1 + x)1 / 2 + (1 − x)1 / 2
56. (b) lim Aliter : Apply L-Hospital’s rule,
x →a [ x − a]
(1 + x)1 / 2 − (1 − x)1 / 2 1 1
= f (a)g′(a) − g(a) f ′(a) = 2 × 2 − (−1) (1) = 5. lim = lim + = 1.
x →0 x x →0 2 1 + x 2 1− x
sin x − sin α (x − 1) (x 2 + x + 1) 3
57. (d) lim 68. (b) lim = .
x →α x −α x →1 (x − 1) (x + 6) 7
cos x
lim = cosα , (Apply L-Hospital's rule) a + 2x − 3 x
x →α 1 69. (b) lim
x →a 3a + x − 2 x
 c2 d2 
 1+ + 1+  a + 2x − 3 x a + 2x + 3 x 3a + x + 2 x
= lim × ×
(a 2 − b 2 )  x2 x 2  a2 − b 2 x →a 3a + x − 2 x a + 2x + 3 x 3a + x + 2 x

58. (a) lim 2 = 2 .
x →∞ (c − d 2 )  2  c − d2
a2 b 3a + x + 2 x 2
 1+ + 1+  = lim = .
 x2 x2  x →a 3 ( a + 2 x + 3 x) 3 3
 
Aliter : Apply L-Hospital’s rule.
 π  1 − x −1 / 3 1
 x−  70. (b) lim = .
59. (c) lim 2  2  = −2 x →1 (1 − x −1 / 3 ) (1 + x −1 / 3 ) 2
x →π / 2  π 
 sin  − x   Aliter : Apply L-Hospital’s rule.
 2  (1 + nx + nC2 x 2 + ...higher pow ers of x to x n ) − 1
71. (a) lim =n.
Aliter : Apply L-Hospital’s rule. x →0 x
sin x − x Aliter : Apply L-Hospital’s rule.
60. (d) lim tan 3 x
x →0 x3 72. (c) lim 3 + lim cos x = 3 + 1 = 4 .
x →0 3x x →0
Expand sin x, then
73. (b) Let sin −1 x = y ⇒ x = sin y
x3 x5
− + − ... 1 + sin y − 1 − sin y
3! 5!  1 x2  −1 −1
= lim = lim − + − ...  = = . So lim ( x → 0 ⇒ y → 0)
x →0 x 3 x → 0  3 ! 5 !  3 ! 6 y →0 y
 
Aliter : Apply L-Hospital’s rule.  Now multiply it by 1 + sin y + 1 − sin y and solve
 1 + sin y + 1 − sin y 
d 2  
61. (c) [ a sin a] = 2a sin a + a 2 cos a.
da =1
Aliter : Apply L-Hospital’s rule, Aliter : Apply L-Hospital’s rule.
 x {sec (x + y) − sec x} 
(a + h) 2 sin(a + h) − a 2 sin a 74. (a) lim  + sec (x + y)
lim y →0
 y 
h→0 h
 x  cos x − cos (x + y) 
2 (a + h) sin (a + h) + (a + h) 2 cos (a + h) = lim    + ylim sec (x + y)
= lim
  cos (x + y) cos x 
y →0  y →0
h→0 1
= 2a sin a + a 2 cos a.   y  y 
 x sin  x +  sin   
 2   2
62. (a) lim 
 x−3 
 = lim
{
(x − 3) x − 2 + 4 − x
= 1.
} = lim 
y →0  cos ( x + y) . cos x
.
y 
+ sec x
x →3  2 (x − 3)  
 x − 2 − 4 − x  x →3  2 
Aliter : Apply L-Hospital’s rule. = xtanxsecx + secx = secx(xtanx+1).
 Functions, Limits, Continuity and Differentiability 949
Aliter : Apply L-Hospital’s rule, 2 − (3 / x) + (1 / x 2 )
(x + y) sec (x + y) − x sec x 84. (b) lim = 2.
lim x →∞ 1 − (1 / x 2 )
y →0 y
3 + (2 / x) − (1 / x 2 ) 3
(x + y) sec (x + y) tan (x + y) + sec (x + y) − 0 85. (c) lim = .
= lim x →∞ 2 − (3 / x) − (3 / x 2 ) 2
y →0 1
{Differentiating w.r.t.y assuming x as constant} | x − 2| | 2 − h − 2|
86. (c) lim = lim = −1
= x sec x tan x + sec x. = sec x(x tan x + 1) x → 2− x−2 h→0 2 − h − 2

| x − 2| | 2 + h − 2|
x.(2 x − 1) 2x − 1 x2 and lim = lim =1
75. (b) lim = lim . x → 2+ x−2 h→0 2 + h − 2
x →0 1 − cos x x →0 x 1 − cos x
Hence limit does not exist.
x2
= log 2 . lim = (log 2) . 2 = 2 log 2 = log 4 . ( 2 − sec x) cos x (1 + cot x)
x →0 x 87. (b) lim
2 sin 2 x →π / 4 cot x [ 2 − sec 2 x]
2
2 sin 2 (θ / 2)
1 1
76. (c) lim . = (2)
2
2 sin x (1 + cot x) 2 1
θ →0 θ = lim = = .
Aliter : Apply L-Hospital’s rule.
x →π / 4 ( 2 + sec x) 2+ 2 2

77. 2
(b) lim [ 3 − 4 sin θ ] − 1 = 2. Aliter : Apply L-Hospital’s rule.
θ →0
cos x − cos a  − sin x 
sin 3θ − sinθ sin 3θ sinθ 88. (c) lim = lim   = lim sin 3 x = sin 3 a .
Aliter : lim = lim − lim x → a cot x − cot a x → a  − cosec 2 x 
  x →a
θ →0 sinθ θ →0 sin θ θ →0 sin θ

3  π  π 
= − 1 = 2. 2  3 sin  + h  − cos  + h 
1  6  6 
89. (d) lim
You may also apply L-Hospital rule. h→0 3 h ( 3 cos h − sin h)
tan x − sin x sin x − sin x cos x
78. (a) lim 3
= lim 4  3 π  1 π 
x →0 x x → 0 x 3 cos x  sin  + h  − cos  + h 
3  2  6  2  6 
 x   = lim
sin x  2 sin 2   x 
sin 2 h→0 h ( 3 cos h − sin h)
 2  sin x 2 2 . 1 = 1 .
= lim = lim . .
x →0 x 3 cos x x →0  x cos x  x  2 4  2 4sin h 1 4
    = lim . . = .
  2  h→0 3 h ( 3 cos h − sin h) 3

2 sin 2 (x / 2) 90. (b) Let y = x x ⇒ log y = x log x

79. (a) lim = 0.
x →0 x ∴ lim log y = lim x log x = 0 = log 1 ⇒ lim x x = 1 .
y →0 x →0 x →0
Aliter : Apply L-Hospital’s rule.
−1 40 5
80. (c) Put cos x = y and x → 1 ⇒ y → 0.  1  1
2 +  4 − 
 x   x 2 40 .4 5
1− x 1 − cos y 91. (c) lim = = 2 5 = 32 .
lim −1 2
= lim 2
x →∞
 3
45
2 45
x →1 (cos x) y →0 y 2 + 
 x
(1 − cos y)
Now rationalizing it, we get lim 1
y →0 y 2 (1 + cos y ) 92. (b) Let tan −1 2 x = θ ⇒ x = tan θ and as x → 0, θ → 0
2
1 − cos y 1 1 1 1
= lim . lim = × = . 1
y →0 y 2 y →0 1 + cos y 2 2 4 tan θ
x 2 1
⇒ lim = lim = .
x → 0 tan −1 2 x θ →0 θ 2
 2 tan 2 x 
x  x − 
 2x  2 x 2
81. (b) lim = −2. 2 sin . (x )
2 1
x →0 tan x 93. (a) lim = .
x →0  x 2  2
2 sinθ 4 sin 2 x .  
5 cosθ − 4 
θ 5−2 3  
82. (a) lim = = .
θ →0 tan θ 3+1 4 Aliter : Apply L-Hospital’s rule two times.
3+
θ sin 3 x sin x
94. (c) lim 3. + lim = 3 + 1 = 4.
83. (c) Apply formula of sin C − sin D , x →0 3x x → 0 x
sin(2 + x) − sin(2 − x) 2 cos 2. sin x π θ
i.e., lim = lim 95. (d) π − 2 x = θ ⇒ x = − and as x → (π / 2), θ → 0
x →0 x x →0 x 2 2
sin x Now solve yourself.
= 2 cos 2 . lim = 2 cos 2
x →0 x 1 − cos 6 x 2 sin 2 3 x x . 2 sin 2 3 x
You may also apply L-Hospital rule. 96. (a) lim = lim = lim =0.
x →0 x x →0 x x →0 x2
 950 Functions, Limits, Continuity and Differentiability
m sin x − sin 2 x
97. (b) (formula). 110. (c) lim {(1 − sin x) tan x} = lim
n x →π / 2 x →π / 2 cos x
3
4 sin x cos x − sin 2 x
98. (a) lim = 4. Apply L-Hospital’s rule, we get lim = 0.
x →0 x3 x →π / 2 − sin x
x2 111. (a) It is obvious.
99. (a) lim . lim x = 0.
x →0 sin x 2 x →0 112. (c) On rationalising, we get
100. (d) lim f (x) = 1 = lim f (x) . x2 +1− x2 1
x →1− x →1+ lim = lim = 0.
x →∞ 2 x →∞ 2
tan 3 x x +1 + x x +1 + x
101. (a) lim = ∞.
x →π / 2 x 113. (a) It is a fundamental concept.
2x 2 1 114. (c) Apply L-Hospital’s rule, we get
102. (d) lim = = .
x →0 x ( 3 + x + 3 − x ) 2 3 3 1 1
cos x − − sin x −
Aliter : Apply L-Hospital’s rule two times. 1 − x = lim (1 − x) 2 1
lim =− .
2
e x − cos x
x →0 2x x →0 2 2
103. (a) lim sin x + log (1 − x)
x →0 x2 Aliter : lim
2 x →0 x2
Now expanding e x and cos x, we get
 3 5   2 3 4 
3x 2 1 1  x − x + x − ...   − x − x − x − x − ... 
+ x 4  −  + .......  3! 5!   2 3 4 
   
2! 2 ! 4 ! 3 = lim + lim
 x 2
x 2
lim 2
= x →0 x →0
x →0 x 2
 3 5
Aliter : Apply L-Hospital’s rule,  sin x = x − x + x − .. and
 3! 5!
2 
2 xe x + sin x 2 sin x 1 3
lim = lim e x + lim = 1+ = . x2 x3 
x →0 2x x →0 x →0 2 x 2 2 log (1 − x) = − x − − − .. 
1 2 3 
104. (a) Let f (x) = log x ⇒ f ′ (x) =
x − x2  1 1  x4
Therefore, given function = f ′(a) + kf ′(e) = 1 − x 3  +  − ...
2 3 ! 3 4 1
1 k = lim =− .
 a −1 x →0 x2 2
⇒ + = 1 ⇒ k = e 
a e  a  c + dx

Aliter : Apply L-Hospital’s rule to find both the limits. 1 

c + dx  1 
a+ bx  a+ bx
115. (a) lim  1 +  = lim  1 + 

 = ed / b
1 x →∞  a + bx  x → ∞  a + bx  
2
(1 − cos 2 x) | sin x |  
105. (d) lim = lim
x →0 x x →0 x   1 
a + bx
c + dx d 
| sin x| | sin x|  xlim 1 +  = e and lim
x →∞ a + bx
= .
So, lim = 1 and lim = −1  → ∞  a + bx  b
x →0 + x x →0 − x
Hence limit does not exist. 116. (c) Given limit = lim[(1 + tan x)cosec x × 1 /(1 + sin x)cosec x ]
x →0
106. (b) It is a formula.
eαx − e βx eαx − 1 − e βx + 1 = lim [{ 1 + tan x)cot x } sec x × { 1 /(1 + sin x)cosec x } ]
x →0
107. (d) lim = lim
x →0 x x →0 x 1 1
= e sec 0 . = e . = 1.
eαx − 1 e βx − 1 e e
= α lim − β lim = α .1 − β .1 = α − β .
x →0 αx x →0 βx
117. (b) Given limit = lim (4 n + 5 n )1 / n
n→ ∞
(1 / x) − (1 / a) a− x −1 −1
108. (d) lim = lim = lim = 2 . (1 / n) . (4 / 5)n
x →a x−a x →a ax ( x − a) x →a ax a  (5 / 4 )n 
  4  
n

 x + 2
x+3
= lim 5 1 +     = 5 . e0 = 5 .
109. (b) Let A = lim   n→ ∞   5  
x →∞  x + 1    
 
( x + 3)
   4 n 
1   ( x +1)
x+3 x +1
 1     → 0 as n → ∞ 
= lim  1 +  = lim  1 +   =e  5 
x →∞  x + 1 x →∞ 
 x + 1    

 x +1 1
 1  118. (a) Putting x = , the given limit
 xlim 1 +  =e t
 →∞  x + 1
sin t
−1
and lim
(x + 3)
= lim
{1 + (3 / x)} = 1 . = lim t =
1−1
= 0, which is given in (a).
x →∞ ( x + 1) x →∞ {1 + (1 / x)}

 t →0 t −1 0 −1
 Functions, Limits, Continuity and Differentiability 951
1 sin x
x 2 sin −x 1−
Aliter : lim x x − sin x x
x →∞ 1−| x| 124. (c) lim f (x) = lim = lim
x →∞ x →∞ x + cos 2 x x →∞ cos 2 x
1+
1 1 1  x
x 2  − + ....  − x
x 3! x
3
  1  1−0  sin x cos 2 x 
= lim ,  → 0 = = 1,  → 0, → 0 as x → ∞  .
1− | x | 1+ 0  x x
x →∞  x   
 1  125. (b) Put cos −1 x = y. So if x → −1, y → π
x − + .... − x 
 6x  π − cos −1 x π − y
= lim ∴ lim = lim
x →∞ 1− | x | x → −1 x +1 y →π 1 + cos y
1 1 π y 
− terms containing powers of  − 
= lim 6 x x = 0. π − y π − y  2 2
x →∞ | x | −1 = lim = lim
y →π 2 cos (y / 2) y →π π y  π y 
2 sin  −   − 
 sin x   2 2  2 2
1+  1 1 1
 x + sin x  x
119. (b) lim   = lim   = lim 1 = 1 = lim . = .
x →∞  x − cos x  x →∞  1 − cos x  x →∞ y →π 2  π y  2π
  ( π + y ) sin  − 
 x  2  2 2
π y 
sin x cos x  − 
[  lim and lim both are equal to 0]  2 2
x →∞ x x → ∞ x
1/ x   x+ x+ x −x
 1 + tan x  126. (b) lim  x + x + x − x  = lim
120. (c) Given limit = lim   x →∞ 
  x →∞
x →0  1 − tan x  x+ x+ x + x

