Class 12 Maths NCERT Solutions Chapter 5 Ex 5.1 – Continuity | Boundless Maths
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Ex 5.1 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 5 Ex 5.1 – Continuity

This Class 12 Maths NCERT Solutions Chapter 5 Ex 5.1 page covers all 34 questions, solved step-by-step, showing exactly how CBSE awards method marks for continuity proofs — from the formal definition (left hand limit equals right hand limit equals the function's value) to the algebra of continuous functions that lets you build new continuous functions from ones you already know.

This is the longest exercise in the chapter, and it builds in four clear stages. Questions 1–5 get you comfortable with the definition itself — proving a function is continuous at a given point by checking all three conditions directly, and testing continuity of a piecewise function class 12 style at more than one point. Questions 6–19 are the bulk of the exercise: finding all points of discontinuity for a series of piecewise functions, including the classic |x|/x case and the greatest integer function continuity proof that shows up in almost every CBSE sample paper. Questions 20–25 shift to continuity of trigonometric functions — sums, differences and products of sin x and cos x, plus the secant, cosecant and cotangent functions — using the algebra of continuous functions instead of checking limits from scratch every time. The final block, 26–34, is exam-favourite territory: finding the value of k in continuity problems (or a, b and λ) that makes a piecewise function continuous at a boundary point, a question type that appears in nearly every board paper's 2-mark or 3-mark section, closing with proofs that \cos(x^2), |\cos x| and \sin|x| are continuous everywhere.

34Questions
Easy–HardDifficulty Mix
2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 5 Ex 5.1 — All 34 Questions

1

Prove that the function f(x) = 5x - 3 is continuous at x = 0, at x = -3 and at x = 5.

Easy +
Solution

f(x)=5x-3 is a polynomial function, so it is defined at every real number and its limit at any point equals its value there.

At x=0: \lim_{x\to 0} f(x) = 5(0)-3 = -3 = f(0).

At x=-3: \lim_{x\to -3} f(x) = 5(-3)-3 = -18 = f(-3).

At x=5: \lim_{x\to 5} f(x) = 5(5)-3 = 22 = f(5).

In each case the limit equals the function value, so f is continuous at x = 0, x = -3 and x = 5.
2

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

Easy +
Solution

f(3) = 2(3)^2 - 1 = 17.

\lim_{x\to 3} f(x) = 2(3)^2 - 1 = 17.

Since the limit equals f(3), the function is continuous at x = 3.
3

Examine the following functions for continuity:
(a) f(x)=x-5
(b) f(x)=\dfrac{1}{x-5},\ x\neq 5
(c) f(x)=\dfrac{x^2-25}{x+5},\ x\neq -5
(d) f(x)=|x-5|

Medium +
Solution

(a) A polynomial function — continuous at every real number.

(b) A rational function, defined for all x \neq 5 — continuous at every point of its domain.

(c) For x \neq -5, f(x) = \dfrac{(x-5)(x+5)}{x+5} = x-5, a polynomial on its domain — continuous at every point of its domain.

(d) The modulus of a polynomial — composite of two continuous functions, so continuous everywhere.

Answer: all four functions are continuous on their respective domains.
4

Prove that the function f(x)=x^n is continuous at x=n, where n is a positive integer.

Easy +
Solution

f(x)=x^n is a polynomial function, so it is defined for every real number and every polynomial function is continuous.

In particular, \lim_{x\to n} x^n = n^n = f(n).

Hence f is continuous at x = n.
5

Is the function f defined by f(x)=\begin{cases}x, & x\le 1\\5, & x \gt 1\end{cases} continuous at x=0? At x=1? At x=2?

Medium +
Solution

At x=0: this lies in the region x\le 1 where f(x)=x, a polynomial — continuous.

At x=1: LHL =\lim_{x\to 1^-} x = 1. RHL =\lim_{x\to 1^+} 5 = 5. Since LHL ≠ RHL, f is not continuous at x = 1.

At x=2: this lies in the region x \gt 1 where f(x)=5, a constant function — continuous.

Answer: f is continuous at x = 0 and x = 2, but discontinuous at x = 1.
6

Find all points of discontinuity of f(x)=\begin{cases}2x+3, & x\le 2\\2x-3, & x \gt 2\end{cases}

Medium +
Solution

Each piece is a polynomial, so f is continuous everywhere except possibly at x=2.

