If $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\left| {x + 2} \right|}}{{{{\tan }^{ - 1}}\left( {x + 2} \right)}},x \ne - 2}\\ {2,x = - 2} \end{array}} \right.$ $f\left( x \right)$ is

Continuous but not derivable at $x = - 2$

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JEE • MATHEMATICS • CONTINUITY

Continuity practice · Hard

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Practice format and chapter quiz

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1 Hard

1 of 7

Application +4 | -1

f\left( x \right) = \frac{{{{\log }_\mathrm{e}}\left( {1 + {x^2}\tan x} \right)}}{{\sin {x^3}}},x \ne 0

, is to be continuous at

x = 0

, then

f\left( 0 \right)

must be defined as

View answer and explanation

Correct option: A

Option A (1) is correct because for the function to be continuous at x=0, f(0) must equal the limit of the function as x approaches 0, which is 1.

2 Hard

2 of 7

Analyzing +4 | -1

Let

\left[ t \right]

denote the greatest integer

\le t

and

\lim\limits_{x \to 0} x\left[ {\frac{4}{x}} \right] = A.

Then the function,

f(x) = \left[ {{x^2}} \right]\sin (\pi x)

is discontinuous, when x is equal to

View answer and explanation

Correct option: A

\sqrt {A + 1}

Option A (

\sqrt {A + 1}

) is correct because the function

f(x) = \left[ {{x^2}} \right]\sin (\pi x)

is discontinuous at non-integer values of

\sqrt{A+1}

, where A = 4.

3 Hard

3 of 7

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\left| {x + 2} \right|}}{{{{\tan }^{ - 1}}\left( {x + 2} \right)}},x \ne - 2}\\ {2,x = - 2} \end{array}} \right.

f\left( x \right)

View answer and explanation

Correct option: B

Not continuous

x = - 2

Option B (Not continuous

x = - 2

) is correct because the left-hand limit and right-hand limit of the function at

x = -2

are not equal, indicating a discontinuity.

4 Hard

4 of 7

Analyzing +4 | -1

For

x \in R, f(x)=|\log _{10} 2-\sin x \mid

and

{g}({x})={f}({f}({x}))

, then:

View answer and explanation

Correct option: C

g^{\prime}(0)=\cos \left(\log _{10} 2\right)

Option C (

g^{\prime}(0)=\cos (\log _{10} 2)

) is correct because by applying the chain rule and evaluating the derivative of the composite function g(x) at x=0, we arrive at this result.

5 Hard

5 of 7

Analyzing +4 | -1

f(x)=x+\left[x+\frac{1}{3}\right]+\left[x+\frac{2}{3}\right] \forall x \in R

then the number of points of discontinuity of

f(x)

[-1,1]

, where [.] denotes greatest integer function.

View answer and explanation

Correct option: D

The correct answer is D (6) because the function f(x) has discontinuities at x = -2/3, -1/3, 0, 1/3, 2/3, and 1 within the interval [-1, 1].

6 Hard

6 of 7

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{1}{{\left| x \right|}}\mathrm{for}\,\,\left| x \right| \ge 1}\\ {a{x^2} + b\,\,\mathrm{for}\,\left| x \right| < 1} \end{array}} \right.

f\left( x \right)

is continuous and differentiable at any point, then

View answer and explanation

Correct option: B

a = - \frac{1}{2},b = \frac{3}{2}

Option B is correct because the conditions for continuity and differentiability at x = 1 lead to the equations a + b = 1 and 2a = -1, which are satisfied by a = -1/2 and b = 3/2.

7 Hard

7 of 7

Analyzing +4 | -1

Consider the following two statements:
Statement I: The function

f(x)=\frac{x}{1+|x|}

is differentiable for real

x

.
Statement II: The function

g(x)=x^2|x|

is thrice differentiable for all real

x

.
Then

View answer and explanation

Correct option: B

Statement I is true but statement II is false.

Option B (Statement I is true but statement II is false) is correct because f(x) is differentiable for all real x, while g(x) is not thrice differentiable.

Revision notes and key takeaways

Review CONTINUITY in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 7 single-correct-answer MCQs on CONTINUITY.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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