Chapter 5 – Continuity And Differentiability

Find the value of the constant k so that the function f defined below is  1 – cos 4 x continuous at x = 0, where f ( x) =  2 ,x≠0 .  8x  k , x=0 Solution It is given that the function f is continuous at x = 0. Therefore, lim x→0 f (x) = f (0) 1 – cos 4 x ⇒ lim =k x →0 8x2 2sin 2 2 x ⇒ lim =k x →0 8x2  sin 2 x  ⇒ lim   =k x →0  2x  ⇒ k=1 Thus, f is continuous at x = 0 if k = 1.

Discuss the continuity of the function f(x) = sin x . cos x. Solution Since sin x and cos x are continuous functions and product of two continuous function is a continuous function, therefore f(x) = sin x . cos x is a continuous function. 92 MATHEMATICS  x 3 + x 2 – 16 x + 20  ,x≠2

If f ( x) =  ( x – 2) 2 is continuous at x = 2, find  =  k , x 2 the value of k. Solution Given f (2) = k. x 3 + x 2 – 16 x + 20 Now, lim– f ( x) = lim+ f ( x) = lim x→2 x→2 x→2 ( x – 2) 2 ( x + 5)( x – 2) 2 = lim = lim( x + 5) = 7 x→2 ( x – 2) 2 x→2 As f is continuous at x = 2, we have lim f ( x) = f (2) x →2 ⇒ k = 7.

Show that the function f defined by  1  x sin , x ≠ 0 f ( x) =  x  0, x = 0 is continuous at x = 0. Solution Left hand limit at x = 0 is given by 1 1 lim f ( x) = lim– x sin = 0 [since, –1 < sin < 1] x →0– x →0 x x Similarly lim+ f ( x) = lim+ x sin = 0 . Moreover f (0) = 0. x→0 x→0 x Thus xlim f ( x) = lim+ f ( x) = f (0) . Hence f is continuous at x = 0 → 0– x→0

Given f(x) = . Find the points of discontinuity of the composite x –1 function y = f [f(x)]. Solution We know that f (x) = is discontinuous at x = 1 x –1 Now, for x ≠ 1 , CONTINUITY AND DIFFERENTIABILITY 93 1 x –1  1  = = f   = 1 f (f (x)) x – 1 –1 2 – x , x –1 which is discontinuous at x = 2. Hence, the points of discontinuity are x = 1 and x = 2.

Let f(x) = x x , for all x ∈ R. Discuss the derivability of f(x) at x = 0  x 2 , if x ≥ 0 Solution We may rewrite f as f ( x ) =  2 − x ,if x < 0 f (0 + h) – f (0) – h2 – 0 Now Lf ′ (0) = lim– = lim– = lim− − h = 0 h →0 h h →0 h h →0 f (0 + h) – f (0) h2 – 0 Now Rf ′ (0) = lim+ = lim+ = lim− h = 0 h →0 h h →0 h h →0 Since the left hand derivative and right hand derivative both are equal, hence f is differentiable at x = 0.

Differentiate tan x w.r.t. x Solution Let y = tan x . Using chain rule, we have dy 1 d = . (tan x ) dx 2 tan x dx

If y = tan(x + y), find . Solution Given y = tan (x + y). differentiating both sides w.r.t. x, we have 94 MATHEMATICS dy d = sec2 ( x + y ) ( x + y ) dx dx  dy  = sec2 (x + y) 1+  or [1 – sec2 (x + y] = sec2 (x + y) dy sec 2 ( x + y ) Therefore, = = – cosec2 (x + y). dx 1 − sec 2 ( x + y )

If ex + ey = ex+y, prove that = − e y−x . Solution Given that ex + ey = ex+y. Differentiating both sides w.r.t. x, we have dy  dy  ex + ey = ex+y 1+  dx dx or (ey – ex+y) = ex+y – ex, dy e x + y – e x ex + e y − ex which implies that = = = – ey−x . dx e y − e x + y e − e −e y x y dy  3x − x3  1 1

Find , if y = tan –1  2  ,− <x < . dx  1 − 3x  3 3 −π π Solution Put x = tan θ , where <θ< . 6 6  3tan θ − tan 3 θ  Therefore, y = tan –1    1 − 3tan θ  = tan–1 (tan3 θ ) −π π = 3θ (because < 3θ< ) 2 2 = 3tan–1x CONTINUITY AND DIFFERENTIABILITY 95 dy 3 Hence, = 2 . dx 1 + x { }

