The value of $k$ for which the function $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \neq 0 \\ k(x+1), & x = 0 \end{cases}$ is a continuous function, is: (A) $\frac{1}{4}$ (B) $2$ (C) $\frac{1}{2}$ (D) $0$

The value of k for which the function f(x) = x² sin (1)/(x), x ≠ 0 k(x+1), x = 0 is a continuous function, is:

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CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

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Concept understanding — Continuity At A Point

Continuity at a Point

Imagine drawing the graph of a function and putting your pen down at $x = a$ . If the function is continuous there, you can draw straight through that point without lifting your pen — no jump, no hole, no break. That is the intuition; here is the precision.

The Three-Condition Test

For $f(x)$ to be continuous at $x = a$ , all three must hold. If even one fails, $f$ is discontinuous there.

Important

Continuity at $x = a$ requires:

$f(a)$ is defined,
$\displaystyle \lim_{x \to a} f(x)$ exists (left- and right-hand limits are equal),
$\displaystyle \lim_{x \to a} f(x) = f(a)$ .

Condition 1 says $a$ is in the domain — the pen must have somewhere to land. Condition 2 says the curve approaches a single value from both sides — no jump. Condition 3 says that common approach value actually matches the function's value at $a$ — no misplaced point.

Why All Three Are Needed

$f(x) = \dfrac{x^2 - 1}{x - 1}$ has $\lim_{x \to 1} f(x) = 2$ , yet $f(1)$ is undefined (zero denominator). Condition 1 fails, leaving a hole at $(1, 2)$ .

A piecewise function shows the opposite can be fine:

f(x) = \begin{cases} x + 1 & x < 2 \\ 3 & x = 2 \\ x + 1 & x > 2 \end{cases}

Here $f(2) = 3$ , both one-sided limits equal $3$ , and they match $f(2)$ — so all three hold and $f$ is continuous at $x = 2$ .

Common Pitfalls

Watch out

"Limit exists" does not mean "continuous." The hole example has a limit but no continuity — the limit must equal the function value.

Watch out

"Defined everywhere" does not mean "continuous." A piecewise function can have a value at every point and still jump. Always check the one-sided limits.

A Quick Check …

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For continuity at $x = 0$ , the limit of $x^2 \sin \frac{1}{x}$ as $x \to 0$ must equal the function value $k(0+1) = k$ . Since the limit is $0$ , we need $k = 0$ .

A function is continuous at a point when three conditions align: the function is defined there, the limit exists as we approach that point, and crucially, the limit equals the function's value at that point. This problem tests whether you can recognize that continuity at $x = 0$ creates a bridge between two different expressions.

The function behaves as $x^2 \sin \frac{1}{x}$ everywhere except at zero, where it suddenly switches to $k(x+1)$ . At $x = 0$ , this second piece gives us $f(0) = k(0+1) = k$ . For continuity, we need:

\lim_{x \to 0} f(x) = f(0)

Since we approach zero from the region where $x \neq 0$ , the relevant limit is:

\lim_{x \to 0} x^2 \sin \frac{1}{x} = k

Let me find this limit.

Recognize the bounded oscillation

The sine function satisfies $-1 \leq \sin \frac{1}{x} \leq 1$ for all $x \neq 0$ , no matter how wildly $\frac{1}{x}$ oscillates as $x \to 0$ .
Apply the squeeze theorem

Multiplying the inequality by $x^2$ (which is always non-negative):

-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2

Evaluate the bounding limits As $x \to 0$ :

\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0

Conclude via the squeeze theorem …

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

Showing the 12 most recent of 20 on this concept.

CBSE 2026Set 65/3/11 markMCQ
$k$
›Reveal solutionSolution
For continuity at $x = 0$ , the limit of $x^2 \sin \frac{1}{x}$ as $x \to 0$ must equal the function value $k(0+1) = k$ . Since the limit is $0$ , we need $k = 0$ .

A function is continuous at a point when three conditions align: the function is defined there, the limit exists as we approach that point, and crucially, the limit equals the function's value at that point. This problem tests whether you can recognize that continuity at $x = 0$ creates a bridge between two different expressions.