{ (1 + tan x) } 1 / tan x (tan x ) / x

e x+ x 1 + x −1 / 2 1.
= lim = −1 = e2 . = lim = lim =
x →0 { (1 − tan x)1 / tan x (tan x ) / x
} e x →∞ x →∞ −1 −3 / 2 2
x+ x+ x + x 1+ x +x +1
1/n 2
 x n x − 4 x + 17 − 4 x − 2
127. (c) We have f (x) + g(x) + h(x) =
121. (c) We have lim (x n + yn )1 / n = y lim  1 +   
x 2 + x − 12
n→ ∞ n→ ∞  y 
    2
x − 8 x + 15 (x − 3) (x − 5)
n
= =
x 2 + x − 12 (x − 3) (x + 4)
n
y 1 x
. . 
 x n  
  x  n  y  (x − 3) (x − 5) 2
= y lim 1 +    ∴ lim [ f (x) + g(x) + h(x)] = lim =− .
n→ ∞  y  (x − 3) (x + 4) 7
   
x →3 x →3

2/ x
1 x
n  ax + b x + c x 
.  128. (d) Let y = lim  

 y
n
 n  y  x →0 3
 n   x 
.
  
x
= y lim  1 +     2  ax + b x + c x 
n→∞ 
 y    ⇒ log y = lim log  

  x →0 x  3 
 
log (a x + b x + c x ) − log 3
 x n  = 2 lim
x x →0 x
= ye 0 = y ,  < 1 ⇒   → 0 as n → ∞  .
 y y  Now applying L-Hospital’s rule, we have
 
log y = log (abc) 2 / 3 ⇒ y = (abc) 2 / 3
122. (a) lim a2 x 2 + ax + 1 − a2 x 2 + 1
x →∞ 1+ 2+ x − 3
129. (a) We have lim
ax x →2 x−2
= lim
x →∞
a x + ax + 1 + a 2 x 2 + 1
2 2 1+ 2+ x − 3
= lim
x →2
a a 1 ( 1 + 2 + x + 3 ) (x − 2)
= lim = = .
x →∞ a 1 1 2a 2 2+ x −2
a2 + + + a2 + 2 = lim
x x2 x x →2
( 1 + 2 + x + 3 ) (x − 2)
tan x x x tan x − x
e −e e [e − 1] (x − 2)
123. (a) lim = lim = lim
x →0 tan x − x x →0 tan x − x x →2
( 1 + 2 + x + 3 ) ( 2 + x + 2) (x − 2)
e tan x − x − 1 1 1
= lim e x . lim = e0 × 1 = 1 . = = .
x →0 x →0 tan x − x
(2 3 ) 4 8 3
 952 Functions, Limits, Continuity and Differentiability

2 sin 2 (x 2 / 2) (cos x − 1) (1 − cos x)  x 

1 − cos x 2 136. (c) lim 2 sin a . = −2 sin a . . 
130. (c) We have lim = lim x →0 x sin x x2  sin x 
x →0 1 − cos x x →0 2 sin 2 (x / 2)
2 sin 2 (x / 2)
 sin (x 2 / 2)  = lim − 2 sin a . = − sin a .
  x →0 2
 x   sin x 
1  x2 / 2  x2 / 2 4   
= lim  2
. 2 = 2.  2  x 
2 x →0   sin (x / 2)   x /4 5
    1/ x2 lim (1 + 5 x 2 )1 / 5 x 
2

 x/2    1 + 5x 2  x →0  e5
137. (a) lim  
 = = = e2 .
x →0 1 + 3 x 2 3
e3
m (log x)n 
n ∞   lim (1 + 3 x 2 )1 / 3 x 
2
131. (a) lim x (log x) = lim  Form  x →0  
x →0 + x →0 + x −m  ∞ 

1 [ lim (1 + x)1 / x = e]
n (log x)(n−1) x →0

= lim x (By L-Hospital's rule) (2 x − 3) (3 x − 4)

x →0 + − mx −m−1 138. (d) lim
x →∞ (4 x − 5) (5 x − 6)
n (log x)n−1  ∞
= lim  Form   3 4
x →0 + − mx −m  ∞ x2 2 −  3 − 
x  x 6 3
= lim  = = .
1 x →∞ 2  5  6  20 10
n (n − 1) (log x)(n− 2) x 4 −  5 − 
= lim x (By L-Hospital's rule)  x  x
x →0 + (−m) 2 x −m−1
sin (e x − 2 − 1)
n− 2 139. (d) lim f (x) = lim
n (n − 1) (log x)  ∞ x→2 x → 2 log ( x − 1)
= lim  Form 
x →0 + m 2 x −m  ∞
sin (e t − 1)
....................... = lim , {Putting x = 2 + t}
t →0 log (1 + t)
......................
n! sin (e t − 1) e t − 1 t
= lim =0 = lim . .
x →0 + (−m)n x −m
t →0 et − 1 t log (1 + t)
 
(Differentiating N r and D r n times). t  

sin (e − 1) 1 t  1
log x 1 = lim . + + ...  ×  
132. (a) lim = lim n = 0 (By L-Hospital's rule) t →0 et − 1  1 ! 2 !    1 − 1 t + 1 t 2 − ...  
x →∞ x n x →∞ nx  
 2 3 
log (x − a) ex − ea  0
133. (a) lim x a
= lim x
,  Form  = 1 . 1 . 1 = 1, ( As t → 0, e t − 1 → 0).
x →a log (e − e ) x → a (x − a) e  0
4x
ex ea 140. (c) On rationalization lim
= lim = = 1.
{(x − a) e }
x →∞
x ( x + 8x + 3 + x 2 + 4x + 3
2
x →a + ex ea
4
1  x2 x3  = lim = 2.
1 x− + −.......  x →∞  
log (1+ x ) x  2 3  8
 1+ + 3 4 3
134. (a) (1 + x)1 / x = ex =e 
+ 1+ + 2 
 x x2 x x 
x x 2
x x2

1− + −...... − + − .....
= e 2 3 = ee 2 3
x n − an
141. (b) We know that, lim = n an−1
  x x2 2 
x →a x−a
 1  x x2 
 
= e 1 + − + 
− .....  +  − + − .....  + ...  k
x −5 k
x k − 5k
  2 3  2 !  2 3 
  ∴ lim = k(5)k −1 ; But given, lim = 500 ,
  x →5 x−5 x →5 x − 5

 x 11 2  ∴ k(5)k −1 = 500 ; k (5)k −1 = 4 (5) 4 −1 , ∴ k = 4 .

= e 1 − + x − .... 
 2 24  142. (b) On rationalising, the given limit
ex (1 − x 2 − 1 − x 2 )
(1 + x)1 / x − e + = lim
∴ lim 2 = 11e . x →0
x2 ( 1 − x2 + 1 + x2)
x →0 x2 24
−2 −2
 π   = lim = = −1
135. (b) lim  x tan x −   sec x  x →0
1− x + 1+ x 2 2 1+1
x →π / 2
  2  
143. (a) lim− f (x) = lim f (0 − h) = lim f (0 − h) = 0
2 x sin x − π  0 x →0 h→0 h→0
= lim
2 cos x
, form 0  and lim f (x) = lim f (0 + h) = lim − (0 + h) = 0
x →π / 2   +
x →0 h→0 h→0
[ 2 sin x + 2 x cos x]  
= lim = −1 ,(By L-Hospital’s rule). ∴ lim f (x) = 0 ,  lim f (x) = lim f (x) .
x →π / 2 − 2 sin x x →0  x →0 − x →0 + 
 Functions, Limits, Continuity and Differentiability 953

x e x − log (1 + x)  0  lim f (x) = lim f (0 + h)

144. (d) Let y = lim 2
,  form  x →0 + h→0
x →0 x  0  1
Applying L-Hospital's rule, = lim sin  = (finite number lies between 0 to 1)
h→0  h
1
ex + x ex − 1
y = lim 1 + x ,  0 form   lim− f (x) ≠ lim+ f (x) ; ∴ lim sin  does not exist.
x →0 2x 0 
x →0 x →0 x →0 x
 x3/2 − 8   (x 1 / 2 )3 − (2)3 
1 1  153. (b) y = lim   = lim  
y = lim e x + e x + x e x +  x →4  x − 4 
x →0 2 (1 + x) 2    x →4  ( x − 2)( x + 2) 

1 3 (x 1 / 2 − 2)(x + 4 + 2 x )
y = lim [1 + 1 + 0 + 1] = . ⇒ y = lim
x →0 2 2 x →4 ( x − 2)( x + 2)

4 x 2 + 5x + 8
2
4 (−1 / h) + 5 (−1 / h) + 8 4+4+2 4
(x + 4 + 2 x ) 12
145. (a) lim = lim ⇒ y = lim = = =3.
x → −∞ 4x + 5 h→0 4 (−1 / h) + 5 ( x + 2)
x →4 4 +2 4
Trick : Applying L-Hospital’s rule, we get
(1 / h) 4 − 5h + 8h 2 4 1 3 1/ 2
= lim = =− . x
(1 / h) (− 4 + 5h) −4 2 3
lim 2
h→0
= (4)1 / 2 = 3.
x mx ⋅
1 x→ 4 1 2
 1   1  m
146. (a) Let y = lim  1 +  = lim  1 +  e1 / x e1 / x 1
x→ ∞  mx  x→ ∞  mx  154. (d) lim = lim 1
= lim = e −1 .
x →0 1 
 +1  x →0 x →0 e
 x  e x  ex .e
 1
⇒ y = e1 / m ,  lim  1 +  = e  . x cos x − log(1 + x) 0 
 x→ ∞  x  155. (a) lim ,  form 
  2
x →0 x  0 
147. (d) L.H.L. = lim f (x) = lim (1 − h) = lim 3(1 − h)
x →1− 0 h→0 h→0 Applying L-Hospital’s rule, we have
= lim (3 − 3h) = 3 − 3 . 0 = 3 1
cos x − x sin x −
h→0
x +1 , 0 
R.H.L. = lim f (x) = lim f (1 + h) = lim [ 5 − 3(1 + h)] lim  form 
x →1+ 0 h→0 h→0
x →0 2x 0 
= lim (2 − 3h) = 2 − 3 . 0 = 2 1
− sin x − sin x − x cos x +
h→0
(x + 1) 2 1
Hence lim f (x) does not exist. = lim = .
x→1
x →0 2 2
sin a − tan a cos a − 1 −(1 − cos a)
3
x −8 0  156. (a) lim = lim = lim
148. (a) lim 2
 form 
, a→ 0 sin3 a a→ 0 sin2 a cos a a→ 0 (1 − cos2 a)(cos a)
x →2x −4  0 
 1  1 −1
Applying L-Hospital's rule, we get = lim − =− =
a→ 0  (1 + cos a) cos a  (1 + 1)1 2
3x 2 3× 2× 2
lim = lim =3. n
x→2 2x x→2 2× 2  
n  
 n  1
x 3 − x 2 − 18 0  157. (d) lim   = lim  
149. (d) Let y = lim ,  form  n→∞ n + y  n→∞  y
  1 +
x →3 x−3 0   
n

Applying L-Hospital's rule, we get −1
 y  −n
y 
n
y = lim 3 x 2 − 2 x = (27 − 6) = 21 . = lim  1 +  = lim  1 +   = e − y .
x →3 n→∞  n n→∞  n  

tan −1 x 0  158. (d) L.H.L. = 0 and R.H.L. cannot be found as the function is
150. (d) lim ,  form 
x →0 x  0  not defined for x > 0.
1 159. (a) lim g( f (x)) = lim [ f (x)] 2 + 1 = lim (sin 2 x + 1) = 1 .
x →0 x →0 x →0
2 1 1
= lim 1 + x = lim = =1. 160. (d) Applying L-Hospital’s rule,
x →0 1 x →0 1 + x 2 1+ 0
1
3 4 −1
2+ + 2 1 + log x − x x 1− x
2x 2 + 3 x + 4 x x 2 lim = lim = lim
151. (a) lim = lim = . x →1 1 − 2 x + x 2 x →1 − 2 + 2 x x →1 2 x( x − 1)
x→ ∞ 3x 2 + 3x + 4 x →∞ 3 4 3
3+ + 2 −1 1
x x Again applying L-Hospital’s rule, lim =− .
x →1 4 x − 2 2
152. (d) lim f (x) = lim f (0 − h)
−
x →0 h→0 asin x − 1 asin x − 1 sin x
161. (c) lim sin x
= lim × sin x
 −1  1 x →0 b − 1 x →0 sin x b −1
= lim sin   = lim − sin
h→0  h  h→ 0 h 1 log a
= loge a × = .
= – 1 (finite number lies between – 1 to 1) loge b log b
 954 Functions, Limits, Continuity and Differentiability
 3x / 2 − 3   x/2   a+ x − a− x 
162. (c) lim  x  = lim  3 −3  169. (d) lim  
  x/2 2 2 
x →2
 3 −9  x → 2
 (3 ) − 3  x →0
 x 
1 1  ( a + x − a − x )( a + x + a − x ) 
= lim = .
x →2 3x / 2 + 3 6 = lim  
x →0
 x( a + x + a − x ) 
sin −1 x − tan −1 x 0 
163. (d) lim ,  form   2x  2 1
x →0 x3  0  = lim  = = .
x → 0  x( a + x + a − x )  a+ a a
Applying L-Hospital’s rule,  
1 1 sin2 α − sin2 β
− 170. (d) lim
1 − x2 1 + x2 0 
α →β α2 − β2
lim ,  form 
x →0 3x 2  0  Applying L-Hospital’s rule,
2 sin α cosα sin 2α sin 2β
−1 −2 x 2x lim = lim = .
× + α →β 2α α →β 2α 2β
2 (1 − x 2 )3 / 2 (1 + x 2 )2
= lim 1 1
x →0 6x [log(1+ x )]
171. (d) (1 + x) x = e x
1 1 2  1 1  x2 x3 x4   2 3 
= lim  + = . x− + − +....   1 − x + x − x + .... 
x →0 6  (1 − x 2 )3 / 2 (1 + x 2 )2  2 x  2 3 4 


