Class 12 Maths Continuity and Differentiability NCERT Solutions EX 5.1

Class 12 Maths Continuity and Differentiability is an important chapter in Class 12 Maths that forms the base of calculus. In this chapter, students learn about continuous functions, differentiability, and their applications. Here, you will find NCERT solutions, solved examples, and important questions explained in a simple and clear way to help you understand concepts and score better in exams.

Class 12 Maths Continuity and Differentiability (Solved Questions)

Q1. Prove that the function f(x)=5x−3 is continuous at x=0,−3 and 5.

Solution:
A function $f(x)$f(x) is continuous at $x = a$x=a if
L.H.L = R.H.L = f(a)

Given function:
$f(x) = 5x – 3$

This is a linear polynomial function. We know that all polynomial functions are continuous for all real values of x.

Now checking at given points:

At $x = 0$
$f(0) = 5(0) – 3 = -3$
L.H.L = R.H.L = -3 ⇒ Continuous

At $x = -3$
$f(-3) = 5(-3) – 3 = -15 – 3 = -18$
L.H.L = R.H.L = -18 ⇒ Continuous

At $x = 5$
$f(5) = 5(5) – 3 = 25 – 3 = 22$
L.H.L = R.H.L = 22 ⇒ Continuous

Hence, the function is continuous at x=0,−3 and 5.

Q2. Examine the continuity of the function $f(x) = 2x^2 – 1$ at x=3

Solution:
Given function:
$f(x) = 2x^2 – 1$

This is a polynomial function, so it is continuous for all real numbers.

At $x = 3$x=3:
$f(3) = 2(3)^2 – 1 = 2(9) – 1 = 18 – 1 = 17$f(3)=2(3)2−1=2(9)−1=18−1=17

L.H.L = R.H.L = 17

Hence, the function is continuous at x=3

Q3. Examine the following functions for continuity

(a) f(x)=x−5

This is a polynomial function.
Polynomial functions are continuous everywhere.

Hence, continuous for all real x.

(b) f(x)=1/x−5,  x≠5

This is a rational function.

The denominator becomes zero at $x = 5$, so the function is not defined there.

Hence:

• Continuous for all $x \ne 5$
• Discontinuous at $x = 5$

(c) f(x)=x2/x+5,  x≠−5

This is also a rational function.

Denominator becomes zero at $x = -5$

Hence:

• Continuous for all $x \ne -5$
• Discontinuous at $x = -5$

(d) f(x)=∣x−5∣

Absolute value functions are continuous for all real values of x.

Even at $x = 5$
L.H.L = R.H.L = 0

Hence, continuous for all real x.

Q4. Prove that the function f(x) = $x^n$ is continuous at x=a, where n is a positive integer.

Solution:
We need to check:
$\lim_{x \to a} x^n = f(a)$

We know that powers of x (like $x^n$) are polynomial functions.

All polynomial functions are continuous for all real numbers.

So,
$\lim_{x \to a} x^n = a^n = f(a)$

Thus,
L.H.L = R.H.L = f(a)

Hence, f(x) = $x^n$ is continuous at every real number x=a.

Q5.

Is the function fff defined by$$f(x) = \begin{cases} x, & x \le 1 \\ 5, & x > 1 \end{cases}$$

continuous at x=0, x=1 and x=2?

Solution:

A function is continuous at $x = a$ if
LHL = RHL = f(a)

At x=0:
Since

f(0) = 0

LHL = RHL = 0 ⇒ Continuous

At x=1
LHL = f(1) = 1
RHL = 5

Since LHL ≠ RHL ⇒ Discontinuous at x=1

At x=2
Since $2 > 1$

f(2) = 5

LHL = RHL = 5 ⇒ Continuous

Final Answer:
Continuous at $x = 0$x=0 and $x = 2$x=2
Discontinuous at $x = 1$x=1

Q6.

Find all points of discontinuity of the function$$f(x) = \begin{cases} x^2 + 3, & x \le 2 \\ x^2 – 3, & x > 2 \end{cases}$$

Solution:

Check at $x = 2$

LHL = $2^2 + 3 = 7$

RHL = $2^2 – 3 = 1$

Since LHL ≠ RHL ⇒ Discontinuous

Final Answer:
Discontinuous at $x = 2$

Q7.

Examine continuity of$$f(x) = \begin{cases} |x| + 3, & x \le -3 \\ x^2 – 2, & -3 < x < 3 \\ x + 2, & x \ge 3 \end{cases}$$

Solution:

Check at $x = -3$

LHL = $|−3| + 3 = 6$

RHL = $(−3)^2 – 2 = 7$

Not equal ⇒ Discontinuous at x=−3

Check at $x = 3$x=3:

LHL = $3^2 – 2 = 7$

RHL = $3 + 2 = 5$

Not equal ⇒ Discontinuous at x=3

Final Answer:
Discontinuous at $x = -3$

Q8.

Examine continuity of$$f(x) = \begin{cases} \frac{|x|}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$$

Solution:

For $x > 0$$\frac{|x|}{x} = 1$

For $x < 0$$\frac{|x|}{x} = -1$

At $x = 0$

LHL = −1
RHL = 1

Not equal ⇒ Discontinuous at x=0

Q9.

