(a) Find $k$ so that $f(x) = \begin{cases} \frac{x^2 - 2x - 3}{x+1}, & x \neq -1 \\ k, & x = -1 \end{cases}$ is continuous at $x = -1$. OR (b) Check the differentiability of function $f(x) = x|x|$ at $x = 0$.

(a) Find k so that f(x) = (x² - 2x - 3)/(x+1), x ≠ -1 k — CBSE Class 12 Mathematics 2025 PYQ

Question

k

(OR)

(b) Check the differentiability of function

f(x) = x|x|

x = 0

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

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Part (a)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 …

Part (b)Concept understanding — Differentiability of Absolute Value

Differentiability of the Absolute Value Function

Start with something familiar: the absolute value of $x$ , written $|x|$ , is its distance from zero on the number line. So $|3| = 3$ , $|-5| = 5$ , and $|0| = 0$ . Graphically, it looks like a V-shape — two straight lines meeting at the origin.

Differentiability is about whether a function has a well-defined slope (derivative) at a point. For smooth curves like $x^2$ or $\sin x$ , the slope exists everywhere. But the absolute value function has a sharp corner at $x = 0$ — and that corner is the whole story.

Intuition: Why the corner matters

Walk along $y = |x|$ from left to right. Approaching $x = 0$ from the left, the slope is $-1$ (the line goes downward). Leaving $x = 0$ to the right, the slope is suddenly $+1$ (the line goes upward). At $x = 0$ , there's no single slope — it changes abruptly. That's why $|x|$ is not differentiable at $x = 0$ . Everywhere else — for $x < 0$ and $x > 0$ — the graph is a straight line with constant slope, so $|x|$ is differentiable at every point except $x = 0$ .

Note

A function must be continuous to be differentiable, but continuity alone isn't enough. The absolute value function is continuous at $x = 0$ (no break), yet fails to be differentiable there because of the sharp corner.

The precise statement

Let $f(x) = |x|$ . Then:

For $x > 0$ : $f(x) = x$ , so $f'(x) = 1$ .
For $x < 0$ : $f(x) = -x$ , so $f'(x) = -1$ .
At $x = 0$ : the derivative does not exist, because the left-hand and right-hand derivatives are different numbers.

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

Watch out

A common mistake is to think that because $|x|$ is continuous at $x=0$ , it must be differentiable there. Continuity is necessary for differentiability, but not sufficient. The absolute value function is the classic counterexample.

The bigger picture …

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Part (a): removable factor gives $\lim_{x\to-1}f(x)=-4$ , so $k=-4$ . Part (b): left and right derivatives of $x|x|$ at $0$ are both $0$ , so it is differentiable there with $f'(0)=0$ .

Part (a)

Continuity at a point needs the limit to equal the function value.

Simplify for $x\ne-1$ : $x^2-2x-3=(x-3)(x+1)$ , so

f(x)=\frac{(x-3)(x+1)}{x+1}=x-3.

Limit. $\displaystyle\lim_{x\to-1}f(x)=\lim_{x\to-1}(x-3)=-4.$
Continuity condition. $\displaystyle\lim_{x\to-1}f(x)=f(-1)=k\Rightarrow k=-4.$ …

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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 29 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 V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Just left of $0$ , $|x|=-x$ has slope $-1$ , so the left-hand derivative is $-1$ .

The left-hand derivative at $x=0$ is

$\lim_{h\to0^-}\frac{|0+h|-|0|}{h}=\lim_{h\to0^-}\frac{-h}{h}=-1,$ …
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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 65/1/11 markMCQ
$f(x) = |x| + |x-1|$
›Reveal solutionSolution
The function $f(x) = |x| + |x-1|$ is a sum of two absolute value functions, each continuous everywhere but with a corner at its respective critical point. At $x=0$ and $x=1$ , the left and right derivatives differ, so $f$ is continuous but not differentiable at both points — option (C).