 2 3 4 

=e =e
x tan 2 x − 2 x tan x  x x2 x3
− +

+.... 
164. (c) lim  2 3
−
4 
x →0 (1 − cos 2 x) 2 = e.e  

x(tan 2 x − 2 tan x) 1 x (tan 2 x − 2 tan x)  2 3   x x2 x3 

2 
= lim = lim   − x + x − x + ...  − + − + ...  
x →0 (2 sin2 x)2 x →0 4 sin4 x   2 3 4 

 2
 3 4 
 
= e + + ... 
    1! 2! 
1 2  x3 2 5
x  2 x + (2 x)3 + (2 x)5 + ...  − 2 x + + x + ...   
 3 15  3 15 
= lim
1    
x →0 4 4
 x 2
x 4 
x 4  1 − + + ....   ex 11e 2 
 3! 5! 
= e − + x + ... + ... 
 2 24 
1 8 2 2 1 ex 11e 2
= . −  = = .  
4  3 3 4 2 (1 + x)1 / x − e  e − 2 − 24 x + ...e 
∴ lim = lim  
(1 − cos 2 x) sin 5 x 2 sin 2 x sin 5 x x →0 x x →0
 x 
165. (a) lim = lim
x →0 x 2 sin 3 x x →0 x 2 sin 3 x  
 sin 5 x   e 11e  e
  ⇒ lim  − − x + ...  = − .
 2 sin 2 x   x  x →0  2 24  2
= lim  
x →0  x 2   sin 3 x  (1 + cosπx) −π sinπx
  172. (b) lim 2
= lim
 x  x →1 tan πx x →1 2π tan πx sec 2 πx
[Using L-Hospital’s rule]
 sin 5 x 
2 5 lim  −1 3 1
 sin x  x → 0 5 x  2 × 5 10 = lim cos π x = .
= lim 2   × = = . x →1 2 2
x →0  x   sin 3 x  3 3
3 lim  m m
x → 0 3 x   x   x 
173. (d) lim  cos  = lim 1 +  cos − 1
166. (d) Applying L-Hospital’s rule, m→∞ m m→∞
  m 
ln(cos x) − tan x − sec 2 x −1   x 
m
 x 
m
lim = lim = lim = . = lim 1 −  − cos + 1 = lim 1 − 2 sin 2
x→0 x2 x →0 2x x →0 2 2 m m→∞  2m 
m→∞
  
5x
− 2
 x
x+2  x+2 
 sin
x 

 x + 2−5   5  −5  −5 lim − 2 2m 
 x2


m
167. (c) lim   = lim  1 −  =e  x  m → ∞  x / 2m   4 m2 
x → ∞ x+2  x →∞  x + 2  lim − 2 sin2
m→ ∞ 
m
2m 
   
 
  =e =e
x2
− 2 lim
 
  = e m→ ∞ 4 m = e0 = 1 .
 lim − 5 x = lim − 5 = −5  . x +b x +b
 x →∞ x + 2 x →∞ 2   x +a  a−b 
 1+  174. (c) lim   = lim  1 + 
 x  x →∞  x + b  x →∞  x+b
a− b
 2   x +b 
 1 + x + x + ... − 1  a − b  a− b 
x
e −1  1! 2!  = lim  1 +   = e a− b .
168. (c) lim = lim   =1. x →∞ 
 x +b 
x →0 x x →0 x  
 Functions, Limits, Continuity and Differentiability 955

 acot x − acos x   cot x −cos x − 1  Trick : Applying L-Hospital’s rule,

175. (a) lim   = lim acos x  a 
x →π / 2 cot x − cos x   cot x − cos x  1
  x →π / 2   {f (x)}−1 / 2 f ′(x) f ′(x) x f ′(1). 1
lim 2 = lim = = 2.
 acot x − cos x − 1  x →1 1 −1 / 2 x →1 f ( x) f (1)
= acos(π / 2) lim   = 1. log a = log a . x
x →π / 2 cot x − cos x  2
 
n(n − 1) n(n − 1)
 cos(π cos 2 x).π .2 cos x(− sin x)   n2 − n + 1   n(n − 1) + 1 
176. (b) Limit = lim   184. (b) lim  2  = lim  
2x  n→ ∞ n − n − 1  n → ∞ n(n − 1) − 1 

x →0
   
n(n−1)
 − sin x   1 
= lim π cos(π cos 2 x). cos x. 
x →0  x   1 + n(n − 1)  e
= lim   = −1 = e 2 .
= π (−1).1.(−1) = π . n→∞
 1 
n(n−1)
e
177. (c) lim [ 3 + h] = 3 and lim− [ 3 − h] = 2  1 − n(n − 1) 
h→0 + h→0  
∴ lim [ x] does not exist. 4 x − 9x 0 
x→3 185. (a) y = lim ,  form 
x →0 x(4 x + 9 x ) 0 
178. (d) f (x) = x(x − 1) sin x − (x 3 − 2 x 2 ) cos x − x 3 tan x
Using L-Hospital’s rule,
= x 2 sin x − x 3 cos x − x 3 tan x + 2 x 2 cos x − x sin x
4 x log 4 − 9 x log 9
y = lim
Hence lim f (x) = lim  sin x − x cos x − x tan x + 2 cos x − sin x  x →0 (4 + 9 x ) + x(4 x log 4 + 9 x log 9)
x
x →0 x 2 x →0  x 

2
= 0 − 0 − 0 + 2 −1 = 1 .  2
log 
log e (1 + x) 0  ⇒ y=
log 4 − log 9
⇒ y=  3  = log 2 .
179. (d) lim ,  form 
x →0 3x − 1 0  2 2 3
1 ax − b x ax − b x x
186. (a) lim = lim . x
Using L-Hospital’s rule, lim 1 + x = 1 = log e . x
x →0 e − 1 x →0 x e −1
3
x → 0 3 x log 3 log e 3
e  ax − 1 b x − 1 x
1 = lim  − 
180. (c) lim cos oscillates between −1 and 1.  x
x →0  x  e x − 1
x →0 x
1  a
∴ Limit doesn’t exist. = (log e a − log e b). = log e  
xf (2) − 2 f (x) log e e b
181. (c) y = lim
x →2 x−2 Trick : Apply L-Hospital’s rule.
xf (2) − 2 f (2) + 2 f (2) − 2 f (x)  3 x − x 3   1 − x 2 
⇒ y = lim 187. (d) f (x) = cot −1  2 
and g(x) = cos−1  
x →2 x−2  1 − 3 x   1 + x 2 
−2 f (x) + 2 f (2) + xf (2) − 2 f (2) Put x = tan θ in both equations
⇒ y = lim
x →2 (x − 2)
 3 tan θ − tan 3 θ 
f (θ ) = cot −1   = cot {tan 3θ }
−1
[ f (x) − f (2)] f (2).( x − 2)
⇒ y = lim − 2 + lim  1 − 3 tan 2 θ 
x →2 x−2 x →2 (x − 2)
f (x) − f (2) π  π
⇒ y = −2 lim + f (2) f (θ ) = cot −1 cot  − 3θ  = − 3θ ⇒ f ′(θ ) = −3 .….(i)
x →2 x−2 2  2
⇒ y = −2 lim f ′(x) + f (2) = − 8 + 4 = − 4 .  1 − tan θ  2
x →2 and g(θ ) = cos−1  −1
 = cos (cos 2θ ) = 2θ
 1 + tan 2 θ 
log x n − [ x] log x n [ x]
182. (a) lim = lim − lim = 0 − 1 = −1.
x →∞ [ x] x →∞ [ x] x →∞ [ x] ⇒ g′(θ ) = 2 …..(ii)

Now lim f (x) − f (a)  = lim f (x) − f (a)  1

f ( x) − 1  
183. (a) y = lim x → a g( x) − g(a) x → a x − a  g( x ) − g(a) 
x →1 x −1   
lim 
( ) ( f (x) + 1) ( x + 1) x−a 
x → a
f ( x) − 1
⇒ y = lim
x →1 ( x −1 ) . ( x + 1) . ( f (x) + 1) = f ′(x).
1
g′(x)
1 3
= −3 × = − .
2 2
f ( x) − 1 x +1 sin −1 (x + 2) 0 
⇒ y = lim . 188. (c) y = lim ,  form 
x →1 x −1 f ( x) + 1 x →−2 x 2 + 2 x 0 
f (x) − f (1) x +1 Using L-Hospital’s rule
⇒ y = lim . lim
x →1 x −1 x →1 f ( x) + 1  1 
 
2 2  
1 − (x + 2)2 1 1
⇒ y = f ′(1). ⇒ y = 2.
2
=2 ⇒ y = lim  ⇒ y= =− .
f (1) + 1 x →−2 2x + 2 −4+2 2
 956 Functions, Limits, Continuity and Differentiability
x +1
x +1 .2  sin x 
 x +3  2  2 − 2 sin 2  
189. (a) lim   = lim  1 +  cos(sin x) − 1  2  = −2. 1 = − 1 .
x →∞  x + 1  x →∞  x +1 197. (d) lim = lim
x →0 x2 x →0 x2 4 2
2
 x +1  1
  2  2  2
1 1
 3n n
=  lim  1 +   =e . 198. (b) lim (3 + n
4 n )n lim (4 n ) n
 x →∞  x + 1  
=  n + 1
 
n→ ∞ n→∞
 4 
1/ n 1/ n
190. (b) lim [1 + (−a) x]1 / x = e − a .    
x →0
   
1  1 
191. (d) Applying L-Hospital’s rule, = lim 4 1 + n
= 4 lim 1 + n
1 n→∞
  4  
n→∞
  4  
0−
2− x −3 2 x−3   3     3  
lim = lim
x →7 x 2 − 49 x →7 2x 0
 1
−1 −1
−1 = 4 1 +  = 4 × (1)0 = 4 × 1 = 4 .
= lim = = .  ∞ 
x →7 4 x x − 3 4.7 7 − 3 56
log(3 + x) − log(3 − x)  a b 
192. (c) lim =k 199. (b) Since, lim  1 + + 2  = e 2
x →0 x
x →∞  x x 
2(ax + b)
1 1
+  x2  x
2
By L-Hospital’s rule, lim 3 + x 3 − x = k ⇒ = k . 
∴ lim  1 +
ax + b  ax +b 
= e2
x →0 1 3  
x →∞ 
 x2  
sin nx  tan x   
193. (d) lim n . lim (a − n)n − =0
x →0 nx x →0 x  2(ax + b)
2(ax + b)
⇒ lim e x = e 2 ⇒ lim = 2 ⇒ 2a = 2 ⇒ a = 1
1 x →∞ x →∞ x
⇒ n((a − n)n − 1) = 0 ⇒ (a − n)n = 1 ⇒ a = n + .
n Thus a = 1 and b ∈ R .
f (2h + 2 + h 2 ) − f (2) f ′(2h + 2 + h 2 )(2 + 2h) −1
194. (d) lim 2
= lim 200. (c) Using L-Hospital's rule, lim = 1.
h→0 f (h − h + 1) − f (1) h→0 f ′(h − h 2 + 1)(1 − 2h) π − cosec 2θ
θ→
2
6× 2
= =3.  −4 
4 ×1   x −1    x −1 
 (3 x −1)
3 x −1  
 4   (−4)  − 4  
e x − e− x 201. (b) lim  1 −  = lim  1 +  
195. (c) y = lim x → ∞ x −1 x →∞  x −1
x →0 sin x  
 x x2   x x2    1 
1 + + + ....  − 1 − + − ....   − 4  3−  
 1! 2!   1! 2!  lim  
x 
⇒ y = lim  x →∞   1 
  1−  
x →0 sin x   x 
=e = e −12 .
x x x 3 5 
2 + + + .......... ...   e x − e sin x  0 
 1! 3! 5 !  202. (c) lim  ,  form 
⇒ y = lim  x →0  x − sin x 
  0 
x →0 sin x
Using L-Hospital’s rule three times, then
 x2 x4 
2 1 + + + .......... . e x − e sin x . cos x e x − e sin x cos 2 x + sin x.e sin x
 3! 4!  lim = lim
⇒ y = lim x →0 1 − cos x x →0 sin x
x →0 sin x
e x − esin x . cos3 x + esin x 2 cos x sin x + esin x . cos x sin x + esin x . cos x
x = lim
x →0 cos x
 x2 
lim 2 1 + + .......  = 1.
x →0 
 2!  ⇒ y = 2 = 2
⇒ y= x 2 + 3x + 2 x 2 + 2x + x + 2
sin x 1 203. (d) lim = lim
lim x →1 x 2 + 4 x + 3 x →−1 x 2 + 3 x + x + 3
x →0 x