At x=2: LHL =2(2)+3=7. RHL =2(2)-3=1.

Answer: LHL ≠ RHL, so x = 2 is the only point of discontinuity.
7

Find all points of discontinuity of f(x)=\begin{cases}|x|+3, & x\le -3\\-2x, & -3 \lt x \lt 3\\6x+2, & x\ge 3\end{cases}

Hard +
Solution

At x=-3: LHL =|-3|+3=6. RHL =-2(-3)=6. f(-3)=6. All equal — continuous at x = -3.

At x=3: LHL =-2(3)=-6. RHL =6(3)+2=20. LHL ≠ RHL — discontinuous at x = 3.

Answer: x = 3 is the only point of discontinuity.
8

Find all points of discontinuity of f(x)=\begin{cases}\dfrac{|x|}{x}, & x\neq 0\\0, & x=0\end{cases}

Medium +
Solution

For x \gt 0, \frac{|x|}{x}=1; for x \lt 0, \frac{|x|}{x}=-1.

At x=0: LHL =-1, RHL =1.

Answer: LHL ≠ RHL, so x = 0 is the only point of discontinuity.
9

Find all points of discontinuity of f(x)=\begin{cases}\dfrac{x}{|x|}, & x \lt 0\\-1, & x\ge 0\end{cases}

Medium +
Solution

For x \lt 0, \frac{x}{|x|}=\frac{x}{-x}=-1.

So f(x)=-1 for every real x — for x < 0 it comes from the first rule, and for x ≥ 0 it's given directly as −1. f is really the constant function f(x)=-1.

Answer: f is continuous everywhere; there are no points of discontinuity.
10

Find all points of discontinuity of f(x)=\begin{cases}x+1, & x\ge 1\\x^2+1, & x \lt 1\end{cases}

Medium +
Solution

At x=1: LHL =1^2+1=2. RHL =1+1=2. f(1)=2.

Answer: LHL = RHL = f(1), so f is continuous everywhere — no points of discontinuity.
11

Find all points of discontinuity of f(x)=\begin{cases}x^3-3, & x\le 2\\x^2+1, & x \gt 2\end{cases}

Medium +
Solution

At x=2: LHL =2^3-3=5. RHL =2^2+1=5. f(2)=5.

Answer: LHL = RHL = f(2), so f is continuous everywhere — no points of discontinuity.
12

Find all points of discontinuity of f(x)=\begin{cases}x^{10}-1, & x\le 1\\x^2, & x \gt 1\end{cases}

Medium +
Solution

At x=1: LHL =1^{10}-1=0. RHL =1^2=1.

Answer: LHL ≠ RHL, so x = 1 is the only point of discontinuity.
13

Is the function defined by f(x)=\begin{cases}x+5, & x\le 1\\x-5, & x \gt 1\end{cases} a continuous function?

Easy +
Solution

At x=1: LHL =1+5=6. RHL =1-5=-4.

Answer: LHL ≠ RHL, so f is not continuous — it is discontinuous at x = 1.
14

Discuss the continuity of f(x)=\begin{cases}3, & 0\le x\le 1\\4, & 1 \lt x \lt 3\\5, & 3\le x\le 10\end{cases}

Medium +
Solution

At x=1: LHL =3, RHL =4 — not equal, discontinuous.

At x=3: LHL =4, RHL =5 — not equal, discontinuous.

On each open interval the function is constant, hence continuous there.

Answer: f is discontinuous at x = 1 and x = 3, and continuous at every other point.
15

Discuss the continuity of f(x)=\begin{cases}2x, & x \lt 0\\0, & 0\le x\le 1\\4x, & x \gt 1\end{cases}

Medium +
Solution

At x=0: LHL =2(0)=0, RHL =0, f(0)=0 — continuous.

At x=1: LHL =0, RHL =4(1)=4 — not equal, discontinuous.

Answer: f is continuous everywhere except at x = 1.
16

Discuss the continuity of f(x)=\begin{cases}-2, & x\le -1\\2x, & -1 \lt x\le 1\\2, & x \gt 1\end{cases}

Medium +
Solution

At x=-1: LHL =-2, RHL =2(-1)=-2, f(-1)=-2 — continuous.

