Let $f\left( x \right)$ be a function satisfying $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)$ and $f\left( x \right) = x{\rm{ }}g\left( x \right)$ for all $x,y \in R$, where $g\left( x \right)$ is continuous. Then,

$f'\left( x \right) = g'\left( x \right)$

$f'\left( x \right) = g\left( x \right)$

If $f$ is a periodic function, then

$f^{\prime}$ is periodic but $f^{\prime \prime}$ is not periodic

$f^{\prime \prime}$ is periodic but $f^{\prime}$ is not periodic

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JEE • MATHEMATICS • CONTINUITY

Continuity practice · Medium

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Showing 61-71 of 71 questions

61 Medium

61 of 71

Analyzing +4 | -1

Let

f\left( x \right)

be a function satisfying

f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)

and

f\left( x \right) = x\mathrm{ }g\left( x \right)

for all

x,y \in R

, where

g\left( x \right)

is continuous. Then,

View answer and explanation

Correct option: C

f'\left( x \right) = g\left( 0 \right)

Option C (

f'(x) = g(0)

) is correct because the derivative of f(x) is equal to g(0), derived from the given functional equation and the continuity of g(x).

62 Medium

62 of 71

Analyzing +4 | -1

The values of

a

and

b

so that the function

f(x)= \begin{cases}x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi\end{cases}

is continuous for

0 \leq x \leq \pi

are, respectively,

View answer and explanation

Correct option: A

\frac{\pi}{3}, \frac{\pi}{12}

Option A (

\frac{\pi}{3}, \frac{\pi}{12}

) is correct because these values of a and b make the given piecewise function continuous at x =

\frac{\pi}{4}

and x =

\frac{\pi}{2}

63 Medium

63 of 71

Application +4 | -1

Let

f(x)

be continuous function such that

f(0)=1

and

f(x)-f\left(\frac{x}{7}\right)=\frac{x}{7} \, \forall \, x \in R

, then

f(42)=

View answer and explanation

Correct option: A

Option A (8) is correct because the limit definition of the derivative leads to f(x) = x/6 + 1, and therefore f(42) = 8.

64 Medium

64 of 71

Application +4 | -1

Let

f(x)=\left\{\begin{array}{cc}\frac{\tan x-\cot x}{x-\frac{\pi}{4},} & x \neq \frac{\pi}{4} \\ a, & x=\frac{\pi}{4}\end{array}\right.

The value of

a

so that

f(x)

is continuous at

x=\pi / 4

, is

View answer and explanation

Correct option: B

Option B (4) is correct because it is the limit of the given function as x approaches

\frac{\pi}{4}

, ensuring continuity.

65 Medium

65 of 71

Analyzing +4 | -1

If the function

f

defined on

\left( {\frac{\pi }{6},\,\frac{\pi }{3}} \right)

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}},}&{x \ne \frac{\pi }{4}}\\ {k,}&{x = \frac{\pi }{4}} \end{array}} \right.

is continuous, then

k

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because the limit of the function as x approaches

\frac{\pi}{4}

must equal k for the function to be continuous, and the limit is calculated to be

\frac{1}{2}

66 Medium

66 of 71

Analyzing +4 | -1

Let

f(x)=-1+|x-2|

, and

g(x)=1-|x|

; then the set of all points where

f o g

is discontinuous is :

View answer and explanation

Correct option: D

an empty set

Option D (an empty set) is correct because the composite function fog simplifies to |x|, which is continuous everywhere.

67 Medium

67 of 71

Analyzing +4 | -1

The function

f:R/\left\{ 0 \right\} \to R

given by

f\left( x \right) = \frac{1}{x} - \frac{2}{{{e^{2x}} - 1}}

Can be made continuous at

x = 0

by defining

f\left( 0 \right)

as function

View answer and explanation

Correct option: D

Option D (1) is correct because the limit of the function as x approaches 0 is 1, which is the value needed to make the function continuous.

68 Medium

68 of 71

Analyzing +4 | -1

If the function

f(x)= \begin{cases}2 x, & |x| \leq 1 \\ x^2+a x+b, & |x| > 1\end{cases}

is continuous for all real

x

, then

View answer and explanation

Correct option: C

a=2, b=-1

and

f

is not differentiable at

x=-1,1

Option C (

a=2, b=-1

and

f

is not differentiable at

x=-1,1

) is correct because the function must be continuous at x = -1 and x = 1, and the derivatives do not match at these points.

69 Medium

69 of 71

Understanding +4 | -1

Let a function

f:R \to R

, where

R

is the set of real numbers satisfying the equation

f\left( {x + y} \right) = f\left( x \right) + f\left( y \right),\forall x,y

f\left( x \right)

is continuous at

x = 0

, then

View answer and explanation

Correct option: B

f\left( x \right)

is continuous,

\forall x \in R

Option B (

f(x)

is continuous,

\forall x \in R

) is correct because the given functional equation and continuity at x=0 imply continuity everywhere.

70 Medium

70 of 71

Understanding +4 | -1

f

is a periodic function, then

View answer and explanation

Correct option: A

f^{\prime}

and

f^{\prime \prime}

are also periodic

Option A (

f^{\prime}

and

f^{\prime \prime}

are also periodic) is correct because the derivative of a periodic function is also periodic with the same period.

71 Medium

71 of 71

Analyzing +4 | -1

Let

x \neq-1

. For each positive integer

n

, let

\begin{aligned} f_n(x)= & \frac{x}{x+1}+\frac{x^2}{(x+1)\left(x^2+1\right)}+\frac{x^4}{(x+1)\left(x^2+1\right)\left(x^4+1\right)} \\ & +\cdots+\frac{x^{2^{n-1}}}{(x+1)\left(x^2+1\right) \cdots\left(x^{2 x^{2-1}}+1\right)} \end{aligned}

for each positive integer

n

. Suppose

f(x)=\lim \limits_{{n \rightarrow \infty}} f(x)

. Then

View answer and explanation

Correct option: D

f

is continuous for all

x \neq-1

Option D (

f

is continuous for all

x \neq -1

) is correct because by simplifying the expression for

f_n(x)

and taking the limit as n approaches infinity, the resulting function

f(x)

is continuous for all

x \neq -1

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