NCERT Exemplar Solutions Class 12 Maths Continuity and Differentiability — Short Answer Q1–24

NCERT Exemplar Problems, Class 12 Mathematics — Continuity and Differentiability, Short Answer questions 1 to 24. All 24 questions solved.

These questions follow one test again and again. A function $f$ is continuous at $x=a$ when the left-hand limit, the right-hand limit and $f(a)$ are all equal. It is differentiable at $a$ when, in addition, the left-hand and right-hand derivatives agree. Questions 1 to 19 are about continuity, Questions 20 to 23 ask about differentiability at a corner, and Question 24 uses the definition of the derivative on a functional equation.

Key insight. Always compute three numbers: the left-hand limit, the right-hand limit and the value at the point. Most answers here fall out of simplifying the expression first (cancelling a common factor, using a standard limit such as $\dfrac{1-\cos 2x}{x^2}\to 2$, or squeezing between bounds).

Question 1

Examine the continuity of the function $f(x) = x^3 + 2x^2 – 1$ at $x = 1$.

Solution. $f$ is a polynomial, and the limit of a polynomial at a point equals its value there. $\lim_{x\to1} f(x) = 1 + 2 – 1 = 2 = f(1)$.

Question 2

Find whether $f(x) = \begin{cases} 3x+5, & x \ge 2 \\ x^2, & x < 2 \end{cases}$ is continuous or discontinuous at $x = 2$.

Solution. Left-hand limit: $\lim_{x\to2^-} x^2 = 4$. Right-hand limit: $\lim_{x\to2^+}(3x+5) = 11$. The two sides do not meet, so the limit does not exist.

Question 3

Find whether $f(x) = \begin{cases} \dfrac{1-\cos 2x}{x^2}, & x \ne 0 \\ 5, & x = 0 \end{cases}$ is continuous or discontinuous at $x = 0$.

Solution. Use $1 – \cos 2x = 2\sin^2 x$. Then $\dfrac{1-\cos 2x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2 \to 2$. The limit is $2$, but $f(0) = 5$, so the limit and the value differ.

Question 4

Find whether $f(x) = \begin{cases} \dfrac{2x^2-3x-2}{x-2}, & x \ne 2 \\ 5, & x = 2 \end{cases}$ is continuous or discontinuous at $x = 2$.

Solution. Factorise: $2x^2 – 3x – 2 = (2x+1)(x-2)$. For $x \ne 2$ the fraction is $2x+1$, so the limit as $x\to2$ is $5$. Since $f(2) = 5$ as well, the function is continuous.

Question 5

Find whether $f(x) = \begin{cases} \dfrac{|x-4|}{2(x-4)}, & x \ne 4 \\ 0, & x = 4 \end{cases}$ is continuous or discontinuous at $x = 4$.

Solution. For $x > 4$, $|x-4| = x-4$ and the fraction is $\tfrac12$. For $x < 4$, $|x-4| = -(x-4)$ and the fraction is $-\tfrac12$. So the left-hand limit is $-\tfrac12$ and the right-hand limit is $\tfrac12$. They differ, so there is a jump.

Question 6

Find whether $f(x) = \begin{cases} |x|\cos\dfrac1x, & x \ne 0 \\ 0, & x = 0 \end{cases}$ is continuous or discontinuous at $x = 0$.

Solution. $\cos\frac1x$ oscillates but stays between $-1$ and $1$, so $-|x| \le |x|\cos\frac1x \le |x|$. Both bounds tend to $0$ as $x\to0$, so by the squeeze theorem the limit is $0 = f(0)$.

Question 7

Find whether $f(x) = \begin{cases} |x-a|\sin\dfrac{1}{x-a}, & x \ne a \\ 0, & x = a \end{cases}$ is continuous or discontinuous at $x = a$.

Solution. Since $\left|\sin\frac{1}{x-a}\right| \le 1$, we get $0 \le |x-a|\left|\sin\frac{1}{x-a}\right| \le |x-a|$. The right side tends to $0$ as $x\to a$, so the limit is $0 = f(a)$ by the squeeze theorem.

Question 8

Find whether $f(x) = \begin{cases} \dfrac{e^{1/x}}{1+e^{1/x}}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ is continuous or discontinuous at $x = 0$.

Solution. As $x\to0^+$, $e^{1/x}\to\infty$, and dividing top and bottom by $e^{1/x}$ gives $\dfrac{1}{e^{-1/x}+1}\to 1$. As $x\to0^-$, $e^{1/x}\to 0$, so the fraction tends to $\dfrac{0}{1+0} = 0$. The one-sided limits are $0$ and $1$, so the limit does not exist.

Question 9

Find whether $f(x) = \begin{cases} \dfrac{x^2}{2}, & 0 \le x \le 1 \\ 2x^2 – 3x + \dfrac32, & 1 < x \le 2 \end{cases}$ is continuous or discontinuous at $x = 1$.

Solution. Left-hand limit: $\frac{1}{2}$, which also equals $f(1)$. Right-hand limit: $2 – 3 + \frac32 = \frac12$. All three agree.