Trick : Applying L-Hospital’s rule, (x + 1)(x + 2) x+2 1

= lim = lim = .
x →−1 ( x + 1)(x + 3) x →−1 x + 3 2
1
e0 + 0
e x − e− x e x + e− x e = 1+1 = 2 . 2
1
lim = lim = 204. (d) lim log(1 + x) = lim 2 log(1 + x) x
x →0 sin x x →0 cos x cos 0 1 x →0 x x →0
196. (b) Using L–Hospital’s rule,
 1 
3 1 = lim 2 log e e = 2  lim(1 + x) x = log e e = 1
3. + 3. x →0
 x →0 
3 cos x + 3 sin x 2 2 = 1 .
lim =
x →π / 6 6 6 3 Trick : Using L Hospital’s rule.
 Functions, Limits, Continuity and Differentiability 957
x +1 x +1
1  1 1 1 1 1
 3x − 4  3  3x + 2 − 6  3 213. (a) lim  1 −  +  −  +  −  + ....
205. (b) lim   = lim   n→ ∞ 2  3 3 5 5 7
x →∞ 3 x + 2  x →∞ 3x + 2 
−6 x +1  1 1 
. +  − 
 + 1) 
x +1 3x+2  3x+2 3
 6  3  6  −6   (2n − 1) (2n
= lim  1 −  = lim  1 − 
x →∞ 3x + 2  x →∞  3x + 2   1 1  1
  = lim 1− = .
n→ ∞ 2 2n + 1  2
−2( x +1)
 −2(x + 1) −2   x2 + 1 
= lim e 3x+2 = e−2 / 3 ,  xlim = .
x →∞  →∞ 3 x + 2 3 214. (c) lim  − 2 x − β  = 0
x →∞ x + 1
 
  1  4 
 x1 +  x  3 +   x 2 (1 − α ) − x(α + β ) + 1 − b
(x + 1)(3 x + 4) x  x  ⇒ lim =0
206. (d) lim = lim   x →∞ x +1
x →∞ x 2 (x − 8) x →∞  3 8 
 x 1 −   Since the limit of the given expression is zero, therefore
  x  degree of the polynomial in numerator must be less
  than denominator.
1  4 
 1 +   3 +   ∴ 1 − α = 0 and α + β = 0 ⇒ α = 1 and β = −1 .
1 x  x 
= lim  =0. f (x)
x →∞  x  8 
∫ 4 t dt
3
  1 −  
  x  6 4( f (x))3 × f ' (x)
215. (b) lim (0 / 0 form) = lim
207. (d) In closed interval of x = 0 at right hand side [x] = 0 and x→2 x−2 x→2 1
at left hand side [ x] = −1. Also [0]=0. Therefore 3
= 4( f (2)) × f ' (2) = 18 .
 sin [ x]  1 2 n 
 (−1 ≤ x < 0) 216. (b) lim  + + ..... +
function is defined as f (x) =  [ x] n→∞  1 − n 2 
1 − n2 1 − n2 
 0 (0 ≤ x < 1)
 Σn 1 n2 + n 1
= lim = lim =− .
sin [ x] n→∞ 1 − n 2
2 n→∞ 1 − n 2 2
∴Left hand limit = lim f (x) = lim
x →0 − x →0 − [ x] 12 + 22 + 32 + ..... + n2 Σn 2
217. (c) Given limit = lim = lim
sin (−1) n→ ∞ 1+ n 3 n → ∞ 1 + n3
= sin 1c
=
−1  1 1
Right hand limit = 0. Hence limit doesn't exist. 1 +   2 + 
1 n (n + 1) (2n + 1) 1 n n
= lim = lim
 1 n  n→ ∞ 6 1 + n3 n→ ∞ 6  1 
(10)n   − 1  3 + 1
1 − (10)n  10   1 n 
208. (c) lim = lim =−
n→ ∞ 1 + (10)n +1 n→ ∞  1  10 1 2 1
(10)n+1  1 + n+1  =.1. =  .
 10  6 (1)  3 
∴α = 1 . S − Sn an+1
218. (a) We have lim n+1 = lim =0
log(1 + x ) 3
3 x /(1 + x ) 2 3 n→ ∞ n n→ ∞ n (n + 1)
209. (b) lim = lim ∑k
x →0 3
sin x x → 0 3 sin2 x cos x k =1 2
[By using L- Hospital rule] (Since n → ∞, numerator → a while denominato r → ∞)
 1  x 2 1  1 1 4 + 3an
= lim    .  = .(1)2 . = 1 . 219. (a) We have an+1 =
x → 0  1 + x 3  sin x  cos x  1 + 0 1 3 + 2an
 
4 + 3an
4θ (tan θ − sin θ ) 4θ sin θ (1 − cos θ ) ⇒ lim an+1 = lim
210. (b) lim = lim n→ ∞ n→ ∞ 3 + 2an
θ →0 (1 − cos 2θ )2 θ →0 4 sin4 θ cos θ
4 + 3a
 θ  2 sin2 θ / 2 ⇒ a= ⇒ 2a 2 = 4 ⇒ a = 2
= lim  3 + 2a
θ → 0 sin θ  sin2 θ cos θ
a ≠ − 2 because each an > 0, therefore lim an = a > 0.
2 sin2 θ / 2 1 220. (b) We know that
= lim
θ → 0 (2 sin(θ / 2) cos(θ / 2)2 ) cos θ
sin 2 n A
cos A cos 2 A cos 4 A.... cos 2 n−1 A =
1 1 1 2 n sin A
= lim = .
θ → 0 2 cos2 (θ / 2). cos θ 2 x
Taking A = , we get
211. (d) Applying L-Hospital’s rule. 2n
xn 1  x   x  x x sin x
212. (c) lim = lim = 1. cos  n  cos  n−1  ... cos  cos   =
n→ ∞ n 1  n→ ∞  1  2  2  4  2  x 
x 1 + n  1 + n  2 n sin n 
 x   x  2 
 958 Functions, Limits, Continuity and Differentiability
x x  x   x 
∴ lim cos   cos  ... cos  n−1  cos  n  Continuity
n→ ∞  2 4 2  2 
sin x sin x (x / 2 n ) sin x 1. (d) Here f (2) = 0
= lim = lim = .
n→ ∞ n  x  n→∞ x sin (x / 2 n ) x lim f (x) = lim f (2 − h) = lim | 2 − h − 2 | = 0
2 sin  n  x → 2− h→0 h→0
2 
lim f (x) = lim f (2 + h) = lim | 2 + h − 2 | = 0
  1 n  x → 2+ h→0 h→0
1 −    Hence it is continuous at x = 2 .
1 1 1 1 1   2  
221. (c) y = lim + 2 + 3 + ....... + n = lim 2. (b) f (π / 2) = 3 . Since f (x) is continuous at x = π / 2
n→ ∞ 2 2 2 2 n→ ∞ 2  1
1 −   k cos x  π  k
 2 ⇒ lim   = f   ⇒ = 3 ⇒ k = 6.
x →π / 2  π − 2 x   2 2
 1
lim 1 − n  = 1 − 0 = 1 3. (b) Since limit of a function is a + b as x → 0, therefore to
n→ ∞  2 
be continuous at a function, its value must be
 1 2 3 n  a + b at x = 0 ⇒ f (0) = a + b.
222. (a) lim  2 + 2 + 2 + ....... + 2 
n→ ∞  n n n n 
4. (a) For continuous lim f (x) = f (2) = k
x →2
n
(n + 1)
 1 + 2 + 3 + ...... + n  2 x + x 2 − 16 x + 20
3
= lim   = lim ⇒ k = lim
n→ ∞  n2  n→ ∞ n 2 x →2 (x − 2) 2
1 n+1 1 1 1 (x 2 − 4 x + 4) (x + 5)
= lim = lim 1 + = = lim =7 .
2 n → ∞ n 2 n → ∞ n 2 x→2 (x − 2)2
1 − n2 (1 − n)(1 + n) 2(1 − n) 5. (a) Here lim f (x) = k, lim f (x) = −k and f (0) = k
223. (a) lim = lim = lim x →0 + x →0 −
n→∞ Σn n→∞ 1 n→∞ n
n(n + 1) But f (x) is continuous at x = 0, therefore k must be
2
zero.
1 
= lim 2  − 1 = 2(0 − 1) = −2 . 6. (b) lim f (x) = f (0) = lim (1 + x)1 / x = e.
n→∞  n  x →0 x →0

1 − 2 + 3 − 4 + 5 − 6 + ..... − 2n 1
224. (b) lim 7. (d) Since lim f (x) ≠ f   .
n→ ∞ n2 + 1 + 4 n2 − 1 x →1 / 2  2
 1 2 3   8. (b) f (a) = 0
n  − + − .....  − 2
n n n   = −2 = −2 .
= lim  lim f (x) = lim 
 x2   (a − h) 2
− a  = lim 

− a = 0
n→∞  1 1  1+ 2 3 x → a− a
n 1 + + 4− 
x → a−
  h→0  a 
 n 2 n 
2
 (a + h) 2 
and lim f (x) = lim a − =0
(x + 1)10 + (x + 2)10 + ...... + (x + 100 )10 x → a+ h→0  a 
225. (d) lim 
x →∞ x 10 + 10 10
Hence it is continuous at x = a .
 1
10
 2
10
 100  
10
9. (c) f (0) = 0
x10  1 +  +  1 +  + ... +  1 +  
 x  x  x  
= lim  = 100 . lim f (x) = lim e −1 / h = 0 and lim f (x) = lim e1 / h = ∞
x →0 − h→0 x →0 + h→0
x →∞
10
 10  10
x 1 + 10  Hence function is discontinuous at x = 0 .
 x 
 x 2 − 4 x + 3 
1 + 2 + 3 + ..... + n 10. (c) f ( x) =   , for x ≠ 1
226. (b) We have, lim  x 2 − 1 
n→ ∞ n 2 + 100
 1 = 2 , for x = 1
n 2 1 + 
n(n + 1)  n 1 x 2 − 4x + 3 (x − 3)
= lim = lim = . f (1) = 2, f (1+) = lim = lim = −1
n→∞ 2(n 2 + 100 ) n→ ∞ 2 100  2 2
(x + 1)
2n  1 + 2 
x →1+ x −1 x →1+

 n  2
x − 4x + 3
x f (1−) = lim = −1 ⇒ f (1) ≠ f (1−)
x2 −1
∫
2 x →1−
cos t dt
0
227. (b) lim Hence the function is discontinuous at x = 1.
x →0 x
x +1
Applying L- Hospital rule, we get 11. (b) f (x) = . Hence the points are 3, – 4.
(x − 3) (x + 4)
x

lim
∫ 0
cos t 2dt
= lim
cos x 2
= 1.
12. (c) f (0+) = f (0−) = 2 and f (0) = 2
x →0 x x →0 1 Hence f (x) is continuous at x = 0.
 Functions, Limits, Continuity and Differentiability 959
1 1 25. (c) lim f (x) = sin −1 (0) = 0 and f (0) = 0
13. (c) lim f (x) = x 2 sin , but − 1 ≤ sin ≤ 1 and x → 0 x →0
x →0 +
x x
Hence f (x) is continuous at x = 0.
Therefore, lim f (x) = 0 = lim f (x) = f (0)
x →0 + x →0 −
2 sin 2 x 2
Hence f (x) is continuous at x = 0. 26. (b) lim f (x) = lim = = k.
x →0 x →0 2x . 5 5
14. (d) f (0−) = lim k(2 x − x 2 ) = 0 ; f (0+) = lim cos x = 1 27. (c) lim f (x) = 0 and lim f (x) = 1 + 1 = 2.
x →0 − x →0 + x →1+ x →1−

∴ f (0) = cos x = 1 Hence f (x) is discontinuous at x = 1 .

Hence no value of k can make f (0−) = 1. 28. (d) lim f (x) = −2 and f (−1) = − 2.
x →1−
−h −h
15. (c) f (0) = 0 ; f (0−) = lim = lim =0 1 1
h→0 e −1 / h + 1 h→0 1 29. (b) lim f (x) = and lim f (x) = and f (2) = 1.
1+ x → 2− 2 x → 2+ 2
e1 / h
h 30. (b) Obviously lim f (x) = f (b) = 0 .
x →b
f (0+) = lim 1/ h
= 0.
h→0 e +1 31. (b) lim f (x) = −1, lim f (x) = 1, f (a) = 1.
[ ]
x → a− x → a+
1 / 2x 2 2
16. (b) lim f (x) = lim (1 + 2 x) =e .
x →0 − x →0 32. (c) lim f (x) = 1, f (1) = 2.
x →1
17. (d) lim f (x) = lim 21 / h = ∞
x →0 + h→0 33. (a) lim f (x) = 3, lim f (x) = 3 and f (2) = 3 .
x → 2− x → 2+
−1 / h 1
lim f (x) = lim 2 = lim =0.  3π 
x →0 − h→0 h→0 21 / h 34. (c) Here f   = 1 and lim f ( x) = 1
 4  x → 3π / 4 −
 sin x 2 
18. (c) f (0) = 0, lim f (x) = lim f (x) = lim x  =0. 2  3π  π
x →0 + x →0 − x →0  x 2 
  lim f (x) = lim 2 sin  + h  = 2 sin = 1 .
x → 3π / 4 + h→0 9 4  6
19. (c) Clearly from curve drawn of the given function f (x) is
3π
discontinuous at x = 0. Hence f (x) is continuous at x = .
4
(0,1/4)
π −π π  π
O 35. (a) lim f (x) = , lim f (x) = and f   = .
x →π / 2− 2 x →π / 2+ 2  2 2
(0,–
 2 sin 2 2 x 
20.
36. (a) lim f (x) = lim  4 = 8
(a) It is obvious. x →0 − x →0 − (2 x) 2 
 
21. (b) lim f (x) = lim x + 1 = 2 = k.
x →1 x →1
lim f (x) = lim 16 + x + 4 = 8 . Hence a = 8 .
22. (b) (i) When 0 ≤ x < 1 x →0 + x →0 +

f (x) doesn't exist as [x] = 0 here. 37. (d) lim f (x) = a − b, lim f (x) = 2 ⇒ a − b = 2
x →1− x →1+
(ii) Also lim f (x) and lim f (x) does not exist.
x →1+ x →1− All the given sets of a, b make f (x) continuous at x=1.
Hence f (x) is discontinuous at all integers and also in 38. (b) lim f (x) = 1 + 1 = 2, lim f (x) = 0, f (0) = 2 .
x →0 − x →0 +
(0, 1).
39. (b) lim f (x) = lim (x + 2) (x 2 + 4) = 32, f (2) = 16 .
sin 2 ax x →2 x →2
23. (b) lim f (x) = a 2 = a 2 and f (0) = 1.
x →0 (ax) 2 40. (b) lim f (x) = 1, lim f (x) = 6.
x →1− x →1+
Hence f (x) is discontinuous at x = 0 , when a ≠ 0 .
(x − 2) (x + 5) −7 7
24. (b) lim f (x) = 0 41. (b) lim f (x) = = = .
x →0 − x→ − 5 (x + 5) (x − 3) − 8 8
f (0) = 0, lim f (x) = −4 42. (d) By definition of continuity, we know that
x →0 +
lim f (x) = f (3) = lim f (x)
f (x) discontinuous at x = 0. x →3+ x →3−

and lim f (x) = 1 and lim f (x) = 1, f (1) = 1 ⇒ lim f (x) = 4 or lim 3 − h + λ = 4
x →1− x →1+ x →3− h→0

Hence f (x) is continuous at x = 1 . ⇒ 3 + λ = 4 ⇒ λ = 1.

