Class 12 Continuity & Diff PYQs

Q1 2024

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If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{\sin 5x}{3x}$ for $x \neq 0$ and $f(0) = k$, find $k$. 1 Mark

$k = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 5x}{3x}$.

Multiply and divide by 5: $\lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{5}{3}$.

$= 1 \cdot \frac{5}{3} = \frac{5}{3}$.

k = 5/3

Q2 2023

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Find the value of $k$ so that $f(x) = \begin{cases} kx + 1 & x \le 5 \\ 3x - 5 & x > 5 \end{cases}$ is continuous at $x = 5$. 2 Marks

LHL $= f(5) = 5k + 1$.

RHL $= \lim_{x \to 5^+} (3x - 5) = 3(5) - 5 = 10$.

For continuity, $LHL = RHL \Rightarrow 5k + 1 = 10 \Rightarrow 5k = 9 \Rightarrow k = 9/5$.

k = 9/5

Q3 2022 (Term 1 MCQ)

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The function $f(x) = [x]$, greatest integer function, is continuous at: 1 Mark

(a)4

(b)-2

(c)1

(d)1.5

The greatest integer function $[x]$ is discontinuous at all integral points.

4, -2, and 1 are integers.

1.5 is a non-integer point, so the function is continuous at $x=1.5$.

1.5

Q4 2020

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Find the value of $k$ for which $f(x) = \begin{cases} \frac{\sin^2(\lambda x)}{x^2} & x \neq 0 \\ 1 & x = 0 \end{cases}$ is continuous at $x = 0$. [Assume $\lambda$ is related to $k$ or find $\lambda$ in terms of $k$]

$\lim_{x \to 0} \frac{\sin^2(\lambda x)}{x^2} = \lim_{x \to 0} \left(\frac{\sin(\lambda x)}{\lambda x} \cdot \lambda\right)^2 = \lambda^2$.

For continuity, $\lambda^2 = f(0) = 1 \Rightarrow \lambda = \pm 1$.

λ = ±1

Q5 2019

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Find the value of $k$ so that $f(x) = \begin{cases} kx^2 & x \le 2 \\ 3 & x > 2 \end{cases}$ is continuous at $x = 2$. 2 Marks

LHL $= f(2) = k(2)^2 = 4k$.

RHL $= \lim_{x \to 2^+} (3) = 3$.

For continuity, $4k = 3 \Rightarrow k = 3/4$.

k = 3/4

Q6 2018

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Find the value of $k$ so that $f(x) = \begin{cases} k(x^2 - 2x) & x < 0 \\ \cos x & x \ge 0 \end{cases}$ is continuous at $x = 0$. 2 Marks

LHL $= \lim_{x \to 0^-} k(x^2 - 2x) = k(0) = 0$.

RHL $= f(0) = \cos 0 = 1$.

Since $0 \neq 1$, there is no value of $k$ for which the function is continuous.

No value of k exists

Q7 2017 (Delhi)

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Determine $k$ so that $f(x) = \begin{cases} \frac{(k-1)(x-8)}{x^2-64} & x \neq 8 \\ 2 & x = 8 \end{cases}$ is continuous at $x = 8$. 2 Marks

$\lim_{x \to 8} \frac{(k-1)(x-8)}{(x-8)(x+8)} = \lim_{x \to 8} \frac{k-1}{x+8} = \frac{k-1}{16}$.

For continuity, $\frac{k-1}{16} = 2 \Rightarrow k-1 = 32 \Rightarrow k = 33$.

k = 33

Q8 2016 (Delhi)

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Find $p$ and $q$ for which $f(x) = \begin{cases} \frac{1-\sin^3 x}{3\cos^2 x} & x < \pi/2 \\ p & x = \pi/2 \\ \frac{q(1-\sin x)}{(\pi-2x)^2} & x > \pi/2 \end{cases}$ is continuous at $x = \pi/2$. 4 Marks

LHL $= \lim_{x \to \pi/2^-} \frac{(1-\sin x)(1+\sin x + \sin^2 x)}{3(1-\sin^2 x)} = \lim_{x \to \pi/2^-} \frac{1+\sin x + \sin^2 x}{3(1+\sin x)} = \frac{1+1+1}{3(1+1)} = \frac{1}{2}$.

So $p = 1/2$.

RHL $= \lim_{h \to 0} \frac{q(1-\sin(\pi/2+h))}{(\pi-2(\pi/2+h))^2} = \lim_{h \to 0} \frac{q(1-\cos h)}{4h^2}$.

$= \lim_{h \to 0} \frac{q(2\sin^2 h/2)}{4h^2} = \frac{2q}{4} \cdot \frac{1}{4} = \frac{q}{8}$.

$"q/8 = 1/2 \Rightarrow q = 4$.

p = 1/2, q = 4

Q9 2025 Board

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Assertion (A): $f(x) = \begin{cases} \frac{x^3 - 8}{x - 5} & x < 5 \\ k & x = 5 \end{cases}$ is continuous at $x = 5$ for $k = 5/2$. Reason (R): For a function $f$ to be continuous at $x = a$, $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$. 1 Mark

(a)Both (A) and (R) are true and (R) is the correct explanation of (A).

(b)Both (A) and (R) are true but (R) is not the correct explanation of (A).

(c)(A) is true but (R) is false.

(d)(A) is false but (R) is true.

Reason (R) is the standard definition of continuity. So (R) is true.

For Assertion (A): $\lim_{x \to 5} \frac{x^3 - 8}{x - 5}$ is of form $117/0$, which is infinite. The limit does not exist.

Thus (A) is false regardless of the value of $k$.

(d) (A) is false but (R) is true.

Q10 2017 (All India)

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Determine 'k' for which $f(x) = \begin{cases} \frac{1-\cos kx}{x \sin x} & x \neq 0 \\ 1/2 & x = 0 \end{cases}$ is continuous at $x = 0$. 3 Marks

$\lim_{x \to 0} \frac{1-\cos kx}{x \sin x} = \lim_{x \to 0} \frac{2\sin^2(kx/2)}{x \sin x}$.

Multiply and divide by terms to use $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$= \lim_{x \to 0} \frac{2 \left(\frac{\sin(kx/2)}{kx/2}\right)^2 \cdot (k^2x^2/4)}{x \left(\frac{\sin x}{x}\right) \cdot x}$.

$= \frac{2 k^2}{4} = \frac{k^2}{2}$.

For continuity, $k^2/2 = 1/2 \Rightarrow k^2 = 1 \Rightarrow k = \pm 1$.

k = ±1

Q11 2015 (All India)

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If $f$ is continuous at $x = 0$, where $f(x) = \frac{\sin ax}{\sin bx}, x \neq 0$, find $f(0)$. 1 Mark

$f(0) = \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \lim_{x \to 0} \frac{\frac{\sin ax}{ax} \cdot ax}{\frac{\sin bx}{bx} \cdot bx}$.

$= \frac{a}{b} \cdot \frac{\lim_{x \to 0} (\sin ax/ax)}{\lim_{x \to 0} (\sin bx/bx)} = \frac{a}{b} \cdot 1 = a/b$.

a/b

Continuity Basics PYQs