Question 10

Find whether $f(x) = |x| + |x-1|$ is continuous or discontinuous at $x = 1$.

Solution. $|x|$ and $|x-1|$ are both continuous everywhere (each is a composition of the continuous modulus function with a polynomial), and a sum of continuous functions is continuous.

Question 11

Find $k$ so that $f(x) = \begin{cases} 3x-8, & x \le 5 \\ 2k, & x > 5 \end{cases}$ is continuous at $x = 5$.

Solution. The left-hand limit equals $f(5) = 3(5) – 8 = 7$. The right-hand limit is the constant $2k$. Continuity needs $2k = 7$.

Question 12

Find $k$ so that $f(x) = \begin{cases} \dfrac{2^{x+2}-16}{4^x-16}, & x \ne 2 \\ k, & x = 2 \end{cases}$ is continuous at $x = 2$.

Solution. Write $2^{x+2} – 16 = 4(2^x – 4)$ and $4^x – 16 = (2^x-4)(2^x+4)$. For $x \ne 2$ the factor $2^x – 4$ cancels, leaving $\dfrac{4}{2^x+4}$. As $x\to2$ this tends to $\dfrac{4}{8} = \dfrac12$, so $k$ must equal $\frac12$.

Question 13

Find $k$ so that $f(x) = \begin{cases} \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & -1 \le x < 0 \\ \dfrac{2x+1}{x-1}, & 0 \le x \le 1 \end{cases}$ is continuous at $x = 0$.

Solution. The value is $f(0) = \dfrac{1}{-1} = -1$, and the right-hand limit is also $-1$. For the left-hand limit, rationalise: $\dfrac{(1+kx)-(1-kx)}{x\left(\sqrt{1+kx}+\sqrt{1-kx}\right)} = \dfrac{2k}{\sqrt{1+kx}+\sqrt{1-kx}} \to \dfrac{2k}{2} = k$. Setting $k = -1$ makes all three agree.

Question 14

Find $k$ so that $f(x) = \begin{cases} \dfrac{1-\cos kx}{x\sin x}, & x \ne 0 \\ \dfrac12, & x = 0 \end{cases}$ is continuous at $x = 0$.

Solution. Write $1 – \cos kx = 2\sin^2\frac{kx}{2}$. Then $$\frac{1-\cos kx}{x\sin x} = \frac{2\sin^2(kx/2)}{(kx/2)^2}\cdot\frac{k^2x^2}{4}\cdot\frac{1}{x\sin x} \to 2\cdot\frac{k^2}{4}\cdot 1 = \frac{k^2}{2}.$$ Continuity needs $\frac{k^2}{2} = \frac12$, so $k^2 = 1$.

Question 15

Prove that $f(x) = \begin{cases} \dfrac{x}{|x|+2x^2}, & x \ne 0 \\ k, & x = 0 \end{cases}$ remains discontinuous at $x = 0$, regardless of the choice of $k$.

Solution. For $x > 0$, $|x| = x$ and $f(x) = \dfrac{x}{x+2x^2} = \dfrac{1}{1+2x} \to 1$. For $x < 0$, $|x| = -x$ and $f(x) = \dfrac{x}{-x+2x^2} = \dfrac{1}{-1+2x} \to -1$. The right-hand limit is $1$ and the left-hand limit is $-1$, so $\lim_{x\to0}f(x)$ does not exist. Continuity at $0$ needs a limit equal to $f(0) = k$, and no single value of $k$ can equal both $1$ and $-1$.

Question 16

Find the values of $a$ and $b$ such that $f(x) = \begin{cases} \dfrac{x-4}{|x-4|} + a, & x < 4 \\ a+b, & x = 4 \\ \dfrac{x-4}{|x-4|} + b, & x > 4 \end{cases}$ is continuous at $x = 4$.

Solution. For $x < 4$, $\frac{x-4}{|x-4|} = -1$, so the left-hand limit is $a – 1$. For $x > 4$ it is $+1$, so the right-hand limit is $b + 1$. Continuity needs $a – 1 = a + b = b + 1$. The first equality gives $b = -1$; the second gives $a = 1$.

Question 17

Given $f(x) = \dfrac{1}{x+2}$, find the points of discontinuity of the composite function $y = f(f(x))$.

Solution. First, $f$ itself is undefined at $x = -2$, so $f(f(x))$ is not defined there. Next, $$f(f(x)) = \frac{1}{\frac{1}{x+2}+2} = \frac{x+2}{1+2(x+2)} = \frac{x+2}{2x+5},$$ which is undefined when $2x + 5 = 0$, that is $x = -\frac52$. Everywhere else the composite is a quotient of continuous functions.

Question 18

Find all points of discontinuity of $f(t) = \dfrac{1}{t^2+t-2}$, where $t = \dfrac{1}{x-1}$.

Solution. The denominator factorises as $t^2 + t – 2 = (t+2)(t-1)$, so $f$ breaks down when $t = 1$ or $t = -2$. Also $t = \frac{1}{x-1}$ itself is undefined at $x = 1$. Now $t = 1$ gives $x – 1 = 1$, so $x = 2$; and $t = -2$ gives $x – 1 = -\frac12$, so $x = \frac12$.