Also lim f (x) = 4(2) 2 − 3 . 2 = 10 1
x → 2− 43. (d) If x → 0, then the value of sin passes through [–1,
x
f (2) = 10 and lim f (x) = 3(2) + 4 = 10 1] infinitely many ways, therefore limit of the function
x → 2+
does not exist at x = 0. Hence there is no value of k for
Hence f (x) is continuous at x = 2.
which the function is continuous at x = 0.
 960 Functions, Limits, Continuity and Differentiability
44. (d) As we are given f (x) = sin x , if x ≠ nπ 51. (c) If function f (x) is continuous at x = 0, then
i.e., x ≠ 0, π , 2π ,.... = 2 otherwise f (0) = lim f (x)
x→0
2
∴ lim+ g { f (x)} = lim+ g { sin x} = lim+ (sin x + 1) = 1  1
x →0 x →0 x →0 Given f (0) = k ; f (0) = k = lim x  sin 
x →0  x
Similarly, lim− g {f (x)} = 1.
x →0
 1 
4 −h−4 f (0) = k = 0,  − 1 ≤ sin ≤ 1 ; ∴ k = 0 .
45. (d) lim f (x) = lim f (4 − h) = lim +a  x 
x →4 − h→0 h→0 | 4 − h − 4 |
52. (a) If f is continuous at x = 0 , then
h lim f (x) = lim+ f (x) = f (0) ⇒ f (0) = lim f (x)
= lim− + a = a − 1.
h→0 h x →0 − x →0 x →0 −

4 +h−4 π
= lim f (x) = lim f (4 + h) = lim +b = b +1 cos [ 0 − h]
x →4 + h→0 h→0 | 4 + h − 4 | k = lim f (0 − h) = lim 2
h→0 h→0 [ 0 − h]
and f (4) = a + b
π π
Since f (x) is continuous at x = 4 cos [ −h] cos [ −h − 1]
k = lim 2 = lim 2
Therefore lim f (x) = f (4) = lim f (x) h→0 [ −h] h→0 [ −h − 1]
x →4 − x →4 +

⇒ a − 1 = a + b = b + 1 ⇒ b = −1 and a = 1.  π
cos  − 
 2 ;
46. (d) For any x ≠ 1, 2 we find that f (x) is the quotient of k = lim k=0.
h→0 −1
two polynomials and a polynomial is everywhere
53. (c) Clearly the function is defined only in the interval [1, ∞)
continuous. Therefore f (x) is continuous for all
hence option (b) cannot even apply. For
x ≠ 1, 2. Check continuity at x = 1, 2.
x > 2, y = 3 x − 2 which is a straight line, hence
47. (c) Since f (x) is continuous at x = 0, therefore continuous. Further y = 4 at x = 2 . Hence, the
(27 − 2 x) 1/ 3
−3  0 function is continuous at x = 2 also (but not at x = 2
f (0) = lim f (x) = lim ,  Form  only).
x →0 x →0 9 − 3 (243 + 5 x)1 / 5  0
54. (a) f (x) is continuous at every point of its domain,
1
(27 − 2 x) − 2 / 3 (−2) ⇒ lim− f (x) = lim+ f (x) = f (1)
= lim 3 = 2. x →1 x →1
x →0 3
− (243 + 5 x) − 4 / 5 (5) ⇒ 5 ×1 − 4 = 4 ×1 + 3× b ×1
5
⇒ 1 = 4 + 3b ⇒ 3b = −3 ⇒ b = −1 .
1
48. (a) lim (cos x)1 / x = k ⇒ lim log (cos x) = log k 55. (c) For continuity at all x ∈ R, we must have
x →0 x →0 x
 π
1 f  −  = lim − (−2 sin x) = lim ( A sin x + B)
⇒ lim lim log cos x = log k  2  x →(−π / 2) x →(−π / 2)+
x →0 x x →0
⇒ 2 = −A + B .....(i)
1
⇒ lim × 0 = log e k ⇒ k = 1 . π 
x →0 x and f   = lim ( A sin x + B) = lim (cos x)
 2 x →(π / 2)− x →(π / 2)+
49. (a) Since lim f (x) = lim f (x) = f (2) = 1
x → 2− x → 2+
⇒ 0 = A+ B .....(ii)
Also it is continuous for all values of x, less than 2 and From (i) and (ii), A = −1 and B = 1 .
greater than 2.
x 2 − 10 x + 25
50. (c) Given function is continuous at all point in (− ∞, 6) and 56. (a) f (5) = lim f (x) = lim
x→5 x →5 x 2 − 7 x + 10
at x = 1, x = 3 function is continuous.
If function f (x) is continuous at x = 1, then (x − 5) 2 5−5
= lim = = 0.
x →5 ( x − 2)( x − 5) 5−2
π
lim f (x) = lim+ f (x) ⇒ 1 + sin = a+b
x →1− x →1 2 57. (c) For continuity at x =0, we must have
x cot x
∴ a+b = 2 .....(i)  1

f (0) = lim f (x) = lim(x + 1)cot x = lim(1 + x) x 
If at x = 3, function is continuous, then x→0 x →0 x →0  

3π
lim− f (3) = lim+ f (x) ⇒ 3a + b = 6 tan 
lim 
x 

x →3 x →3 12  1  x →0  tan x 

∴ 3a + b = 6 .....(ii) = lim(1 + x) x  = e1 = e .
x →0  

From (i) and (ii), a = 2, b = 0 .
58. (a) It is obvious.
 Functions, Limits, Continuity and Differentiability 961
59. (a) It is obvious that | x | is continuous for all x. x x x x x
2 cos2 − 2 sin cos cos − sin
| 0 + h | −0 68. (c) f ( x) = 2 2 2 = 2 2
Now, Rf ′(x) = lim =1 x x x x x
h→0 h 2 cos2 + 2 sin cos cos + sin
2 2 2 2 2
| 0 − h | −0
L f ′(x) = lim = −1 π x  π
h→0 −h = tan  −  at x = π , f (π ) = − tan = −1 .
 4 2 4
Hence f (x) =| x | is not differentiable at x = 0 .
 1 − cos x
π  ,x ≠ 0
60. (c) f (x) is continuous at x = , then 69. (a) f (x) =  x continuous at x = 0
2  k
 , x = 0
1 − sin x  0 
lim f (x) = f (0) or λ = lim ,  form 
x →π / 2 x →π / 2 π − 2x  0  2. sin 2 x / 2
lim+ f (x) = f (0) ⇒ lim =k
Applying L-Hospital’s rule, x →0 x →0 x
− cos x cos x 2 sin 2 x / 2 x
λ = lim ⇒ λ = lim = 0. ⇒ lim . = k ⇒ k = 0.
x →π / 2 − 2 x →π / 2 2 2
x →0
x 4
61. (a) Since f (x) is continuous at x = 0, therefore  
 2
sin π x
lim f (x) = f (0) ⇒ lim =k 70. (a) It is obvious.
x →0 x →0 5x
 sinπ x  π π π  1
⇒ lim  . = k ⇒ (1). = k ⇒ k = . e x − 1
x →0 πx  5 5 5  ,x≠0
71. (d) Given f (x) =  1
62. (d) If f (x) is continuous at x = 0, then e + 1
x
0 ,x = 0
2− x + 4 0  
f (0) = lim f (x) = lim ,  form 
x →0 x →0 sin 2 x 0  1
ex −1 e∞ − 1
 1  ⇒ lim = = −1
−  +
x →0
1
e∞ + 1
 2 x+4 
Using L–Hospital’s rule, f (0) = lim   =−1 . ex +1
x →0 2 cos 2 x 8 1
−
1

63. (c) At no point, function is continuous. ex −1 1−e x 1 − e −∞

⇒ lim 1
= 1
= =1
x −9 2 −
x →0 1 + e∞
64. (b) lim f (x) = lim = lim(x + 3) = 6 ex +1 1+ ex
x →3 x →3 x − 3 x →3

and f (3) = 2(3) + k = 6 + k So, lim f (x) exists at x = 0 , but at x = 0 it is not

x→ 0
 f is continuous at x = 3 ; ∴ 6 + k = 6 ⇒ k = 0 . continuous.
−1
 1  72. (b) We have f (x) = 2 x − 1, if x > 2, f (x) = k,
65. (b) f (x) =  x 2 + e 2− x  and f (2) = k
  If x = 2 and x 2 − 1, if x < 2 , function is continuous.
If f (x) is continuous from right at x = 2 then ∴ lim f (x) = f (2) ⇒ lim(2 x − 1) = k ⇒ k = 3 .
x →2 x →2
lim+ f (x) = f (2) = k
x →2  2 x − sin x  −1
0 
73. (b) f (x) = lim   = f (0) ,  form 
x →0 2 x + tan −1 x 
−1
 1     0 
⇒ lim+  x 2 + e 2− x  = k ⇒ k = lim f (2 + h)
x →2   h→0 Applying L-Hospital’s rule,

−1  1 
 1  2− 
⇒ k = lim (2 + h)2 + e 2−(2+h)   1 2 
h→0   f (0) = lim 
− x  = 2−1 = 1
  x →0  1  2+1 3
[
⇒ k = lim 4 + h 2 + 4 h + e −1 / h
h→0
] −1 2+

2
1+ x 


1 sin −1 x
⇒ k = [ 4 + 0 + 0 + e −∞ ] −1 ⇒ k =. 2−
4 x 2−1 1
Trick : f (0) = lim = = .
66. (d) By L-Hospital’s rule lim f (x) is 2. Therefore, for f (x) to x →0 tan −1 x 2+1 3
x→0 2+
x
be continuous, the value of function should be 2.
| x|
1 + kx − 1 − kx 74. (c) | x | is continuous at x=0 and is also
67. (c) L.H.L. = lim− =k x
x →0 x
discontinuous at x = 0
R.H.L. = lim+ (2 x 2 + 3 x − 2) = −2 | x|
x →0 ∴ f (x) =| x | + is discontinuous at x = 0 .
Since it is continuous, L.H.L = R.H.L ⇒ k = −2 . x
 962 Functions, Limits, Continuity and Differentiability
 2 x − 2− x   (2 x + 2 − x ) log e 2   1 − cos 4 x
75. (e) f (0) = lim f (x) = lim  = lim    ,x ≠ 0
x →0 x →0 x  x →0
 1 79. (b) f (x) =  8 x 2
  
k ,x = 0

= (20 + 20 ) loge 2
If f (x) is continous function at point x = 0 then
= (1 + 1) loge 2 lim [ f (x) = lim [ f (x)] lim[ f (x)] = lim [ f (0 + h)]
x →0 + x →0 − x →0 h→ 0 −
= 2 loge 2 = loge 4 .
1 − cos 4 h 2 sin 2 2h sin 2 2h
= lim[ f (h)] = lim = lim = lim
h→0 h→0 8h 2 h→0 8h 2 h→0 4 h 2
2x 2 + 7 9x 2 + 7
76. (c) f ( x) = 2 2 2
x (x + 3) − 1(x + 3) (x − 1)(x + 3)  sin 2h 
= lim  = (1)2 = 1
h→ 0 2h 
2x 2 + 7
= 1 − cos 4(−h)
(x − 1)(x + 1)(x + 3) lim f (x) = lim[ f (0 − h) = lim[ f (−h)] = lim
x →0 − h→ 0 h→ 0 h→ 0 8(−h)2
Hence points of discontinuity are
1 − cos 4 h
x = 1 , x = −1 and x = −3 only. = lim =1
h→ 0 8h2
1 f (0) = 1 ⇒ k = 1 .
77. (a) f (x) = x p sin , x ≠ 0 and f (x) = 0, x = 0
x
ex ; x ≤ 0
Since at x = 0 , f (x) is a continuous function 
80. (b, d) f (x) = 1 − x; 0 < x ≤ 1
1 x − 1 ; x > 1
∴ lim f (x) = f (0) = 0 ⇒ lim x p sin = 0 ⇒ p > 0 . 
x →0 x →0 x
f (0 + h) − f (0) 1− h−1
Rf ' (0) = lim = lim = −1
f (x) − f (0) h→ 0 h h→ 0 h
f (x) is differentiable at x = 0 , if lim exists
x →0 x−0 f (0 − h) − f (0) e− h − 1
Lf ' (0) = lim = lim =1
1 h→ 0 −h h→ 0 −h
x p sin −0
⇒ lim x exists So, it is not differentiable at x = 0 .
x →0 x−0 Similarly, it is not differentiable at x = 1 .
1 But it is continous at x = 0 , 1.
⇒ lim x p −1 sin exists
x →0 x
Differentiability
⇒ p − 1 > 0 or p > 1
1. (d) Statement (d) is true, because differentiable function is
1
p −1 always continuous.
If p ≤ 1 , then lim x sin  does not exist and at
x →0 x 2. (d) As L f ′ (2) ≠ Rf ′ (2) .
x = 0 f (x) is not differentiable. 3. (b) f (0 + 0) = lim f (x) = lim f (0 + h)
h→0 h→0
∴ for 0 < p ≤ 1 f(x) is a continuous function at x = 0 1 / 0+h
e − e −1 / 0 + h e1 / h − e −1 / h
but not differentiable. = lim (0 + h) = lim h =0
h→0 e1 / 0 + h + e −1 / 0 + h h→ 0 e1 / h + e −1 / h
 1− | x | 1 ,x < 0
 , x ≠ −1  e −1 / h − e1 / h
78. (d) f (x) =  1 + x and f (x) =  1 − x and f (0 − 0) = lim f (0 − h) = lim − h =0
1 1 + x , x ≥ 0 h→0 e −1 / h + e1 / h
h→0
 , x = −1 
and f (0) = 0 ; ∴ f (0 + 0) = f (0 − 0) = f (0)
1 , x<0 Hence f is continuous at x = 0.