At x=1: LHL =2(1)=2, RHL =2, f(1)=2 — continuous.

Answer: f is continuous at every real number.
17

Find the relationship between a and b so that f(x)=\begin{cases}ax+1, & x\le 3\\bx+3, & x \gt 3\end{cases} is continuous at x=3.

Medium +
Solution

LHL at x=3: 3a+1. RHL: 3b+3. f(3)=3a+1.

For continuity, 3a+1=3b+3, i.e. 3a-3b=2.

Answer: the required relationship is 3a − 3b = 2, i.e. a = b + 2/3.
18

For what value of \lambda is the function f(x)=\begin{cases}\lambda(x^2-2x), & x\le 0\\4x+1, & x \gt 0\end{cases} continuous at x=0? What about continuity at x=1?

Hard +
Solution

LHL at x=0: \lambda(0-0)=0. RHL: 4(0)+1=1. f(0)=0.

For continuity we'd need 0=1, which is impossible — so no value of λ makes f continuous at x = 0.

At x=1, f is given by the polynomial 4x+1 for all x near 1 (since x = 1 > 0), so f is continuous at x = 1 for every value of λ.

Answer: no value of λ works at x = 0; f is continuous at x = 1 for every λ.
19

Show that the function defined by g(x)=x-[x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Hard +
Solution

Let n be any integer. As x\to n^-, [x]=n-1, so g(x)\to n-(n-1)=1.

As x\to n^+, [x]=n, so g(x)\to n-n=0.

LHL (1) ≠ RHL (0) at every integer n.

Answer: g is discontinuous at every integer, proved.
20

Is the function defined by f(x)=x^2-\sin x+5 continuous at x=\pi?

Easy +
Solution

f is a sum/difference of continuous functions (a polynomial, sine, and a constant), so f is continuous everywhere.

In particular, \lim_{x\to \pi} f(x) = \pi^2-\sin\pi+5 = \pi^2+5 = f(\pi).

Answer: yes, f is continuous at x = π.
21

Discuss the continuity of:
(a) f(x)=\sin x+\cos x
(b) f(x)=\sin x-\cos x
(c) f(x)=\sin x\cdot \cos x

Easy +
Solution

Sine and cosine are both continuous functions on \mathbb{R}.

(a), (b) Sum and difference of continuous functions are continuous — both are continuous everywhere.

(c) Product of continuous functions is continuous — continuous everywhere.

Answer: all three functions are continuous at every real number.
22

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Medium +
Solution

Cosine is continuous at every real number (proved analogously to sine).

Cosecant =\dfrac{1}{\sin x} is a quotient of continuous functions, so continuous wherever \sin x\neq 0, i.e. for all x\neq n\pi.

Secant =\dfrac{1}{\cos x} is continuous wherever \cos x\neq 0, i.e. for all x\neq (2n+1)\frac{\pi}{2}.

Cotangent =\dfrac{\cos x}{\sin x} is continuous wherever \sin x\neq 0, i.e. for all x\neq n\pi.

Answer: each function is continuous on its full domain.
23

Find all points of discontinuity of f(x)=\begin{cases}\dfrac{\sin x}{x}, & x \lt 0\\x+1, & x\ge 0\end{cases}

Hard +
Solution

At x=0: LHL =\lim_{x\to 0^-}\dfrac{\sin x}{x}=1 (standard limit). RHL =0+1=1. f(0)=1.

Elsewhere, each piece is continuous on its own domain.

Answer: all three values coincide at x = 0, so f is continuous everywhere — there are no points of discontinuity.
24

Determine if f defined by f(x)=\begin{cases}x^2\sin\dfrac{1}{x}, & x\neq 0\\0, & x=0\end{cases} is a continuous function.

Hard +
Solution

Since -1\le \sin\frac{1}{x}\le 1, we have -x^2 \le x^2\sin\frac{1}{x}\le x^2.

As x\to 0, both bounds \to 0, so by the squeeze principle \lim_{x\to 0} x^2\sin\frac{1}{x}=0.

This equals f(0)=0.

Answer: yes, f is continuous everywhere, including at x = 0.
25

Examine the continuity of f, where f(x)=\begin{cases}\sin x-\cos x, & x\neq 0\\-1, & x=0\end{cases}

Medium +
Solution

\lim_{x\to 0}(\sin x - \cos x) = \sin 0 - \cos 0 = 0-1=-1.