If y = sin–1 x 1 − x − x 1 − x and 0 < x < 1, then find 2 dy . { } Solution We have y = sin–1 x 1 − x − x 1 − x , where 0 < x < 1. Put x = sinA and x = sinB { Therefore, y = sin–1 sin A 1 − sin B − sin B 1 − sin A 2 2 } = sin–1 { sin A cos B − sin Bcos A} = sin–1 { sin(A − B)} = A – B Thus y = sin–1 x – sin–1 x Differentiating w.r.t. x, we get = 1− x − . ( x) 1− ( x) dx 2 2 dx 1 1 = − 1− x 2 2 x 1− x . dy π

If x = a sec3 θ and y = a tan3 θ , find at θ = . dx 3 Solution We have x = a sec3 θ and y = a tan3 θ . Differentiating w.r.t. θ , we get dx d = 3a sec 2 θ (sec θ) = 3a sec3 θ tan θ dθ dθ dy d and = 3a tan 2 θ (tan θ) = 3a tan 2 θ sec2 θ . dθ dθ dy d θ 3a tan 2 θ sec 2 θ tan θ = = = = sin θ Thus dx dx 3a sec3 θ tan θ sec θ . dθ 96 MATHEMATICS  dy  π 3 Hence,  dx  at θ= π = sin 3 = 2 . dy log x

If xy = ex–y, prove that = . dx (1+ log x) 2 Solution We have xy = ex–y . Taking logarithm on both sides, we get y log x = x – y ⇒ y (1 + log x) = x i.e. y = 1+ log x Differentiating both sides w.r.t. x, we get  1 (1+ log x).1 − x   dy  x log x . = = dx (1+ log x) 2 (1+ log x) 2 d2y cos x

If y = tanx + secx, prove that = . dx 2 (1 − sin x) 2 Solution We have y = tanx + secx. Differentiating w.r.t. x, we get = sec2x + secx tanx 1 sin x 1 + sin x 1+ sin x = 2 + 2 = 2 = (1+ sin x)(1− sin x) . cos x cos x cos x dy 1 thus = 1– sin x . Now, differentiating again w.r.t. x, we get d 2 y – ( – cos x ) cos x 2 = (1– sin x ) 2 = dx (1– sin x) 2  3π 

If f (x) = |cos x|, find f ′   . CONTINUITY AND DIFFERENTIABILITY 97 π Solution When < x < π, cosx < 0 so that |cos x| = – cos x, i.e., f (x) = – cos x ⇒ f ′ ( x) = sin x.  3π  3π 1 Hence, f ′   = sin   =  4  4 2  π

If f (x) = |cos x – sinx|, find f ′   . π Solution When 0 < x < , cos x > sin x, so that cos x – sin x > 0, i.e., f (x) = cos x – sin x ⇒ f ′ ( x) = – sin x – cos x  π π π 1 Hence f ′   = – sin – cos = − (1 + 3) .

Verify Rolle’s theorem for the function, f (x) = sin 2x in 0,  .  2  π Solution Consider f (x) = sin 2x in 0,  . Note that:  2  π

• (i)The function f is continuous in 0,  , as f is a sine function, which is  2 always continuous.  π  π
• (ii)f ′ (x) = 2cos 2x, exists in  0,  , hence f is derivable in  0,  . 2  2  π  π
• (iii)f (0) = sin0 = 0 and f   = sinπ = 0 ⇒ f (0) = f   . 2 2  π Conditions of Rolle’s theorem are satisfied. Hence there exists at least one c ∈  0,  such that f ′(c) = 0. Thus π π 2 cos 2c = 0 ⇒ 2c = ⇒ c= . 2 4 98 MATHEMATICS

Verify mean value theorem for the function f (x) = (x – 3) (x – 6) (x – 9) in [3, 5]. Solution

• (i)Function f is continuous in [3, 5] as product of polynomial functions is a polynomial, which is continuous.
• (ii)f ′(x) = 3x2 – 36x + 99 exists in (3, 5) and hence derivable in (3, 5). Thus conditions of mean value theorem are satisfied. Hence, there exists at least one c ∈ (3, 5) such that f (5) − f (3) f ′ (c ) = 5−3 8−0 ⇒ 3c2 – 36c + 99 = =4 ⇒ c = 6± . Hence c = 6 − (since other value is not permissible).

If f (x) = ,x≠ cot x − 1 4  π π find the value of f   so that f (x) becomes continuous at x = . 4 4 2 cos x − 1 π Solution Given, f (x) = ,x≠ cot x − 1 4 2 cos x − 1 lim f ( x) = lim Therefore, x→ π x→ π cot x −1 4 4 ( 2 cos x − 1) sin x = lim π cos x − sin x x→ ( 2 cos x − 1) . ( 2 cos x + 1) . ( cos x + sin x) .sin x = lim x→ π ( 2 cos x + 1) ( cos x − sin x) ( cos x + sin x) CONTINUITY AND DIFFERENTIABILITY 99 2cos 2 x − 1 cos x + sin x = x → π cos 2 x − sin 2 x 2 cos x +1 ( lim . . sin x ) cos 2 x  cos x + sin x  = x → π cos 2 x  2 cos x +1  ( lim . . sin x )   ( cos x + sin x ) sin x = x→ π 2 cos x + 1 1  1 1   +  2 2 2 1 = = 1 2 2. +1 lim f ( x) = Thus, π 2 x→ π 1 π If we define f   = , then f (x) will become continuous at x = . Hence for f to be   4 2 4 π  π 1 continuous at x = , f   = . 4 4 2  1  ex − 1  , if x ≠ 0