Question 19

Show that $f(x) = |\sin x + \cos x|$ is continuous at $x = \pi$.

Solution. The functions $\sin x$ and $\cos x$ are continuous everywhere, so $g(x) = \sin x + \cos x$ is continuous. The modulus function is continuous, and a composition of continuous functions is continuous. Hence $f = |g|$ is continuous at every point, in particular at $\pi$. Directly: $\lim_{x\to\pi} f(x) = |\sin\pi + \cos\pi| = |-1| = 1 = f(\pi)$.

Question 20

Examine the differentiability of $f(x) = \begin{cases} x[x], & 0 \le x < 2 \\ (x-1)x, & 2 \le x < 3 \end{cases}$ at $x = 2$.

Solution. Just left of $2$, $[x] = 1$, so $f(x) = x$ and $f(x)\to 2$; also $f(2) = (2-1)(2) = 2$, and just right of $2$, $f(x) = x^2 – x \to 2$. So $f$ is continuous at $2$ with $f(2) = 2$. Now the one-sided derivatives:

• Left: $f(x) = x$, so $\text{LHD} = 1$.
• Right: $f(x) = x^2 – x$, so $\text{RHD} = 2(2) – 1 = 3$.

Since $1 \ne 3$, the function has a corner at $x=2$.

Question 21

Examine the differentiability of $f(x) = \begin{cases} x^2\sin\dfrac1x, & x \ne 0 \\ 0, & x = 0 \end{cases}$ at $x = 0$.

Solution. Use the definition: $$f'(0) = \lim_{h\to0}\frac{f(h)-f(0)}{h} = \lim_{h\to0}\frac{h^2\sin(1/h)}{h} = \lim_{h\to0} h\sin\frac1h = 0,$$ because $\left|h\sin\frac1h\right| \le |h|\to0$ (squeeze). The same limit from both sides is $0$.

Question 22

Examine the differentiability of $f(x) = \begin{cases} 1+x, & x \le 2 \\ 5-x, & x > 2 \end{cases}$ at $x = 2$.

Solution. The function is continuous at $2$: the left value is $1 + 2 = 3$ and the right limit is $5 – 2 = 3$, with $f(2) = 3$. The slope on the left is $\frac{d}{dx}(1+x) = 1$ and on the right is $\frac{d}{dx}(5-x) = -1$. The one-sided derivatives disagree.

Question 23

Show that $f(x) = |x-5|$ is continuous but not differentiable at $x = 5$.

Solution. Continuity: $\lim_{x\to5}|x-5| = 0 = f(5)$. Differentiability: $\dfrac{f(5+h)-f(5)}{h} = \dfrac{|h|}{h}$, which is $-1$ for $h < 0$ and $+1$ for $h > 0$. The left derivative is $-1$ and the right derivative is $1$, so the derivative does not exist at $5$.

Question 24

A function $f:\mathbb{R}\to\mathbb{R}$ satisfies $f(x+y) = f(x)f(y)$ for all $x, y\in\mathbb{R}$, and $f(x)\ne0$. Suppose $f$ is differentiable at $x = 0$ and $f'(0) = 2$. Prove that $f'(x) = 2f(x)$.

Solution. First find $f(0)$. Putting $x = y = 0$ gives $f(0) = f(0)^2$, so $f(0) = 0$ or $1$. Since $f$ is never $0$, $f(0) = 1$. Now for any $x$, $$f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to0}\frac{f(x)f(h)-f(x)}{h} = f(x)\lim_{h\to0}\frac{f(h)-f(0)}{h} = f(x)\,f'(0).$$ The last limit is exactly the derivative at $0$, which is given as $2$. Therefore $f'(x) = 2f(x)$.

Common mistakes

• Comparing only the limit with the value, or only the two sides. Continuity needs the left limit, the right limit and $f(a)$ all equal. Questions 3 and 4 differ only in whether the limit matches the stated value.
• Forgetting the points where the inner function is undefined. In a composite such as Questions 17 and 18, the inner function’s own discontinuity ($x=-2$, $x=1$) counts alongside the zeros of the outer denominator.
• Using $\frac{|h|}{h}=1$. The ratio is $-1$ on the left and $+1$ on the right, which is exactly why $|x-5|$ fails to be differentiable (Question 23) and why one-sided limits matter in Questions 5 and 8.
• Assuming continuity gives differentiability. Questions 20, 22 and 23 are continuous at the point and still have a corner there.
• Skipping $f(0)=1$ in Question 24. Without ruling out $f(0)=0$ using $f\ne0$, the step $f(h)-1$ is not justified.

Practise next

• Exemplar Short Answer Q25–43, the differentiation questions that follow these.
• Exemplar Short Answer Q44–64, parametric and implicit differentiation.
• NCERT textbook Exercise 5.1 for continuity at a point and finding $k$.
• NCERT textbook Exercise 5.2 for continuity of composite functions and differentiability.