The function behaves as $x^2 \sin \frac{1}{x}$ everywhere except at zero, where it suddenly switches to $k(x+1)$ . At $x = 0$ , this second piece gives us $f(0) = k(0+1) = k$ . For continuity, we need:
$\lim_{x \to 0} f(x) = f(0)$
Since we approach zero from the region where $x \neq 0$ , the relevant limit is:
$\lim_{x \to 0} x^2 \sin \frac{1}{x} = k$
Let me find this limit.

Recognize the bounded oscillation

The sine function satisfies $-1 \leq \sin \frac{1}{x} \leq 1$ for all $x \neq 0$ , no matter how wildly $\frac{1}{x}$ oscillates as $x \to 0$ .

Apply the squeeze theorem

Multiplying the inequality by $x^2$ (which is always non-negative):

$-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

Evaluate the bounding limits As $x \to 0$ :

$\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0$

Conclude via the squeeze theorem …
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CBSE 2026Set ANNUAL1 mark
$Q. Prove that the function f(x) = 5x - 3 is continuous at x = -3.$
›Reveal solutionSolution

A function $f$ is continuous at $x=a$ if $\displaystyle\lim_{x\to a} f(x) = f(a)$ ; check this directly for the linear function $f(x)=5x-3$ at $a=-3$ .

Concept: $f$ is continuous at $x=a$ when: (i) $f(a)$ is defined, (ii) $\displaystyle\lim_{x\to a}f(x)$ exists, and (iii) the two are equal.

Working:

$f(-3) = 5(-3) - 3 = -15 - 3 = -18$

$\lim_{x\to -3} f(x) = \lim_{x\to -3}(5x-3) = 5(-3) - 3 = -18$

…

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CBSE 2025Set 65/4/11 markMCQ
$f$
›Reveal solutionSolution
The function has a jump at $x=1$ because the left-hand limit ( $1$ ) and the right-hand limit ( $5$ ) do not match, so it is discontinuous only at $x=1$ . The correct option is (B).

Continuity at a point means three things must hold: the function is defined there, the limit exists there, and the limit equals the function value. For a piecewise function, the only place where things can go wrong is at the boundary between the pieces — here, at $x=1$ . Everywhere else, the function is just a simple rule (either $x$ or the constant $5$ ), so it's automatically continuous.

Let’s check each candidate point.

At $x=0$

For $x \le 1$ , the rule is $f(x)=x$ . Since $0 \le 1$ , we have $f(0)=0$ .

The left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0$ .

The right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$ (because near $0$ , $x$ is still $\le 1$ ).

So the limit exists and equals $0$ , which matches $f(0)$ . Continuous here.

At $x=1$ — the critical boundary

Left-hand limit: as $x$ approaches $1$ from below, $x \le 1$ , so $f(x)=x$ . Hence

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1.$

Right-hand limit: as $x$ approaches $1$ from above, $x > 1$ , so $f(x)=5$ . Hence

$\lim_{x \to 1^+} f(x) = 5.$

The left and right limits are different ( $1 \neq 5$ ), so the two-sided limit does not exist.

The function value is $f(1)=1$ (since $1 \le 1$ ). …
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CBSE 2025Set IX1 mark
$f(x) = |x|$
›Reveal solutionSolution
Left limit, right limit and the value all equal $0$ , so $|x|$ is continuous at $0$ .

Concept. $f$ is continuous at $x=a$ iff $\displaystyle\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$ .

Here $f(x)=|x|=\begin{cases}-x,&x<0\\ x,&x\ge0\end{cases}$

Left-hand limit: $\displaystyle\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(-x)=0.$

Right-hand limit: $\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0.$ …
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CBSE 2025Set ANNUAL1 markMCQ
$f(x) = |x| - |x + 1|$
(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 $x = -1$
›Reveal solutionSolution

$|x|$ and $|x+1|$ are each continuous everywhere, and the difference of two continuous functions is continuous.

