Q: Let f(x)= cases e^ x-1 , & x<0 \\ x^2-5x+6, & x 0 cases and g(x)=f(|x|)+|f(x)|. If the number of points where g is not continuous and is not differentiable are and respectively, then + is equal to ___

We are given the function: f(x) = cases e^ x-1 , & x We need to analyze the continuity and differentiability of g(x) = f(|x|) + |f(x)|. For x 0. Thus, f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6. Also, for x 0, so |f(x)| = e^ x-1 . Therefore, for x For x 0, |x| = x. Thus, f(|x|) = f(x) = x^2 - 5x + 6. Therefore, for x 0, g(x) = x^2 - 5x + 6 + |x^2 - 5x + 6|. Let us check the continuity of g(x) at x = 0: _ x 0^- g(x) = _ x 0^- (x^2 + 5x + 6 + e^ x-1 ) = 6 + 1 e _ x 0^+ g(x) = _ x 0^+ (x^2 - 5x + 6

Q: The number of points, at which the function f(x) = \ 6x, 2 + 3x^2\ + |x - 1| | |x^2 - 1 4 | |, x (- , ), is not differentiable, is _____.

Let f(x) = g(x) + h(x), where g(x) = \ 6x,\ 2 + 3x^2\ h(x) = |x - 1| | |x^2 - 1 4 | | We find the points of non-differentiability of each part in x (- , ). Non-differentiability of g(x): g(x) = \ 6x,\ 2 + 3x^2\ is non-differentiable where the two curves intersect with different slopes. 6x = 2 + 3x^2 3x^2 - 6x + 2 = 0 x = 6 36 - 24 6 = 6 2 3 6 = 1 1 3 Both values lie in (- , ). At these points, the derivatives of the two inner functions are 6 and 6x; since x 1, the slopes differ. So g(x) contribu

Q: The number of points in the interval [2, 4], at which the function f(x) = [x^2 - x - 1 2 ], where [ ] denotes the greatest integer function, is discontinuous, is _______.

Let g(x) = x^2 - x - 1 2 . Differentiating with respect to x, we get g'(x) = 2x - 1. For x [2, 4], g'(x) > 0, which implies that g(x) is strictly increasing in the interval [2, 4]. The values of g(x) at the endpoints are: g(2) = 2^2 - 2 - 1 2 = 1.5 g(4) = 4^2 - 4 - 1 2 = 11.5 The function f(x) = [g(x)] is discontinuous at all points where g(x) is an integer, as g(x) is strictly monotonic and crosses these integer values. The integers between 1.5 and 11.5 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Since

Question 1

For the function f(x) = e^ |x| - |x|, x R , consider the following statements: Statement I: f is differentiable for all x R . Statement II: f is increasing in (- , - 2 ). In the light of the above sta

Accepted Answer

For Statement I: The given function is f(x) = e^ |x| - |x|. For x > 0, f(x) = e^ x - x. Differentiating with respect to x, we get: f'(x) = e^ x x - 1 The right-hand derivative at x = 0 is: f'(0^+) = e^ 0 0 - 1 = 1(1) - 1 = 0 For x Differentiating with respect to x, we get: f'(x) = -e^ - x x + 1 The left-hand derivative at x = 0 is: f'(0^-) = -e^ - 0 0 + 1 = -1(1) + 1 = 0 Since f'(0^+) = f'(0^-) = 0, the function f(x) is differentiable at x = 0. For all other x R (x 0), the function is a composit

Question 2

Let f(x) = cases 1 3 , & x /2 \ b(1- x) ( -2x)^2 , & x > /2 cases . If f is continuous at x= /2, then the value of _ 0 ^ 3b-6 |x^2+2x-3|\,dx is:

Accepted Answer

Since f(x) is continuous at x = /2, the right-hand limit must equal the value of the function at x = /2. _ x /2^+ f(x) = f( /2) _ x /2^+ b(1- x) ( -2x)^2 = 1 3 Let x = /2 + h. As x /2^+, h 0^+. _ h 0^+ b(1- ( /2+h)) ( -2( /2+h))^2 = 1 3 _ h 0^+ b(1- h) (-2h)^2 = 1 3 _ h 0^+ b(1- h) 4h^2 = 1 3 Using the standard limit _ h 0 1- h h^2 = 1 2 , we get: b 4 1 2 = 1 3 b 8 = 1 3 b = 8 3 The upper limit of the integral is 3b - 6 = 3 ( 8 3 ) - 6 = 2. The integral to evaluate is I = _ 0 ^ 2 |x^2+2x-3|\,dx.

