The value of $k$ for which function $f(x) = \begin{cases} x^2, & x \geq 0 \\ kx, & x < 0 \end{cases}$ is differentiable at $x = 0$ is : (a) $1$ (b) $2$ (c) any real number (d) $0$

The value of k for which function f(x) = x², x ≥ 0 kx, — CBSE Class 12 Mathematics 2023 PYQ

Question

k

(a)

1

(b)

2

(d)

0

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

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A piecewise function is differentiable at a point only if it is continuous there and the left and right derivatives match. For $f(x)$ at $x=0$ , continuity forces $k$ to be anything, but matching derivatives forces $k=0$ .

Differentiability is a stronger condition than continuity. For a function to be differentiable at a point, two things must happen: the function must be continuous there, and the derivative must exist (meaning the left-hand and right-hand derivatives must be equal).

Let's check both conditions systematically for $f(x)$ at $x = 0$ .

Checking Continuity at $x = 0$

For continuity at $x = 0$ , we need:

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)

From the right: As $x \to 0^+$ , we use $f(x) = x^2$ , so $\lim_{x \to 0^+} f(x) = 0^2 = 0$ .
From the left: As $x \to 0^-$ , we use $f(x) = kx$ , so $\lim_{x \to 0^-} f(x) = k \cdot 0 = 0$ .
At the point: $f(0) = 0^2 = 0$ (since $0 \geq 0$ , we use the first piece).

All three equal $0$ regardless of $k$ , so $f$ is continuous at $x = 0$ for any value of $k$ .

Checking Differentiability at $x = 0$

Now we compute the left-hand derivative (LHD) and right-hand derivative (RHD) using the definition:

f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}

Right-hand derivative ( $h \to 0^+$ ): …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Every differentiable function is continuous, but the converse is not true.$
›Reveal solutionSolution

True - differentiable $\Rightarrow$ continuous, converse false.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): Every differentiable function is continuous. Reason (R): Every continuous function is differentiable. Choose the correct option:$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Differentiability implies continuity, but the converse is false.

A is TRUE: Every differentiable function is continuous — this is a standard theorem.

…

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CBSE 2023Set 65/3/11 markMCQ
$k$
(a) $1$
(b) $2$
(c) any real number
(d) $0$
›Reveal solutionSolution
A piecewise function is differentiable at a point only if it is continuous there and the left and right derivatives match. For $f(x)$ at $x=0$ , continuity forces $k$ to be anything, but matching derivatives forces $k=0$ .

Differentiability is a stronger condition than continuity. For a function to be differentiable at a point, two things must happen: the function must be continuous there, and the derivative must exist (meaning the left-hand and right-hand derivatives must be equal).

Let's check both conditions systematically for $f(x)$ at $x = 0$ .

Checking Continuity at $x = 0$

For continuity at $x = 0$ , we need:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$

From the right: As $x \to 0^+$ , we use $f(x) = x^2$ , so $\lim_{x \to 0^+} f(x) = 0^2 = 0$ .

From the left: As $x \to 0^-$ , we use $f(x) = kx$ , so $\lim_{x \to 0^-} f(x) = k \cdot 0 = 0$ .

At the point: $f(0) = 0^2 = 0$ (since $0 \geq 0$ , we use the first piece).

All three equal $0$ regardless of $k$ , so $f$ is continuous at $x = 0$ for any value of $k$ .

Checking Differentiability at $x = 0$

Now we compute the left-hand derivative (LHD) and right-hand derivative (RHD) using the definition:
$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$
Right-hand derivative ( $h \to 0^+$ ): …
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CBSE 2020Set HE8231 mark
$Q. Fill in the blank: Every differentiable function is ______.$
›Reveal solutionSolution

Every differentiable function is continuous.

If a function $f$ is differentiable at a point $x=a$ , then

$\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=f'(a)\text{ exists (finite)}.$

Writing $f(a+h)-f(a)=\dfrac{f(a+h)-f(a)}{h}\cdot h$ and taking the limit as $h\to0$ :

$\lim_{h\to0}[f(a+h)-f(a)]=f'(a)\cdot 0=0,$

which means $\displaystyle\lim_{h\to0}f(a+h)=f(a)$ — exactly the condition for continuity at $x=a$ .

…

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CBSE 2019Set ANNUAL1 mark
$f'(2^{+}) = 0$
›Reveal solutionSolution

Existence of a finite one-sided derivative at a point forces one-sided continuity there; since both $f'(2^+)$ and $f'(2^-)$ exist (and equal $0$ ), $f$ is continuous at $x=2$ .

By definition, $f'(2^+) = \displaystyle\lim_{h\to 0^+}\dfrac{f(2+h)-f(2)}{h}$ .

For this limit to exist and be finite (here, equal to $0$ ), the numerator $f(2+h)-f(2)$ must itself tend to $0$ as $h\to 0^+$ — because $f(2+h)-f(2) = h\cdot\dfrac{f(2+h)-f(2)}{h} \to 0\cdot 0 = 0$ .

So $\displaystyle\lim_{h\to0^+} f(2+h) = f(2)$ , i.e. $f$ is right-continuous at $x=2$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. If f$
(x) is differentiable at x = a, which of the following statement may be false ?
(a) f
(x) is continuous at x = a
(b) lim x→a f
(x) exist
(c) lim h→0⁻ [f(a+h) − f(a)]/h = lim h→0⁺ [f(a+h) − f(a)]/h
(d) The second derivative of f
(x) i.e. f″
(x) exist at x = a
›Reveal solutionSolution
Differentiability at a point guarantees continuity and the existence of f′(a), but never guarantees f″(a) exists.

If $f$ is differentiable at $x=a$ , then by definition the limit $f'(a) = \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}$ exists, which forces:

(a) $f$ continuous at $a$ — always true (differentiability $\Rightarrow$ continuity).

(b) $\lim_{x\to a} f(x)$ exists — always true, again because differentiability implies continuity.

(c) $\lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h} = \lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$ — always true, since a two-sided derivative existing means the left-hand and right-hand derivatives are equal. …
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Q.The value of $k$ for which function $f(x) = \begin{cases} x^2, & x \geq 0 \\ kx, & x < 0 \end{cases}$ is differentiable at $x = 0$ is :

Concept understanding — Continuity Differentiability Relationship

How Continuity and Differentiability Are Related

Why differentiability forces continuity

The converse is FALSE

Putting it together

Checking Continuity at $x = 0$

Checking Differentiability at $x = 0$

Checking Continuity at $x = 0$

Checking Differentiability at $x = 0$

Q.The value of kkk for which function f(x)={x2,x≥0kx,x<0f(x) = \begin{cases} x^2, & x \geq 0 \\ kx, & x < 0 \end{cases}f(x)={x2,kx,​x≥0x<0​ is differentiable at x=0x = 0x=0 is :

Checking Continuity at x=0x = 0x=0

Checking Differentiability at x=0x = 0x=0

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Q.The value of $k$ for which function $f(x) = \begin{cases} x^2, & x \geq 0 \\ kx, & x < 0 \end{cases}$ is differentiable at $x = 0$ is :

Checking Continuity at $x = 0$

Checking Differentiability at $x = 0$

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