1 1 At remaining points f (x) is obviously continuous.
, 0≤x<
1 , x<0  2 Thus it is everywhere continuous.
 
f (2 x) =  1 − [ 2 x] ⇒ f (2 x) =  1 f (0 − h) − f (0)
 1 + [ 2 x] , x > 0 0 , ≤ x ≤1
2
Again, L f ′(0) = lim
  h→0 −h
 1 3
− 3 , 1≤ x < e −1 / h − e1 / h
2 h. −0
= lim e −1 / h + e1 / h = −1
1 h→0 −h
⇒ f (x) , for all values of x where x < a continous
2 e1 / h − e −1 / h
1 h
function and for x = and x =1 f (x) be a f (0 + h) − f (0) e1 / h + e −1 / h = 1
2 R f ′(0) = lim = lim
h→0 h h→0 h
discontinous function.
 L f ′(0) ≠ R f ′(0)
∴ f is not differentiable at x = 0 .
 Function, Limits , Continuity and Differentiability 963

4. (d) lim f (x) = lim f (3 − h) = lim | 3 − h − 3 | = 0   x2

x →3− h→0 h→0  x2  = x, x > 0
lim f (x) = lim f (3 + h) = lim | 3 + h − 3 | = 0
 , x≠0  x
9. (b) We have f (x) = | x | = 0, x = 0
x →3+ h→0 h→0
 0, x = 0  x 2
  = − x, x < 0
 lim− f (x) = lim+ f (x) = f (3)  − x
x →3 x →3
|x – 3| We have lim f (x) = lim − x = 0, lim f (x) = lim x = 0
Hence f is continuous at x = 3 x →0 − x →0 x →0+ x →0

f (3 − h) − f (3) O x=3 and f (0) = 0.

Now L f ′(3) = lim
h→0 −h So f (x) is continuous at x = 0.
| 3 − h − 3| − 0 h Also f (x) is continuous for all other values of x.
= lim = lim = −1
h→0 −h h→0 − h
Hence, f (x) is continuous everywhere. Clearly,
f (3 + h) − f (3) | 3 + h − 3| − 0 Lf ' (0) = −1 and Rf ' (0) = 1. Therefore f (x) is not
R f ′(3) = lim = lim =1
h→0 h h→ 0 h differentiable at x = 0.
 L f ′(3) ≠ R f ′(3) . Hence f is not differentiable at x = 3 . f (1 + h) − f (1)
10. (b) We have Rf ' (1) = lim
h→0 h
Trick : Can be seen by graph it is continuous but
tangent is not defined at x = 3 .
= lim
{(1 + h) 3
}
−1 − 0
=3
2 h→0 h
5. (a,c,d) x ≤ x ⇒ x (1 − x) ≤ 0 ⇒ x (x − 1) ≥ 0
Lf ' (1) = lim
f (1 − h) − f (1)
= lim
{(1 − h) − 1} − 0 = 1
x : x≤0 h→0 −h h→0 −h

⇒ x ≤ 0 or x ≥ 1 ; ∴ h(x) =  x 2 : 0 < x < 1 ∴ Rf ' (1) ≠ Lf ' (1) ⇒ f (x) is not differentiable at x = 1.
 x : x ≥1
Now, f (1 + 0) = lim f (1 + h) = 0
h→0
h (x) is continuous for every x but not differentiable at
and f (1 − 0) = lim f (1 − h) = 0
 1 x<0 h→0
 ∴ f (1 + 0) = f (1 − 0) = f (0) ⇒ f (x)
 not exists x =0 is continuous at
x = 0 and 1. Also h′(x) =  2x 0 < x < 1 x = 1. Hence at x = 1, f (x) is continuous and not

 not exists x =1 differentiable.
 1 x >1 11. (d) Here, when −1 ≤ x ≤ 1, 0 ≤ x sin πx < 1

∴ h′(x) = 1 for all x > 1 . ⇒ f (x) = [ x sin πx] = 0 for −1 ≤ x ≤ 1,

6. (d) It is obvious. i.e., f (x) is constant function (equal to zero) in [−1, 1].
⇒ f (x) is differentiable in (−1, 1) .
x , 0 ≤ x ≤ 1
7. (a) f (x) = 
1 , 1 < x ≤ 2 12. (d) Since | x − 3 | = x − 3, if x ≥ 3 = − x + 3, if x < 3

lim f (x) = lim f (1 − h) = lim (1 − h) = 1 ∴ The given function can be defined as

x →1− h→0 h→0
1 2 3 13
lim+ f (x) = lim f (1 + h) = 1 4 x − 2 x + 4 , x < 1
O x=1 
x →1 h→0 f ( x) =  3 − x, 1 ≤ x < 3
Hence function is continuous in (0, 2).  x − 3, x ≥ 3

Now lim f (x) = lim (0 + h) = 0 = f (0) 
+ h→0
x →0 Now proceed to check the continuity and
lim f (x) = lim (2 − h) = 1 = f (2) differentiability at x = 1.
x → 2− h→0
13. (b) Given f (x) is differentiable at x = 0 . Hence, f (x) will
Hence function is continuous in [0, 2] be continuous at x = 0 .
Clearly, from graph it is not differentiable at x = 1. ∴ lim− (e x + ax) = lim+ b(x − 1)2
x →0 x →0
8. (a) Since this function is continuous at x = 0
Now for differentiability ⇒ e + a × 0 = b(0 − 1)2 ⇒ b = 1
0
….. (i)

f (x) =| x | = | 0 | = 0 and f (0 + h) = f (h) = | h | But f (x) is differentiable at x = 0 , then

d x d
f (0 + h) − f (0) | h| L f ′(x) = Rf ′(x) ⇒ (e + ax) = b(x − 1)2
∴ lim = lim = −1 dx dx
h→0 − h h→0 − h
⇒ e x + a = 2b(x − 1)
f (0 + h) − f (0) | h|
and lim = lim =1.
h→ 0 + h h → 0 + h At x = 0, e 0 + a = −2b ⇒ a + 1 = −2b ⇒ a = −3
Therefore it is continuous and non-differentiable. ⇒ (a, b) = (−3, 1) .
 964 Functions, Limits, Continuity and Differentiability
14. (d) It can be easily seen from the graph of f (x) =| sin x | f (x) − f (0)
Rf ′(x) = lim+
that it is every where continuous but not differentiable x →0 x−0
at integer multiples of π and at x = 0 . f (0 + h) − f (0) (0 + h)2 − 0
Y = lim = lim =0
h→0 h h→0 h
⇒ L f ′(x) = Rf ′(x)
Hence f (x) is differentiable at x = 0 .
X′ X
–π 0 π 2π 0 , x < 0
–2π 3π π π 3π Now f ′(x) =  ; lim− f ′(x) = lim f ′(0 − h) = 0
− −
2 2 2 2
 2 x , x ≥ 0 x →0 h→0

and lim+ f ′(x) = lim f ′(0 + h) = lim 2(0 + h) = 0

Y′ x →0 h→0 h→0
e − x , x ≥ 0
15. (c) We have, f (x) =  x ⇒ lim− f ′(x) = lim+ f ′(x) = 0
 e , x < 0 x →0 x →0

Hence f ′(x) is continuous function at x = 0 .

Clearly, f (x) is continuous and differentiable for all
f (x) − f (0) f (0 − h) − f (0)
non zero x. Now L f ′′(x) = lim = lim
x →0 −
x−0 h→ 0 −h
Now lim f (x) = lim e x = 1 , lim f (x) = lim f (x)e − x = 1
x →0 − + x →0 +
x →0 x →0 0−0
= lim =0
0
Also, f (0) = e = 1 . So, f (x) is continuous for all x. h→0 − h

f (x) − f (0) f (0 + h) − f (0)

 d  R f ′′(x) = lim+ = lim
(LHD at x = 0) =  (e x ) =1 x →0 x−0 h→0 h
 dx  x =0
2(0 + h) − 0 2h
 d  = lim = lim = 2 ⇒ L f ′′(x) ≠ Rf ′′(x)
(RHD at x = 0) =  (e − x ) = −1 h→0 h h→0 h
 dx  x =0 Hence f ′(x) is not differentiable at x = 0 .
So, f (x) is not differentiable at x = 0 .
20. (c)  f is continuous at x = 0 , ∴ f (0 − ) = f (0 + ) = f (0) = −1
−| x|
Hence f (x) = e is everywhere continuous but not Also L f ′(0) = Rf ′(0)
differentiable at x = 0 . f (0 − h) − f (0) f (0 + h) − f (0)
16. (c) lim− 1 + (2 − h) = 3 , lim+ 5 − (2 + h) = 3 , f (2) = 3 ⇒ lim = lim
h→0 h→0
h→0 −h h→0 h
Hence, f is continuous at x = 2  2 
 ah + bh − 1 + 1 
5 − (2 + h) − 3  e − 2h − 1 + 1  2
Now Rf ′(x) = lim = −1 ⇒ lim  = lim 
h→0 h h→0 −h  h→0 h 
   
1 + (2 − h) − 3  
L f ′(x) = lim =1
h→0 −h  − 2e −2h 
⇒ lim  = lim a + bh 
 Rf ′(x) ≠ L f ′(x) ; ∴ f is not differentiable at x = 2 .  h→0
h→0
 −1  2 
[k − h] sinπ (k − h) − [k] sinπk
17. (a) f ′(k − 0) = lim ⇒ 2 = a+ 0 ⇒ a = 2, b any number.
h→0 −h
21. (b) A continuous function may or may not be
(−1)k −1 (k − 1) sinπh − k × 0 differentiable. So (b) is not true.
= lim
h→0 −h 1 1
22. (d) lim f (x) = x 2 sin  , but − 1 ≤ sin  ≤ 1 and x → 0
(−1)k −1 (k − 1) sinπh x →0 x x
= lim = (−1)k .(k − 1)π .
h→0 −h ∴ lim+ f (x) = 0 = lim− f (x) = f (0)
x →0 x →0
f (2 + h) − f (2)
18. (d) Rf ′(2) = lim Therefore f (x) is continuous at x = 0 . Also, the
h→0 h
1
2(2 + h) − 1 − (4 − 1) 4 + 2h − 1 − 3 function f (x) = x 2 sin is differentiable because
= lim = lim =2 x
h→0 h h→0 h
1
f (2 − h) − f (2) 2− h+1− 3 h 2 sin −0
and L f ′(2) = lim = lim =1. h h2 sin(1 / − h)
h→0 −h h→0 −h Rf ′(x) = lim = 0 , L f ′(x) = lim = 0.
h→0 h h→ 0 −h
Thus f ′(2) does not exist. f (1 + h) − f (1)
23. (b) By definition, f ′(1) = lim
 0, x < 0 h→0 h
19. (c) f ( x) =  2 ; lim− f (x) = lim f (0 − h) = 0
 x , x ≥ 0 x →0 h→0 1  −1 
− 
 1 1
+ 

and lim+ f (x) = lim f (0 + h) = lim(0 + h)2 = 0 2(1 + h) − 5  3   2h − 3 3 
x →0 h→0 h→0
= lim = lim
h→0 h h→0 h
⇒ lim− f (x) = lim+ f (x) = f (0)
x →0 x →0  3 + 2h − 3   2h 
= lim  = lim 
Hence f (x) is continuous function at x = 0 . h→0 3h(2h − 3)  h→0 3h(2h − 3) 
f (x) − f (0) f (0 − h) − 0 0−0 2 2 −2
L f ′(x) = lim = lim = lim =0 = lim = = .
x →0 −
x−0 h→ 0 −h h→ 0 − h h→0 3(2h − 3) 3(−3) 9
 Function, Limits , Continuity and Differentiability 965
d  x  1 ⇒ f ' (x) = 2 for at least one x ∈ (1,3) .
24. (b) Let x < 0 ⇒| x | = − x ⇒ f (x) =  =
dx  1 − x  (1 − x) 2 1 1 1
32. (b) f (1) = f   = f   = ...... = lim f   = 0
⇒ [ f ′(x)] x =0 = 1 . Again x > 0 ⇒ | x | = x  2 3 n→ ∞  n 

d  x  1 Since there are infinitely many points in x ∈ (0, 1)

f ( x) =  = ⇒ [ f ′(x)] x =0 = 1
dx  1 + x  (1 + x)2 1
where f (x) = 0 and lim f   = 0 ⇒ f (0) = 0
⇒ f ′(0) = 1 . n→ ∞  n 

f (1 − h) − f (1) And since there are infinitely many points in the

25. (d) L f ′ (1) = lim
h→ 0 −h neighbourhood of x = 0 such that
m (1 − h) 2 − m m [1 + h 2 − 2h − 1] ⇒ f (x) remains constant in the neighbourhood of
= lim = lim
h→0 −h h→0 −h x = 0 ⇒ f ' (0) = 0 .
f (1 + h) − f (1) 33. (b) f (1) = −3 ; f ' (x) ≥ 9 for all x ∈ (1,5) ; ∴ f (5) ≥ 36 .
= lim m (2 − h) = 2m and R f ′ (1) = lim
h→0 h→0 h 34. (c) f (x) = 1 + sin(3 x)g(x)
2 (1 + h) − m f ' (x) = 3 cos 3 x g(x) + sin 3 x g' (x) = f (x) cos 3 x .
= lim . For
h→0 h
differentiability, L f ′(1) = R f ′ (1) .  1 ∀x < 0