Question 3

Let f(x) = cases x^3 + 8 ; & x Then the number of points, where the function g f is discontinuous, is __________.

Accepted Answer

The possible points of discontinuity for the composite function g(f(x)) are the points where f(x) is discontinuous and the points where f(x) is equal to a point of discontinuity of g(x). First, we find the points of discontinuity of f(x) and g(x). For f(x), the only possible point of discontinuity is at x = 0. _ x 0^- f(x) = _ x 0^- (x^3 + 8) = 8 _ x 0^+ f(x) = _ x 0^+ (x^2 - 4) = -4 Since _ x 0^- f(x) _ x 0^+ f(x), f(x) is discontinuous at x = 0. For g(x), the only possible point of discontinui

Question 4

Let f(x)= cases e^ x-1 , & x<0 \ x^2-5x+6, & x 0 cases and g(x)=f(|x|)+|f(x)|. If the number of points where g is not continuous and is not differentiable are and respectively, then + is equal to ___

Accepted Answer

We are given the function: f(x) = cases e^ x-1 , & x We need to analyze the continuity and differentiability of g(x) = f(|x|) + |f(x)|. For x 0. Thus, f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6. Also, for x 0, so |f(x)| = e^ x-1 . Therefore, for x For x 0, |x| = x. Thus, f(|x|) = f(x) = x^2 - 5x + 6. Therefore, for x 0, g(x) = x^2 - 5x + 6 + |x^2 - 5x + 6|. Let us check the continuity of g(x) at x = 0: _ x 0^- g(x) = _ x 0^- (x^2 + 5x + 6 + e^ x-1 ) = 6 + 1 e _ x 0^+ g(x) = _ x 0^+ (x^2 - 5x + 6

Question 5

The number of points, at which the function f(x) = \ 6x, 2 + 3x^2\ + |x - 1| | |x^2 - 1 4 | |, x (- , ), is not differentiable, is _____.

Accepted Answer

Let f(x) = g(x) + h(x), where g(x) = \ 6x,\ 2 + 3x^2\ h(x) = |x - 1| | |x^2 - 1 4 | | We find the points of non-differentiability of each part in x (- , ). Non-differentiability of g(x): g(x) = \ 6x,\ 2 + 3x^2\ is non-differentiable where the two curves intersect with different slopes. 6x = 2 + 3x^2 3x^2 - 6x + 2 = 0 x = 6 36 - 24 6 = 6 2 3 6 = 1 1 3 Both values lie in (- , ). At these points, the derivatives of the two inner functions are 6 and 6x; since x 1, the slopes differ. So g(x) contribu

Question 6

The number of points in the interval [2, 4], at which the function f(x) = [x^2 - x - 1 2 ], where [ ] denotes the greatest integer function, is discontinuous, is _______.

Accepted Answer

Let g(x) = x^2 - x - 1 2 . Differentiating with respect to x, we get g'(x) = 2x - 1. For x [2, 4], g'(x) > 0, which implies that g(x) is strictly increasing in the interval [2, 4]. The values of g(x) at the endpoints are: g(2) = 2^2 - 2 - 1 2 = 1.5 g(4) = 4^2 - 4 - 1 2 = 11.5 The function f(x) = [g(x)] is discontinuous at all points where g(x) is an integer, as g(x) is strictly monotonic and crosses these integer values. The integers between 1.5 and 11.5 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Since

Question 7

Let f(x)= _ 0 ( x-x^ ( 2 ) (x-1) 1+x^ ( 2 ) (x-1) ), x R . Consider the following two statements : (I) f(x) is discontinous at x=1. (II) f(x) is continous at x=-1. Then,

Accepted Answer

Taking the limit as 0 , we note x^ (2/ ) 0 for |x| 1 . This gives: f(x) = cases x & x 1^- \ - (x-1) (x-1) & x 1^+ cases Continuity at x = 1 : RHL = _ x 1^+ - (x-1) (x-1) = -1 LHL = _ x 1^- x = -1 , f(1) = -1 f(x) is continuous at x = 1 . So Statement (I) is false. Continuity at x = -1 : f(x) = cases - (x-1) -(x-1) & x -1^- \ x & x -1^+ cases RHL = _ x -1^+ x = -1 LHL = _ x -1^- - (x-1) -(x-1) = 2 -2 Since LHL RHL, f(x) is discontinuous at x = -1 . So Statement (II) is also false. Hence, neithe