$g(x)=|x|$ is continuous on all of $\mathbb{R}$ , and $h(x)=|x+1|$ (a shifted absolute value) is also continuous on all of $\mathbb{R}$ . Since $f(x)=g(x)-h(x)$ is a difference of two functions continuous eve …

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CBSE 2025Set ANNUAL1 mark
$f$
›Reveal solutionSolution

A function is continuous at a point when its limit there equals its value; check both.

Here $f(x)=2x+3$ (a polynomial), and we test $x=1$ .

Value: $f(1)=2(1)+3=5.$

Limit: $\displaystyle\lim_{x\to1}(2x+3)=2(1)+3=5.$

Since $\displaystyle\lim_{x\to1}f(x)=5=f(1)$ , the function is continuous at $x=1$ .

…

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CBSE 20241 markMCQ
$f(x) = \begin{cases} \frac{\sqrt{3} \cos x + \sin x}{3x + \frac{\pi}{2}}, & x \neq -\frac{\pi}{3} \\ k, & x = -\frac{\pi}{3} \end{cases}$
›Reveal solutionSolution

Continuity requires $k = \lim_{x\to -\pi/3} f(x)$ . The quotient is a $\tfrac{0}{0}$ form at $x=-\tfrac{\pi}{3}$ , and the limit evaluates to $\tfrac{2}{3}$ — option (A).

For $f$ to be continuous at $x = -\tfrac{\pi}{3}$ , we need
$k = \lim_{x \to -\pi/3} \frac{\sqrt{3}\cos x + \sin x}{3\left(x + \tfrac{\pi}{3}\right)}.$
Simplify the numerator. Writing it as a single sine:

$\sqrt{3}\cos x + \sin x = 2\left(\tfrac{\sqrt{3}}{2}\cos x + \tfrac{1}{2}\sin x\right) = 2\sin\!\left(x + \tfrac{\pi}{3}\right).$ …
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CBSE 2024Set 65/1/11 markMCQ
$f(x) = \begin{cases} x^2+3, & x \neq 0 \\ 1, & x=0 \end{cases}$
›Reveal solutionSolution
The function is a parabola with a hole at $x=0$ and a single isolated point at $(0,1)$ . Because the limit as $x\to 0$ is $3$ , not $1$ , the function is discontinuous at $x=0$ — but it is continuous and differentiable everywhere else. The correct option is (C).

The key to this problem is understanding what continuity and differentiability mean at a point, and then checking the one point where the definition changes.

Continuity at a point $x = a$ requires three things to match: the function value $f(a)$ , the left-hand limit $\lim_{x \to a^-} f(x)$ , and the right-hand limit $\lim_{x \to a^+} f(x)$ . If any one of these differs, the function is discontinuous there.

Differentiability at a point requires continuity first — and then the left and right derivatives must also be equal. So if a function is discontinuous at a point, it cannot be differentiable there.

Here, the function is defined by two pieces: for every $x$ except $0$ , it behaves like $x^2 + 3$ (a smooth parabola shifted up by 3). At $x = 0$ alone, it jumps to the value $1$ . That single point is the only place where anything unusual can happen.

Let’s check systematically.

Check continuity at $x = 0$ For $x \neq 0$ , $f(x) = x^2 + 3$ . As $x$ approaches $0$ from either side, $x^2$ approaches $0$ , so

$\lim_{x \to 0} f(x) = 0^2 + 3 = 3.$
But $f(0) = 1$ . Since $3 \neq 1$ , the limit does not equal the function value.

Watch out

A common mistake is to think that because the formula $x^2+3$ is continuous everywhere, the whole function is continuous. But the definition at $x=0$ overrides that — the function is piecewise-defined, and the value at the breakpoint must match the limit.

Hence $f$ is discontinuous at $x = 0$ .

Check continuity for $x \neq 0$

For any $a \neq 0$ , near $a$ the function is simply $f(x) = x^2 + 3$ , which is a polynomial. Polynomials are continuous everywhere. So $f$ is continuous at every $x \neq 0$ .

Check differentiability at $x = 0$ …
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CBSE 2024Set EX1 mark
$f(x) = \begin{cases} x + 2 & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$
›Reveal solutionSolution

The limit as $x\to 0$ is $2$ (from $x+2$ ), but the defined value is $f(0)=1$ . Limit $\ne$ value, so $f$ is discontinuous at $0$ .