The key to this problem is understanding what absolute value does to differentiability. The function $|x|$ has a V-shaped graph — it is continuous everywhere, but at $x=0$ the slope changes abruptly from $-1$ to $+1$ . That sharp corner means the derivative does not exist at $x=0$ , even though the function is perfectly continuous there. The same logic applies to $|x-1|$ at $x=1$ .

When you add two such functions, the sum inherits the continuity of each piece. But at a point where either term has a corner, the sum may also have a corner — unless the slopes happen to cancel, which they do not here.

Let’s check each point carefully.

Continuity at $x=0$ Compute the left-hand limit, right-hand limit, and the function value. For $x < 0$ : $|x| = -x$ , $|x-1| = -(x-1) = 1-x$ , so

$f(x) = -x + (1-x) = 1 - 2x.$
As $x \to 0^-$ , $f(x) \to 1$ .

For $x > 0$ but $x < 1$ : $|x| = x$ , $|x-1| = 1-x$ , so
$f(x) = x + (1-x) = 1.$
As $x \to 0^+$ , $f(x) \to 1$ .

Also $f(0) = |0| + |0-1| = 0 + 1 = 1$ .

Since left limit = right limit = function value, $f$ is continuous at $x=0$ .

Differentiability at $x=0$ The left-hand derivative uses the expression for $x<0$ : $f(x) = 1 - 2x$ , so $f'(x) = -2$ . Hence

$f'_-(0) = -2.$
The right-hand derivative uses the expression for $0 < x < 1$ : $f(x) = 1$ , so $f'(x) = 0$ . Hence
$f'_+(0) = 0.$
Since $-2 \neq 0$ , the left and right derivatives are different. Therefore $f$ is not differentiable at $x=0$ .

Watch out

A common mistake is to think that because $|x|$ alone is not differentiable at $0$ , the sum must also fail — which is true here, but you must check the actual slopes. If the slopes from both terms happened to match on both sides, the sum could become differentiable. Always compute the left and right derivatives explicitly. …
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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 X11 markMCQ
$y = f(x)$
(a) Statement 1 is true, Statement 2 is false
(b) Statement 1 is false, Statement 2 is true
(c) Both Statements 1 and 2 are true
(d) Both Statements 1 and 2 are false
›Reveal solutionSolution

One-sided derivatives / differentiability at a corner — correct option (a).

The left-hand derivative is the slope of the left branch from $(0,1)$ to $(1,0)$ : $\dfrac{0-1}{1-0} = -1$ , so Statement 1 is true. The right branch rises from $(1,0)$ to $(2,1)$ with slope $\dfrac{1-0}{2-1} = +1$ . Since the left and right derivatives differ ( $-1 \neq +1$ ), the …
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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
$f(x) = |x|$
(a) continuous but not differentiable
(b) differentiable but not continuous
(c) continuous and differentiable
(d) neither continuous nor differentiable
›Reveal solutionSolution

|x| has a 'corner' at x = 0 — no jump (continuous) but a sharp change in slope (not differentiable).

Continuity: $\lim_{x\to0^-}|x| = 0$ , $\lim_{x\to0^+}|x|=0$ , and $f(0)=0$ . All three agree, so f is continuous at x = 0.

Differentiability: Left-hand derivative: $\lim_{h\to0^-}\dfrac{|0+h|-|0|}{h} = \lim_{h\to0^-}\dfrac{-h}{h} = -1$ .

…

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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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Q.(a) Find kkk so that f(x)={x2−2x−3x+1,x≠−1k,x=−1f(x) = \begin{cases} \frac{x^2 - 2x - 3}{x+1}, & x \neq -1 \\ k, & x = -1 \end{cases}f(x)={x+1x2−2x−3​,k,​x=−1x=−1​ is continuous at x=−1x = -1x=−1.

Part (a)

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Q.(a) Find $k$ so that $f(x) = \begin{cases} \frac{x^2 - 2x - 3}{x+1}, & x \neq -1 \\ k, & x = -1 \end{cases}$ is continuous at $x = -1$ .

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