Show that the function f given by f ( x) =  1 e +1  0, if x = 0 is discontinuous at x = 0. Solution The left hand limit of f at x = 0 is given by e x −1 0 −1 lim f ( x) = lim− = = −1 x →0− x→0 1 0 +1 . e x +1 100 MATHEMATICS e x −1 Similarly, lim f ( x) = lim+ 1 x → 0+ x→0 e x +1 1− 1 −1 lim ex 1− e x 1− 0 x → 0+ 1 = xlim −1 = =1 = 1+ → 0+ 1+ 0 1+ e x Thus lim− f ( x) ≠ lim f ( x), therefore, lim f ( x) does not exist. Hence f is discontinuous x →0 + x →0 x →0 at x = 0.  1 − cos 4 x  , if x < 0 x2   , if x = 0

Let f ( x) =  a  x  , if x > 0  16 + x − 4 For what value of a, f is continuous at x = 0? Solution Here f (0) = a Left hand limit of f at 0 is 1 − cos 4 x 2sin 2 2 x lim− f ( x) = lim− = lim x→0 x→0 x2 x →0− x2  sin 2 x  = lim − 8   2 x  = 8 (1)2 = 8. 2 x→0 and right hand limit of f at 0 is lim f ( x) = lim+ x → 0+ x→0 16 + x − 4 x ( 16 + x + 4) = xlim → 0+ ( 16 + x + 4)( 16 + x − 4) CONTINUITY AND DIFFERENTIABILITY 101 = lim x → 0+ x ( 16 + x + 4) 16 + x − 16 = lim+ x→0 ( 16 + x + 4) = 8 Thus, lim+ f ( x) = lim− f ( x) = 8 . Hence f is continuous at x = 0 only if a = 8. x→0 x→0

Examine the differentiability of the function f defined by  2 x + 3, if − 3 ≤ x < − 2  f ( x) =  x + 1 , if − 2 ≤ x < 0  x + 2 , if 0 ≤ x ≤1  Solution The only doubtful points for differentiability of f (x) are x = – 2 and x = 0. Differentiability at x = – 2. f (–2 + h) − f (–2) Now L f ′ (–2) = lim− h→ 0 h 2(–2 + h) + 3 − (–2 + 1) 2h = lim− = lim− = lim− 2 = 2 . h→ 0 h h→ 0 h h→ 0 f (–2 + h) − f (–2) and R f ′ (–2) = lim+ h→ 0 h –2 + h + 1 − ( −2 + 1) = lim− h→ 0 h h −1 − (–1) h = lim− = lim+ = 1 h→ 0 h h→ 0 h Thus R f ′ (–2) ≠ L f ′ (–2). Therefore f is not differentiable at x = – 2. Similarly, for differentiability at x = 0, we have f (0 + h) − f (0) L (f ′(0)= lim− h→ 0 h 0 + h + 1 − (0 + 2) = lim− h→ 0 h h −1  1 = hlim = lim− 1 −  h→0− h → 0  h which does not exist. Hence f is not differentiable at x = 0. 102 MATHEMATICS  1 − x2 

Differentiate tan   x -1  ( )  with respect to cos-1 2 x 1 − x 2 , where  1  x ∈  2  . ,1  1 − x2  Solution Let u = tan-1    x  and v = cos ( -1 2 x 1 − x 2 . ) = We want to find dv dv  1 − x2  π π Now u = tan  -1  . Put x = sinθ.  <θ<  .  x  4 2  1 − sin 2 θ  Then u = tan -1  = tan-1 (cot θ)  sin θ   π  π π = tan-1  tan  − θ   = − θ = − sin x –1  2  2 2 du −1 Hence dx = . 1− x2 Now v = cos–1 (2x 1− x2 ) π = – sin–1 (2x 1− x2 ) π π – sin–1 (2sinθ 1− sin θ ) = − sin (sin 2θ) 2 –1 = 2 2 π π = – sin–1 {sin (π – 2θ)} [since < 2 θ < π] 2 2 CONTINUITY AND DIFFERENTIABILITY 103 π −π = − (π − 2θ) = + 2θ 2 2 −π ⇒ v= + 2sin–1x dv 2 ⇒ = . dx 1− x2 du −1 = = 1 − x 2 = −1 . Hence dv dv 2 2 dx 1 − x2

The function f (x) =  x  k , if x = 0 is continuous at x = 0, then the value of k is

• (A)3
• (B)2
• (C)1
• (D)1.5 Solution (B) is the Correct answer.