Question 8

Let [t] denote the greatest integer less than or equal to t. If the function f(x)= \ array cl b^ 2 ( 2 [ 2 ( x+ x) x ] ), & x 0 \ a &, x=0 array . is continuous at x=0, then a^ 2 +b^ 2 is equal to

Accepted Answer

For x > 0: x - 1 2 2x = x(1 - x). _ x 0^+ x(1- x) x^3 = _ x 0^+ x x 1- x x^2 = 1 1 2 = 1 2 . So a = 1 2 . For x f(x) = b^2 ( 2 1 ) = b^2. For continuity: b^2 = a = 1 2 . a^2 + b^2 = 1 4 + 1 2 = 3 4 .

Question 9

If the function f(x)= e^ x (e^ x-x -1 )+ _ e ( x+ x)-x x-x is continuous at x=0, then the value of f(0) is equal to

Accepted Answer

f(0) = _ x 0 e^ x - e^x + ( x + x) - x x - x Applying L'hospital rule f(0) = _ x 0 e^ x ^2 x - e^x + x - 1 ^2 x - 1 f(0) = _ x 0 e^ x ( ^2 x - 1) + (e^ x - e^x) + x - 1 ^2 x f(0) = _ x 0 ( e^ x + e^x ( e^ x - x - 1 ) ^2 x + 1 x + 1 ) f(0) = 1 + 0 + 1 2 = 3 2

Question 10

Let , R be such that the function f(x)= cases 2 (x^ 2 -2 )+2 x &, x be differentiable at all x R . Then 34( + ) is equal to

Accepted Answer

f(x) = cases 2 x^2 + 2 x - 4 & ; x f(1^-) = 2 - + 3, f(1) = -2 + 2 2 - + 3 = 2 - 2 4 - 3 + 3 = 0 ...(1) f'(1^+) = 4 + 2 , f'(1^-) = + 3 4 + 2 = + 3 3 + 2 - 3 = 0 ...(2) Solving (1) & (2) We get = 3 17 , = 21 17 34( + ) = 34 27 17 = 48

Question 11

If f(x)= \ array cc a|x|+x^ 2 -2( |x|)( |x|) x &, x 0 \ b &, x=0 array . is continuous at x=0, then a+b is equal to

Accepted Answer

Note 2 |x| |x| = (2|x|). For x > 0: f(x) = a + x - 2x x x 0^+ a - 2. For x For continuity: a - 2 = 2 - a a = 2, and b = a - 2 = 0. a + b = 2.

Question 12

Let f(x)= cases a x^ 2 +2 a x+3 4 x^ 2 +4 x-3 &, x - 3 2 , 1 2 \ ~b &, x=- 3 2 , 1 2 cases be continuous at x=- 3 2 . If f f(x)= 7 5 , then x is equal to:

Accepted Answer

4x^2+4x-3 = (2x+3)(2x-1). For continuity at x = -3/2, numerator must vanish there: a(9/4) - 3a + 3 = 0 a = 4. f(x) = (2x+1)(2x+3) (2x+3)(2x-1) = 2x+1 2x-1 for x -3/2. f f(x) = f\! ( 2x+1 2x-1 ) = 6x+1 2x+3 . 6x+1 2x+3 = 7 5 30x+5 = 14x+21 x = 1.

Question 13

Let [ ] denote the greatest integer function, and let f(x)= \ 2 x, x^ 2 \ . Let S = \ x (-2,2) . : the function g (x)=|x| [x^ 2 ] is discontinuous at .x \ . Then _ x S f(x) equals

Accepted Answer

g ( x )=| x | [ x ^2 ] Points of discontinuity of g ( x ) in (-2,2) are ( 1, 2 , 3 ) S=\ -1,1,- 2 , 2 ,- 3 , 3 \ f(x)= \ 2 x, x^2 \ _ x S f(x)=- 2 +1-2+2- 6 + 6 =1- 2

Continuity and Differentiability — Previous Year Questions & Solutions