Examine continuity of$$f(x) = \begin{cases} \frac{x}{|x|}, & x \ne 0 \\ 1, & x = 0 \end{cases}$$

Solution:

For $x > 0$

For $x < 0$

At $x = 0$

LHL = −1
RHL = 1

Not equal ⇒ Discontinuous at x=0

Q10.

Examine continuity of$$f(x) = \begin{cases} x^2 + 1, & x \ge 1 \\ x + 1, & x < 1 \end{cases}$$

Solution:

At $x = 1$

LHL = $1 + 1 = 2$

RHL = $1^2 + 1 = 2$

Equal ⇒ Continuous

Q11.

Examine continuity of$$f(x) = \begin{cases} x^2 – 3, & x \le 2 \\ x + 1, & x > 2 \end{cases}$$

Solution:

At $x = 2$

LHL = $4 – 3 = 1$

RHL = $2 + 1 = 3$

Not equal ⇒ Discontinuous at x=2

Q12.

Examine continuity of$$f(x) = \begin{cases} x^2 – 1, & x \le 1 \\ x, & x > 1 \end{cases}$$

Solution:

At $x = 1$

LHL = $1 – 1 = 0$
RHL = 1

Not equal ⇒ Discontinuous at x=1

Q13.

Is the function defined by$$f(x) = \begin{cases} x + 5, & x \le 1 \\ x – 5, & x > 1 \end{cases}$$

continuous at x=1?

Solution:

At $x = 1$x=1:

LHL = $1 + 5 = 6$
RHL = $1 – 5 = -4$

Not equal ⇒ Discontinuous at x=1

Q14.

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

Solution:

Check continuity at boundary points $x = 1$ and $x = 3$

At x=1:
LHL = 3
RHL = 4

Since LHL ≠ RHL ⇒ Discontinuous at x=1

At x=3:
LHL = 4
RHL = 5

Since LHL ≠ RHL ⇒ Discontinuous at x=3

Final Answer:
Discontinuous at $x = 1$ and $x = 3$

Q15.

Discuss the continuity of the function f, where$$f(x) = \begin{cases} x^2, & x < 0 \\ 0, & 0 \le x \le 1 \\ 4x, & x > 1 \end{cases}$$

Solution:

Check at $x = 0$$x = 1$

At x=0
LHL = $0^2 = 0$

RHL = 0

Equal ⇒ Continuous at x=0

At x=1
LHL = 0
RHL = $4(1) = 4$

Not equal ⇒ Discontinuous at x=1

Final Answer:
Continuous at $x = 0$

Discontinuous at $x = 1$

Q16.

Discuss the continuity of the function f, where$$f(x) = \begin{cases} 2 – x, & x \le -1 \\ x^2, & -1 < x \le 1 \\ 2, & x > 1 \end{cases}$$

Solution:

Check at $x = -1$ , x=1

At x=−1:
LHL = $2 – (-1) = 3$

RHL = $(-1)^2 = 1$

Not equal ⇒ Discontinuous at x=−1

At x=1
LHL = $1^2 = 1$

RHL = 2

Not equal ⇒ Discontinuous at x=1

Final Answer:
Discontinuous at $x = -1$

Q17.

Find the relationship between aaa and bbb so that the function fff defined by$$f(x) = \begin{cases} ax + 1, & x \le 3 \\ bx + 3, & x > 3 \end{cases}$$

is continuous at x=3

Solution:

For continuity at $x = 3$x=3:
LHL = RHL

LHL = $a(3) + 1 = 3a + 1$

RHL = $b(3) + 3 = 3b + 3$

Equate:
$3a + 1 = 3b + 3$

$3a – 3b = 2$

Divide by 3:$$a – b = \frac{2}{3}$$

Q18.

For what value of λ is the function$$f(x) = \begin{cases} \lambda(x – 2), & x \le 0 \\ x^2 + 1, & x > 0 \end{cases}$$

continuous at x=0? Also check continuity at x=1.

Solution:

Continuity at x=0:

LHL = $\lambda(0 – 2) = -2\lambda$

RHL = $0^2 + 1 = 1$

For continuity:$$-2\lambda = 1 \Rightarrow \lambda = -\frac{1}{2}$$

Continuity at x=1:

Since $x > 0$, function becomes $x^2 + 1$, which is a polynomial ⇒ continuous

Final Answer:
$\lambda = -\frac{1}{2}$

Function is continuous at $x = 1$

Q19.

Show that the function g(x)=x−[x]is discontinuous at all integral points.

Solution:

Here $[x]$[x] is the greatest integer function.

Let $x = n$x=n, where $n$n is an integer.

LHL = $n – (n-1) = 1$

RHL = $n – n = 0$

Since LHL ≠ RHL ⇒ Discontinuous at every integer

Final Answer:
Function is discontinuous at all integers

Q20.

Is the function f(x)=x2−sin⁡x+5 continuous at x=π ?

Solution:

All given functions:

• $x^2$→ continuous
• $\sin x$→ continuous

Sum/difference of continuous functions is continuous

Hence $f(x)$ is continuous everywhere

Final Answer:
Continuous at $x = \pi$

Q21.