The function f (x) = [x], where [x] denotes the greatest integer function, is continuous at

• (A)4
• (B)–2
• (C)1
• (D)1.5 Solution (D) is the correct answer. The greatest integer function[x] is discontinuous at all integral values of x. Thus D is the correct answer.

The number of points at which the function f (x) = x –[ x] is not continuous is

• (A)1
• (B)2
• (C)3
• (D)none of these 104 MATHEMATICS Solution (D) is the correct answer. As x – [x] = 0, when x is an integer so f (x) is discontinuous for all x ∈ Z.

The function given by f (x) = tanx is discontinuous on the set

• (A){ nπ : n ∈Z }
• (B){ 2nπ : n ∈Z }  π   nπ 
• (C)(2n + 1) : n ∈Z
• (D) : n ∈Z  2   2  Solution C is the correct answer.

Let f (x)= |cosx|. Then,

• (A)f is everywhere differentiable.
• (B)f is everywhere continuous but not differentiable at n = nπ, n ∈Z . π
• (C)f is everywhere continuous but not differentiable at x = (2n + 1) , n∈ Z .
• (D)none of these. Solution C is the correct answer.

The function f (x) = |x| + |x – 1| is

• (A)continuous at x = 0 as well as at x = 1.
• (B)continuous at x = 1 but not at x = 0.
• (C)discontinuous at x = 0 as well as at x = 1.
• (D)continuous at x = 0 but not at x = 1. Solution Correct answer is A.

The value of k which makes the function defined by  1 sin , if x ≠ 0 f ( x) =  x , continuous at x = 0 is  k , if x = 0

• (A)8
• (B)1
• (C)–1
• (D)none of these Solution (D) is the correct answer. Indeed lim sindoes not exist. x→ 0 x

The set of points where the functions f given by f (x) = |x – 3| cosx is differentiable is CONTINUITY AND DIFFERENTIABILITY 105

• (A)R
• (B)R – {3}
• (C)(0, ∞)
• (D)none of these Solution B is the correct answer.

Differential coefficient of sec (tan–1x) w.r.t. x is x x

• (A)
• (B)1+ x 2 1+ x 2
• (C)x 1+ x 2
• (D)1+ x 2 Solution (A) is the correct answer. –1  2 x   2x  du

If u = sin  2  and v = tan –1  2  , then is  1 + x   1 − x  dv 1 1– x 2

• (A)
• (B)x
• (C)
• (D)1 2 1+ x 2 Solution (D) is the correct answer.

The value of c in Rolle’s Theorem for the function f (x) = ex sinx, x ∈[0, π] is π π π 3π

• (A)
• (B)
• (C)
• (D)6 4 2 4 Solution (D) is the correct answer.

The value of c in Mean value theorem for the function f (x) = x (x – 2), x ∈ [1, 2] is 3 2 1 3

• (A)
• (B)
• (C)
• (D)2 3 2 2 Solution (A) is the correct answer.

Match the following COLUMN-I COLUMN-II  sin 3 x  , if x ≠ 0

• (A)If a function f ( x) =  x (a) |x|  k , if x = 0  2 is continuous at x = 0, then k is equal to 106 MATHEMATICS
• (B)Every continuous function is differentiable (b) True
• (C)An example of a function which is continuous (c) 6 everywhere but not differentiable at exactly one point
• (D)The identity function i.e. f (x) = x ∀ x∈R is a (d) False continuous function Solution A → c, B → d, C → a, D → b Fill in the blanks in each of the Examples 37 to 41.

The number of points at which the function f (x) = log | x | is discontinuous is ________. Solution The given function is discontinuous at x = 0, ± 1 and hence the number of points of discontinuity is 3.  ax + 1if x ≥1

If f ( x) =  is continuous, then a should be equal to _______.  x + 2if x <1 Solution a = 2

The derivative of log10x w.r.t. x is ________. Solution ( log10 e ) . –1  x + 1  –1  x –1  dy

If y = sec   + sin   , then is equal to ______.  x −1   x +1  dx Solution 0.

The deriative of sin x w.r.t. cos x is ________. Solution – cot x State whether the statements are True or False in each of the Exercises 42 to 46.

For continuity, at x = a, each of xlim f ( x ) and lim– f ( x) is equal to f (a). → a+ x→ a Solution True.

y = |x – 1| is a continuous function. Solution True.

A continuous function can have some points where limit does not exist. Solution False.

|sinx| is a differentiable function for every value of x. CONTINUITY AND DIFFERENTIABILITY 107 Solution False.

cos |x| is differentiable everywhere. Solution True.