35. (d) f (x) =  π
But for any value of m, R f ′ (1) = L f ′(1) not possible. 1 + sin, ∀ 0 ≤ x < 2

26. (c) (gof ) (x) = g[ f (x)] = g [1 − cos x] = e1− cos x , for x ≤ 0
 0, ∀ x < 0 (LHD)
(gof )′ (x) = e1− cos x . sin x, for x ≤ 0 ∴ f ' ( x) = 
cos x, 0 ≤ x ≤ π / 2, (RHD)
(gof )′ (0) = 0 .
 0 , x<0
f (1 + h) − f (1) ∴ f ' (0) =  , ∴ f ' (0) does not exist.
27. (a) f ' (1) = lim ; As function is differentiable so cos 0 = 1
h→ 0 h
f (1 + h) 36. (d) f (x) = x 2 − 2 x + 4 ; f ' (x) = 2 x − 2
it is continous as it is given that lim = 5 and
h→ 0 h At x = c , f ' (c) = 2c − 2
f (1 + h)
hence f (1) = 0 . Hence f ' (1) = lim = 5. f (5) = 5 2 − 2(5) + 4 = 19 ; f (1) = 12 − 2(1) + 4 = 3
h→ 0 h
f (5) − f (1) 19 − 3 16
f (x) − f (y) = f (c) ⇒ = 2c − 2 ⇒ = 2c − 2
28. (d) lim ≤ lim | x − y | or | f ' (x)| ≤ 0 5 −1 5 −1 4
x →y x−y x →y
⇒ 4 = 2c − 2 ⇒ 2c = 6 or c = 3 .
⇒ f ' (x) = 0 ⇒ f (x) is constant, As f (0) = 0 f (x + h) − f (x) f (x) + f (h) − f (x)
37. (d) We have f ' (x) = lim = lim
∴ f (1) = 0 . h→ 0 h h→ 0 h
29. (c) As f (1) = −2 and f ' (x) ≥ 2∀x ∈ [1,6] [ f (x + y) = f (x) + f (y)]
Applying lagrange’s mean value theorem, 2
f (h) h g(h)
f (6) − f (1) = lim = lim = 0.g(0) = 0
= f ' (c) ≥ 2 h→ 0 h h → 0 h
5
[ g is continuous therefore lim g(h) = g(0)] .
⇒ f (6) ≥ 10 + f (1) ⇒ f (6) ≥ 10 − 2 ⇒ f (6) ≥ 8 . h→ 0

| x | −1, | x | −1 ≥ 0 38. (d) Since function | x | is not differentiable at x = 0

30. (b) = 
− | x | +1, | x | −1 < 0 ∴| x 2 − 3 x + 2 | =| (x − 1)(x − 2)|
| x | −1, x ≤ −1 or x ≥ 1 Y Hence is not differentiable at x = 1 and 2
=
− | x | +1, −1 < x < 1 (0,1) Now f (x) = (x 2 − 1)| x 2 − 3 x + 2 | cos(| x |) is not
− x − 1, x ≤ −1 differentiable at x = 2

 x + 1, − 1 < x < 0 For 1 < x < 2 , f (x) = −(x 2 − 1)(x 2 − 3 x + 2) + cos x
= X
− x + 1, 0 ≤ x < 1 (-1, 0) (1, 0)
For 2 < x < 3 , f (x) = +(x 2 − 1)(x 2 − 3 x + 2) + cos x
 x − 1, x ≥1
Lf ' (x) = −(x 2 − 1)(2 x − 3) − 2 x(x 2 − 3 x + 2) − sin x
From the graph. It is clear that f (x) is not differentiable
Lf ' (2) = −3 − sin 2
at x = −1, 0 and 1.
Rf ' (x) = (x 2 − 1)(2 x − 3) + 2 x(x 2 − 3 x + 2) − sin x
31. (b) Let a function be g(x) = f (x) − x 2
Rf ' (2) = (4 − 1)(4 − 3) + 0 − sin 2 = 3 − sin 2
⇒ g(x) has at least 3 real roots which are x = 1 , 2 , 3
Hence Lf ' (2) ≠ Rf ' (2) .
⇒ g' (x) has at least 2 real roots in x ∈ (1, 3)
⇒ g" (x) has at least 1 real roots in x ∈ (1, 3)
 966 Functions, Limits, Continuity and Differentiability
dy h
39. (c) Since = cos x which is defined at x = 0 and no −0
dx f (h) − f (0) 1+ | h| 1
lim = lim = lim =1
other differential coefficient is defined at x = 0 h→ 0 h−0 h→ 0 h h→ 0 1+ | h |

40. (c) It is fundamental concept.

Therefore f is differentiable at x = 0 , so f is
x + 2 , − 1 < x < 3 differentiable in (−∞, ∞) .

41. (d) If f (x) =  5 , x = 3 and f (3) = 5
46. (b) y' = 1 2(1 + x 2 ) − 4 x 2 2(1 − x 2 )
8 − x , x >3 . =
 2 (1 + x 2 ) 2 (1 − x 2 ) 2 .(1 + x 2 )
 2x 
f (x) − f (3) f (3 − h) − f (3) 1− 2

L.H.D = lim = lim 1+ x 
x →3− x−3 h→ 0 −h
 2
(3 − h + 2) − 5 −h  for | x | < 1
= lim = lim =1 2
h→ 0 −h h → 0 −h ⇒ y' =  1 + x
 −2 for | x | > 1
f (x) − f (3) f (3 + h) − f (3)
R.H.D = lim = lim  1 + x 2
x →3+ x−3 h→ 0 h
Hence for | x | = 1 , the derivative does not exist.
8 − (3 + h) − 5 −h
= lim = lim = −1
h→ 0 h h→ 0 h 47. (c) Since the function is defined for x ≥ 0 i.e. not defined
L.H.D ≠ R.H.D f (x) is not differentiable. for x < 0 . Hence the function neither continuous nor
differentiable at x = 0 .
 x, 0≤ x ≤1
42. (c) f ( x) =  48. (c) Function f (x) =| x − 0.5 | + | x − 1| + tan x does not
 2 x − 1, x >1
π
lim f (x) = lim f (1 − h) = lim(1 − h) = 1 have a derivative at the points x = 0.5, 1, ∈ (0, 2) .
x →1 − h→ 0 h→ 0 2
lim f (x) = lim f (1 + h) = lim 2(1 + h) − 1 = 1
x →1+ h→ 0 h→ 0 Critical Thinking Questions
 lim f (x) = lim f (x) = 1
x →1− x →1+
cos 2 x + sin 4 x cos 2 x + sin 2 x(1 − cos 2 x)
∴ Function is continuous at x = 1 . 1. (a) f (x) = 2 4
⇒ f ( x) =
sin x + cos x sin 2 x + cos 2 x(1 − sin 2 x)
f (1 − h) − f (1) (1 − h) − 1
Lf ' (1) = lim = lim =1 sin 2 x + cos 2 x − sin 2 x cos 2 x
h→ 0 −h h→ 0 −h ⇒ f ( x) =
sin 2 x + cos 2 x − sin 2 x cos 2 x
f (1 + h) − f (1) 2 + 2h − 1 − 1
Rf ' (1) = lim = lim =2 ⇒ f (x) = 1 ⇒ f (2002 ) = 1 .
h→ 0 −h h → 0 h
∴ Lf ' (1) ≠ Rf ' (1) 2. (d) f (x + y) = f (x) + f (y)

∴ Function is not differentiation at x = 1 Put x = 1, y = 0 ⇒ f (1) = f (1) + f (0) = 7

f (0 + h) − f (0) 1 + sinh − 1 Put x = 1, y = 1 ⇒ f (2) = 2. f (1) = 2.7
43. (d) Rf ' (0) = lim = lim =1
h→ 0 h h→ 0 h Similarly f (3) = 3.7 and so on
f (0 − h) − f (0) 1−1
f ' (0) = lim = lim =0 n 7n(n + 1)
h→ 0 −h h→ 0 − h ∴ ∑ f (r ) = 7 (1 + 2 + 3 + ..... + n) = .
r =1 2
Hence, f ' (0) does not exist.
 1 1 1 1
44. (c) f (x) possesses derivative at x = 0 , so it is both 3. (c) By verification, f  −  = f  = − 1 = −
continuous and differentiable at x = 0 . Now  2 2
  2 2
f (0 + 0) = 0 , f (0 − 0) = b, f (0) = b , ∴ b = 0 Hence f (| x |) = x .
Also Rf ' (0) = 0, Lf ' (0) = 0, ∀a ∈ R  x 3
∴ f ' (0) = 0 if b = 0 .  3 , for x 3 ≠ 0
| x |
45. (a) Let h(x) = x, x ∈ (−∞, ∞) ; g(x) = 1+ | x |, x ∈ (−∞, ∞)  0 , for x 3 = 0

Here h is differentiable in (−∞, ∞) but | x | is not  x
 , for x ≠ 0
differentiable at x = 0 . 4. (d) Here, f (x) = sgn x 3 = | x |
 0 , for x = 0
Therefore g is differentiable in (−∞,0) ∪ (0, ∞) and 
h(x) x − 1, x < 0
g(x) ≠ 0, Vx ∈ (−∞, ∞) , therefore f (x) = = 
g(x) 1+ | x |  0, x = 0
 1, x > 0
It is differentiable in (−∞,0) ∪ (0, ∞) for x = 0 
Thus, f (x) = sgn x 3 = sgn x, which is neither
continuous nor derivable at 0. Note that
 Function, Limits , Continuity and Differentiability 967
f (0 + h) − f (0) 1− 0 ⇒ (x + 1) 4 = 1 + x ⇒ (x + 1) [(x + 1) 3 − 1] = 0
f ′(0 + ) = lim+ = lim+ →∞
h→0 h h→0 h
⇒ x = −1 or (x + 1) 3 = 1 ⇒ x + 1 = 1, ω , ω 2
− f (0 − h) − f (0) −1 − 0
and f ′(0 ) = lim− = lim− →∞.
h→0 h h→0 h −3+i 3 −3−i 3
+ −
⇒ x = 0, − 1, , .
∴ f ′(0 ) ≠ f ′(0 ) , ∴ f is not derivable at x = 0 . 2 2
5. (d) g( f (x)) ≤ f (g(x)) ⇒ g(| x |) ≤ f [ x] ⇒ [| x |] =| [ x] | . 14. (a) Since f is an even function f (− x) = f (x) , ∀x ∈ (−5, 5).
This is true for x ∈ R. .  x +1
We are given that f (x) = f  
6. (b)  [x] denotes the integral part of x. Hence, after term  x + 2
 1 50   −x + 1   −x + 1 
 2 + 100  , each term will be one. Hence the sum of ⇒ f (− x) = f   ⇒ f ( x) = f  
   − x + 2  − x + 2
given series will be 50.
[ f (− x) = f (x)]
7. (a) Since domain of f and domain of composite function gof
To find the values of x, we set
are same.
− x +1 3± 9−4 3± 5
8. (b) x + 2 ≥ 0 i.e., x ≥ −2 or − 2 ≤ x x= ⇒ x= =
−x+2 2 2
 log10 (1 − x) ≠ 0 ⇒ 1 − x ≠ 1 ⇒ x ≠ 0
 x +1
Also f (x) = f   = f (− x)
Again 1 − x > 0 ⇒ 1 > x ⇒ x < 1  x + 2
All these can be combined as −2 ≤ x < 0 and 0 < x < 1 . To find the values of x, we set
y x x x
9. (d) 2 = 2 − 2 ; y is real, if 2 − 2 > 0 ⇒ 2 > 2 x +1 −3± 9−4 −3± 5
−x= ⇒ x= =
⇒ 1 > x ⇒ x ∈ (−∞, 1) x+2 2 2
Thus the four required values of x are
b2 b2
10. (d) f (x) = (1 + b 2 )x 2 + 2bx + − +1 −3− 5 −3+ 5 3− 5 3+ 5
(1 + b ) 1 + b 2
2 , , , .
2 2 2 2
2
 b  1 1  π  π
= (1 + b 2 )  x +  + ≥ 15. (d) f ′(x) = 2 sin x cos x + 2 sin  x +  cos  x + 
 2
1+b  1+b 2
1 + b2  3  3

1  π  π
∴ m(b) = , so range of m(b) = (0,1] . − sin x cos  x +  − cos x sin x + 
1 + b2  3  3

11. (d) For 7− x

Px − 3 to be defined, 7 − x > 0 ⇒ x < 7  2π   π
= sin 2 x + sin  2 x +  − sin  x + x + 
 3   3
x−3≥0⇒ x ≥ 3; 7−x ≥ x−3⇒ x ≤5
 2π  π   π
∴ x ∈ {3, 4, 5} ⇒ f (3) = 1, f (4) = 3, f (5) = 2 = 2 sin  2 x +  cos   − sin  2 x +  = 0
 3  3  3
So, the range of function = {1, 2, 3} . ⇒ f (x) = k, where k is a constant.
2
12. (d) 2 sin x + 3 sin x − 2 > 0 π  π  5
But f (0) = sin 2 0 + sin 2   + cos 0 cos   =
3 3 4
2 sin 2 x + 4 sin x − sin x − 2 > 0
5
Thus f (x) = , ∀ x ∈ R.
2 sin x (sin x + 2) − 1 (sin x + 2) > 0 4
(sin x + 2) (2 sin x − 1) > 0 5
Therefore, (gof ) (x) = g [ f (x) ] = g   = 1.
4
2 sin x − 1 > 0 ⇒ sin x > 1 / 2
16. (a) g {f (x)} =| sin x |, f {g(x)} = (sin x ) 2
x > π / 6 ⇒ x ∈ (π / 6, ∞) .....(i)
Considering f (x) = sin 2 x, g(x) = x , then
and x 2 − x − 2 < 0 ⇒ x 2 − 2 x + x − 2 < 0
g [ f (x)] = g (sin 2 x) = sin 2 x = | sin x |
x (x − 2) + 1 (x − 2) < 0

(x + 1) (x − 2) < 0 ⇒ x ∈ (−1, 2) .....(ii) f [ g(x)] = f [ x] = (sin x ) 2 .