This equals f(0)=-1.

Answer: f is continuous at x = 0, and hence continuous everywhere.
26

Find k so that f(x)=\begin{cases}\dfrac{k\cos x}{\pi-2x}, & x\neq \frac{\pi}{2}\\3, & x=\frac{\pi}{2}\end{cases} is continuous at x=\frac{\pi}{2}.

Hard +
Solution

Put x=\frac{\pi}{2}+h, so h\to 0 as x\to \frac{\pi}{2}. Then \cos x = \cos(\frac{\pi}{2}+h)=-\sin h and \pi-2x = -2h.

\lim_{h\to 0}\dfrac{k(-\sin h)}{-2h} = \dfrac{k}{2}\lim_{h\to 0}\dfrac{\sin h}{h} = \dfrac{k}{2}.

For continuity, \dfrac{k}{2}=3.

Answer: k = 6.
27

Find k so that f(x)=\begin{cases}kx^2, & x\le 2\\3, & x \gt 2\end{cases} is continuous at x=2.

Easy +
Solution

LHL =k(2)^2=4k. RHL =3.

Set 4k=3.

Answer: k = 3/4.
28

Find k so that f(x)=\begin{cases}kx+1, & x\le \pi\\\cos x, & x \gt \pi\end{cases} is continuous at x=\pi.

Easy +
Solution

LHL =k\pi+1. RHL =\cos\pi=-1.

Set k\pi+1=-1 \Rightarrow k\pi=-2.

Answer: k = −2/π.
29

Find k so that f(x)=\begin{cases}kx+1, & x\le 5\\3x-5, & x \gt 5\end{cases} is continuous at x=5.

Easy +
Solution

LHL =5k+1. RHL =3(5)-5=10.

Set 5k+1=10 \Rightarrow 5k=9.

Answer: k = 9/5.
30

Find the values of a and b such that f(x)=\begin{cases}5, & x\le 2\\ax+b, & 2 \lt x \lt 10\\21, & x\ge 10\end{cases} is a continuous function.

Hard +
Solution

Continuity at x=2: 5=2a+b …(i)

Continuity at x=10: 10a+b=21 …(ii)

Subtracting (i) from (ii): 8a=16 \Rightarrow a=2. Substituting back: b=5-2(2)=1.

Answer: a = 2, b = 1.
31

Show that the function defined by f(x)=\cos(x^2) is a continuous function.

Easy +
Solution

f is the composite g\circ h where g(x)=\cos x and h(x)=x^2, both continuous everywhere.

Answer: a composite of continuous functions is continuous, so f is continuous everywhere.
32

Show that the function defined by f(x)=|\cos x| is a continuous function.

Easy +
Solution

f is the composite g\circ h where g(x)=|x| and h(x)=\cos x, both continuous everywhere.

Answer: a composite of continuous functions is continuous, so f is continuous everywhere.
33

Examine that \sin|x| is a continuous function.

Easy +
Solution

f is the composite g\circ h where g(x)=\sin x and h(x)=|x|, both continuous everywhere.

Answer: a composite of continuous functions is continuous, so sin|x| is continuous everywhere.
34

Find all the points of discontinuity of f defined by f(x)=|x|-|x+1|.

Medium +
Solution

|x| and |x+1| are each composites of continuous functions, hence continuous everywhere.

The difference of two continuous functions is continuous.

Answer: f is continuous everywhere; there are no points of discontinuity.

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Common Questions

FAQs — Class 12 Maths NCERT Solutions Chapter 5 Ex 5.1

How many questions are there in Exercise 5.1?

Exercise 5.1 has 34 questions on continuity — proving and checking continuity at a point, finding points of discontinuity for piecewise functions, and finding unknown constants that make a function continuous.

What concept does Exercise 5.1 test?

It tests the formal definition of continuity — that the left hand limit, right hand limit, and the value of the function at a point must all be equal — along with the algebra of continuous functions (sums, differences, products, quotients and composites).

Where can I find the official NCERT textbook for this exercise?

Exercise 5.1 is from Chapter 5, Continuity and Differentiability, in the NCERT Class 12 Mathematics textbook (Part I), published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the questions exactly as they appear there.

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