Concept. $f$ is continuous at $x=0$ iff $\displaystyle\lim_{x\to 0}f(x)=f(0)$ .

Limit. For $x\ne 0$ , $f(x)=x+2$ , so

$\lim_{x\to 0}f(x)=\lim_{x\to 0}(x+2)=0+2=2.$

Value. By definition $f(0)=1$ .

…

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CBSE 2024Set ANNUAL1 mark
$f(x) = 5x - 3$
›Reveal solutionSolution

Check that the limit at $x=5$ equals the function value there.

Given $f(x) = 5x - 3$ .

Function value: $f(5) = 5(5) - 3 = 25 - 3 = 22.$

Limit: $\displaystyle\lim_{x\to5} f(x) = \lim_{x\to5}(5x-3) = 5(5)-3 = 22.$

Since

$\lim_{x\to5} f(x) = 22 = f(5),$ …

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CBSE 2024Set ANNUAL1 mark
$f(x)$
›Reveal solutionSolution

Standard definition: left-hand limit = right-hand limit = function value at that point.

A function $f(x)$ is said to be continuous at $x=c$ if:

$\lim_{x\to c^-}f(x) = \lim_{x\to c^+}f(x) = f(c)$ …

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CBSE 2023Set 65/1/11 markMCQ
$f(x) = \begin{cases} 3x+5, & x \ge 2 \\ kx^2, & x < 2 \end{cases}$
›Reveal solutionSolution

For a piecewise function to be continuous at the join point $x=2$ , the left-hand limit and right-hand limit must equal the function value at $x=2$ . Equating $k(2)^2$ with $3(2)+5$ gives $4k = 11$ , so $k = \frac{11}{4}$ . The correct option is (D).

The Core Idea: Continuity at a Point

A function is continuous at a point if three things match perfectly — the value from the left, the value from the right, and the actual function value at that point. For a piecewise function like this one, the only place where things could break is at the boundary where the formula changes, which is $x = 2$ .

Think of it like two roads meeting at a junction. For a smooth ride, the elevation of the left road as you approach the junction must exactly match the elevation of the right road as you approach from the other side — and that elevation must also be the height of the junction itself. If they don't match, there's a jump, and the function is discontinuous.

Here, the left piece ( $x < 2$ ) uses $kx^2$ , and the right piece ( $x \ge 2$ ) uses $3x+5$ . The function value at $x=2$ is given by the right piece (since $x \ge 2$ includes $2$ ). So we need the left-hand limit to equal that value.

Step-by-Step Solution

1. Find the function value at $x=2$ .

Since $x=2$ falls in the case $x \ge 2$ , we use $f(x) = 3x+5$ .
$f(2) = 3(2) + 5 = 6 + 5 = 11$
2. Find the left-hand limit as $x \to 2^-$ .

For $x < 2$ , the function is $f(x) = kx^2$ . As $x$ approaches $2$ from the left, we simply substitute $x=2$ into this expression (since $kx^2$ is a polynomial and polynomials are continuous everywhere).
$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} kx^2 = k(2)^2 = 4k$
3. Find the right-hand limit as $x \to 2^+$ .

For $x > 2$ , the function is $f(x) = 3x+5$ . Again, this is a polynomial, so the limit is just the value at $x=2$ .
$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x+5) = 3(2)+5 = 11$
4. Apply the continuity condition.

For $f$ to be continuous at $x=2$ , we need:

$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$ …
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Q.The value of kkk for which the function f(x)={x2sin⁡1x,x≠0k(x+1),x=0f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \neq 0 \\ k(x+1), & x = 0 \end{cases}f(x)={x2sinx1​,k(x+1),​x=0x=0​ is a continuous function, is: (A) 14\frac{1}{4}41​ (B) 222 (C) 12\frac{1}{2}21​ (D) 000

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Q.The value of $k$ for which the function $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \neq 0 \\ k(x+1), & x = 0 \end{cases}$ is a continuous function, is: (A) $\frac{1}{4}$ (B) $2$ (C) $\frac{1}{2}$ (D) $0$

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