17. (a) f (x) = 3 x + 10 and g(x) = x 2 − 1
From (i) and (ii), x ∈ (π / 6, 2) .
⇒ f o g = f (g(x)) = 3(g(x)) + 10
13. (d) Let f (x) = (x + 1) 2 − 1, x ≥ −1. Since f (x) = f −1 (x)
= 3(x 2 − 1) + 10 = 3 x 2 + 7 .....(i)
2
∴ (x + 1) − 1 = 1 + x − 1  f ( −1
( x) = 1 + x − 1 ) 2 2
Let 3 x + 7 = y ⇒ 3 x = y − 7
 968 Functions, Limits, Continuity and Differentiability

y−7 y−7
1/ 2 and lim (1 − x + [ x − 1] + [1 − x])
⇒ x2 = ⇒x=  x →1+
3  3 
= lim (1 − (1 + h) + [1 + h − 1] + [1 − (1 + h)])
h→0
We know that f (x) = y , then x = f −1 (y)
1/ 2 = lim (−h + [ h] + [ −h]) = lim (−h + 0 − 1) = −1
 x −7 h→0 h→0
so ( fog)−1 =   .
 3  ∴ lim f (x) = −1 .
(c) g[ f (x)] = 8 or g(2 x + 3) = 8
x →1
18.
x (1 + a cos x) − b sin x
⇒ (2 x + 3)2 + 7 = 8 ⇒ 2 x + 3 = ±1 ⇒ x = −1, − 2 . 24. (c) lim =1
x →0 x3
 πx    x2 x4 x6   x3 x5 
19. (c) lim (1 − x) tan   . Put 1 − x = y as x → 1, y → 0 x 1 + a  1 − + − + ...  − b  x − + − ... 
x →1  2 ⇒ lim
  2! 4! 6!   3! 5! 
=1
x →0 x3
 πy 
   b a   a b 
π (1 − y) 2 2 2 2 (1 + a − b) + x 2  −  + x 4  −  + ...
Thus lim y tan = lim .   = × 1 = . ⇒ lim  3 ! 2 !   4 ! 5 ! 
= 1 .....(i)
y →0 2 y →0 π π
 y π π 2
tan   x →0 x
 2 If 1 + a − b ≠ 0, then L.H.S. → ∞ as x → 0 while
(1 + x) − (1 − x)  2 + 3 x + 2 − 3 x  R.H.S. =1, therefore 1 + a − b = 0.
20. (b) lim  
x →0 (2 + 3 x) − (2 − 3 x)  1 + x + 1 − x   b a   a b 
x 2  −  + x 4  −  + ...
 3 ! 2 !   4 ! 5 ! 
Now from (i), lim =1
1 2 2  2 2 1 x →0 x2
=  = ,0< < .
3  2  3 3 2 b a
⇒ − = 1 ⇒ b − 3a = 6 . Solving 1+ a−b = 0
Aliter : Apply L-Hospital’s rule, 3! 2 !
and b − 3a = 6, we get a = −5 / 2, b = −3 / 2 .
1 1
+
1+ x − 1− x 2 1+ x 2 1− x 25. (a) Using L-Hospital's rule, we get
lim = lim
x →0 2 + 3x − 2 − 3x x →0 3 3 ax − x a a x log e a − ax a−1
+ − 1 = lim = lim
2 2 + 3x 2 2 − 3x x →a x x − aa x →a x x + a a log e a
1 1 a a log e a − a . a a−1 log e a − 1
+ ⇒ −1 = = .....(i)
2 2 2 2 2
= = = . a a
a + a log e a log e a + 1
3 3 6 3
+
2 2 2 2 Now (i) is satisfied only when a = 1.
26. (b) We have x 1 = 3, x n+1 = 2 + xn
xn x n−1 n! n!
21. (b) lim = lim n = ...... = lim = = 0,
x →∞ ex x →∞ ex x →∞ ex ∞
x 2 = 2 + x1 = 2 + 3 = 5 , x 3 = 2 + x 2 = 2 + 5
where n is any whole number
∴ x1 > x 2 > x 3
( n ! is defined for all positive integers including
zero). It can be easily shown by mathematical induction that
the sequence x 1 , x 2 ,........ x n ,.... is a monotonically
  1  
1/ 2
decreasing sequence bounded below by 2. So it is
22. (b) Given limit = lim sinnπ  1 + 2  
n→ ∞   n   convergent. Let lim x n = x. Then

  xn+1 = 2 + xn ⇒ lim xn+1 = 2 + lim xn ⇒ x = 2 + x

1 1 
= lim sin nπ  1 + 2 − 4 + ... 
n→ ∞
  2n 8n  ⇒ x 2 − x − 2 = 0 ⇒ (x − 2) (x + 1) = 0 ⇒ x = 2

 π π  ( x n > 0 ∀ n; ∴ x > 0)
= lim sin nπ + − + .... 
n→ ∞  2n 8n 3  x
t2 
 1 1  x  
= lim (−1)n sin π  − + ....  = 0. ∫π / 2 t . dt  2  π / 2
n→ ∞  2n 8 n 3
 27. (c) y = lim ⇒ y = lim
x →π / 2 sin (2 x − π ) x →π / 2 sin (2 x − π )
23. (c) We have lim (1 − x + [ x − 1] + [1 − x])
x →1− x π 2 2

 2 − 8 
= lim (1 − (1 − h) + [1 − h − 1] + [1 − (1 − h)])   1 (4 x 2 − π 2 )
h→0 y = lim ⇒ y = lim
x →π / 2 sin (2 x − π ) x →π / 2 8 sin (2 x − π )
= lim (h + [ −h] + [ h]) = lim (h − 1 + 0) = −1
h→0 h→0
 Function, Limits , Continuity and Differentiability 969
1 (2 x − π ) (2 x + π ) x 2 − (sum of roots) x+ (Product of roots) = 0
y = lim
x →π / 2 8 sin (2 x − π )
i.e., x 2 − 17 x + 66 = 0 .
lim (2 x + π )
1 x →π / 2  θ 
y= ,  lim = 1  2x − 1 
8 sin (2 x − π )  θ → 0 sinθ  32. (c) f (x) = [ x] cos  π
lim  2 
x →π / 2 (2 x − π )
Since g(x) = [ x] is always discontinuous at all integral
1 π
y= × 2π = . values of points. Hence f (x) is discontinuous for all
8 4
integral points.
28. (c) y = lim(cos x)cot x
x →0 33. (a) Let f (x) = ln (x), x > 0
Taking log on both sides, f (x) = ln (x) is a continuous function of x for every
⇒ log y = lim cot x log cos x positive value of x.
x →0
x x
f   = ln   = ln (x)– ln (y)=f(x)– f(y).
log cos x 0  y y
⇒ log y = lim ,  form 
x →0 tan x 0 
34. (d) For f (x) to be continuous at x = 0, we should have
Applying L-Hospital’s rule, lim f (x) = f (0) = 12 (log 4) 3
x →0
− tan x
⇒ log y = lim =0 x
x →0 sec 2 x 3  
 4x −1  p px 2
⇒ y = e0 ⇒ y = 1 . lim f (x) = lim   × .
 1
x →0 x →0
 x   x  
 sin  log  1 + x 2 
29. (c) n cannot be negative integer for then the  p  3 
limit = 0
 
 
x x2
2 sin 2
e x − cos x 1 e x − cos x = (log 4) . 1 . p . lim 
3 
2 x →0  1 2 1 4 
Limit = lim 2 2 2
= lim  x − x + ......... 
x →0 2 ( x / 2) x n −
2 x →0 x n− 2 3 18 
(n ≠ 1 for then the limit = 0) = 3 p (log 4) 3 . Hence p = 4.

1 e x + sin x 35. (d) Given f (x) = [ x] 2 − [ x 2 ]

= lim .
2 x →0 (n − 2)x n−3
− 1 < x < 0, f (x) = (−1) 2 − 0 = 1
1 x = 0, f (x) = 0 2 − 0 = 0
So, if n = 3, the limit is which is finite.
2(n − 2)
0 < x < 1, f (x) = 0 2 − 0 = 0
If n = 4, the limit is infinite.
x = 1, f (x) = 1 2 − 1 = 0
f ( x 2 ) − f ( x) 0 
30. (c) lim ,  form 
x → 0 f ( x) − f (0) 0  1< x < 2 , f ( x) = 1 2 − 1 = 0

2 xf ' (x 2 ) − f ' (x) x= 2 , f (x) = 1 2 − 2 = −1

= lim , (using L' Hospital's rule)
x →0 f ' ( x) 2 < x < 3 , f (x) = 1 2 − 2 = −1
2 xf ' (x 2 ) x = 3 , f (x) = 1 2 − 3 = −2
= −1 + lim = −1, f ' (0) ≠ 0,
x →0 f ' ( x)
3 < x < 2, f (x) = 1 2 − 3 = −2
as f is strictly increasing.
x = 2, f (x) = 4 − 4 = 0 ; 2 < x < 5 , f (x) = 4 − 4 = 0
 x 2 − 3, 2 < x < 3
31. (c) f ( x) = 
2 x + 5, 3 < x < 4 x = 5 , f (x) = 4 − 5 = −1
Hence function is discontinuous at all integers except 1.
∴ lim f (x) = lim (x 2 − 3) = 6
x →3− x →3−  1 1
− + 
 | x| x 
and lim+ f (x) = lim+ (2 x + 5) = 11 36. (b) f (0) = 0 and f (x) = xe
x →3 x →3
h
Hence, the required equation will be R.H.L. = lim(0 + h)e − 2 / h = lim =0
h→0 h→0 e2 / h
 970 Functions, Limits, Continuity and Differentiability
1 1
− − 
 h h
L.H.L. = lim(0 − h)e = 0 ; ∴ f (x) is continuous.
h→0

1 1 1 1
− +  − + 
 h h  h h
(0 + h)e − he
Rf ′ ( x) = lim =0
h→0 h
1 1 1 1
− −  − + 
 h h  h h
(0 − h)e − he
L f ′(x) = lim =1
h→0 −h
⇒ L f ′(x) ≠ Rf ′(x) . f (x) is not differentiable at x = 0.
 L g′(0) = R g′(0) then g(x) is differentiable at x = 0
1 − tan x  0 
37. (c) lim f (x) = lim ,  form 
π π 4x − π  0  1
x→
4
x→
4 Now g(x) = x 2 sin
x
− sec 2 x − 2 − 1 1 1 1
lim = = . g′(x) = 2 x sin + x 2 cos × − 2
x→
π 4 4 2 x x x
4
1 1 1
π  π  −1 g′(x) = 2 x sin − cos ⇒ g′(x) = 2 f (x) − cos
∴ For f(x) to be continuous at x = , f  = x x x
4 4 2
So, g′(x) is not differentiable at x = 0 .
 1  2 1
38.
 x sin , x ≠ 0
(a,b) f (x) =  , g(x) =
 x sin , x ≠ 0 39. (a, c) f (x) = max {(1 − x), (1 + x), 2}, ∀ x ∈ (− ∞, ∞).
x  x
 0 , x=0  0 , x=0
1 + x; x >1

f (0 − h) − f (0) f ( x) =  2; − 1 ≤ x ≤ 1
L f ′(0) = lim 1 − x;
h→0 −h  x < −1
1 Since f (x) = 1 − x or 1 + x are polynomial functions
(0 − h) sin (− ) − (0)
h 1 and f (x) = 2 is a constant function.
= lim = lim − sin  
h→0 −h h→0  h
∴ These are continuous at all points .....(i)
= a quantity which lies between – 1 and 1
f (0 + h) − f (0)
R f ′(0) = lim
h→0 h
1
(0 + h) sin −0 –1 1
h 1
= lim = lim sin ∴ f (x) is differentiable at all the points, except at
h→0 h h→0 h
= a quantity which lies between – 1 and 1 x = 1 and at x = −1 .....(ii)
Hence L f ′(0) ≠ R f ′(0) 40. (a) We have, f (x) =| x | + | x − 1|

∴ f (x) is not differentiable at x = 0  − 2 x + 1, x<0 − 2 x + 1, x<0

 
=  x − x + 1, 0 ≤ x < 1 =  1 0 ≤ x <1
f (0 − h) − f (0)  x + x − 1,  2 x − 1,
Now L g′(0) = lim  x ≥ 1  x ≥1
h→0 0−h
1 Clearly, lim− f (x) = 1, lim+ f (x) = 1, lim− f (x) = 1
(0 − h) 2 sin (− ) − 0 x →0 x →0 x →1
h 1
L g′(0) = lim = lim h sin   and lim+ f (x) = 1 . So, f (x) is continuous at x = 0, 1.
h→0 −h h→ 0  h x →1

 1  − 2, x < 0
L g′(0) = 0 ×  − 1 ≤ sin ≤ 1 ⇒ L g′(0) = 0 
 h  Now f ' (x) =  0, 0 ≤ x < 1
 2, x ≥ 1
1 
(0 + h) 2 sin  − 0
f (0 + h) − f (0)  h Here x = 0, f ' (0 + ) = 0 while f ' (0 − ) = −2
and Rg′(0) = lim = lim
h→0 h h→0 h
and at x = 1, f ' (1+ ) = 2 while f ' (1− ) = 0
1  1 
= lim h sin  = 0 ×  − 1 ≤ sin  ≤ 1 = 0 Thus, f (x) is not differentiable at x = 0 and 1.
h→0  h   h 