Discuss the continuity of the following functions:

(a) f(x)=sin⁡x+cos⁡x

Both functions are continuous ⇒ sum is continuous

(b) f(x)=sin⁡x−cos⁡x

Continuous for all real x

Difference of continuous functions

Continuous for all real x

(c) f(x)=sin⁡x⋅cos⁡x

Product of continuous functions

Continuous for all real x

Q22.

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

Solution:

• Cos x → Continuous for all real x
• Cosec x (1/sin x) → Discontinuous where sin x = 0
  ⇒ $x = n\pi$
• Sec x (1/cos x) → Discontinuous where cos x = 0
  ⇒ $x = \frac{\pi}{2} + n\pi$
• Cot x (cos x / sin x) → Discontinuous where sin x = 0
  ⇒ $x = n\pi$

Q23.

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

Solution:

Check at $x = 0$

LHL = $\sin 0 = 0$

RHL = undefined (division by zero)

⇒ Discontinuous at x=0

Q24.

Determine whether the function$$f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & x \ne 0 \\ 0, & x = 0 \end{cases}$$

is continuous.

Solution:

Check at $x = 0$

We know:$$-1 \le \sin\left(\frac{1}{x}\right) \le 1$$

So,$$-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$$

Taking limit as $x \to 0$x→0:$$\lim x^2 = 0$$

By sandwich theorem:$$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 = f(0)$$

Final Answer:
Function is continuous at $x = 0$

Hence continuous everywhere

Q25.

Examine the continuity of the function fff, where$$f(x) = \begin{cases} \sin x – \cos x, & x \ne 0 \\ -1, & x = 0 \end{cases}$$

Solution:

Check at $x = 0$$$\lim_{x \to 0} (\sin x – \cos x) = 0 – 1 = -1$$$$f(0) = -1$$f(0)=−1

Since LHL = RHL = f(0),
function is continuous at x=0

Q26.

Find the values of k so that the function f is continuous at the indicated point in Exercises
26 to 29

$$f(x) = \begin{cases} k \cos x, & x \ne \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$$

Solution:

$$\lim_{x \to \frac{\pi}{2}} k \cos x = k \cdot 0 = 0$$

For continuity:$$0 = 3$$

This is not possible.

Final Answer:
No value of $k$makes the function continuous.

Q27.

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

Solution:

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

RHL = 3

For continuity:$$4k = 3 \Rightarrow k = \frac{3}{4}$$

Q28.

Find k so that the function is continuous at x=π$$f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases}$$

Solution:

LHL = $k\pi + 1$

RHL = $\cos \pi = -1$$$k\pi + 1 = -1 \Rightarrow k\pi = -2 \Rightarrow k = -\frac{2}{\pi}$$

Q29.

Find k so that the function is continuous at x=5:$$f(x) = \begin{cases} kx + 1, & x \le 5 \\ 3x – 5, & x > 5 \end{cases}$$

Solution:

LHL = $5k + 1$

RHL = $3(5) – 5 = 10$$$5k + 1 = 10 \Rightarrow 5k = 9 \Rightarrow k = \frac{9}{5}$$

Q30.

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

Solution:

Continuity at $x = 2$$$5 = 2a + b \quad …(1)$$

Continuity at $x = 10$$$10a + b = 21 \quad …(2)$$

Subtract (1) from (2):$$(10a + b) – (2a + b) = 21 – 5 \Rightarrow 8a = 16 \Rightarrow a = 2$$

Substitute in (1):$$5 = 2(2) + b \Rightarrow b = 1$$

Final Answer:
$a = 2,\; b = 1$

Class-wise Solutions

Class 12:

Class 12 Physics – NCERT Solutions

Class 11:

• Class 11 Physics – NCERT Solutions
• Class 11 Chemistry – NCERT Solutions
• Class 11 Biology – NCERT Solutions
• Class 11 Math – NCERT Solutions

Class 10:

• Class 10 Science – NCERT Solutions
• Class 10 Math – NCERT Solutions
• Class 10 Science – CBSE SQP 2024

Class 9:

• Class 9 Science – NCERT Solutions
• Class 9 Math – NCERT Solutions

Class 8:

• Class 8 Science – Oxford Solutions
• Class 8 Science – NCERT Solutions

Class 7:

• Class 7 Science – Oxford Solutions

Class 6:

• Class 6 Science – Oxford Solutions

Subject-wise Solutions

Physics:

• Class 11 Physics – NCERT Solutions
• Class 12 Physics – NCERT Solutions

Chemistry:

• Class 11 Chemistry – NCERT Solutions
• Class 12 Chemistry – NCERT Solutions

Biology:

• Class 11 Biology – NCERT Solutions

Math:

• Class 11 Math – NCERT Solutions
• Class 10 Math – NCERT Solutions
• Class 9 Math – NCERT Solutions
• Class 8 Math – NCERT Solutions

Science:

• Class 10 Science – NCERT Solutions
• Class 9 Science – NCERT Solutions
• Class 8 Science – Oxford Solutions
• Class 7 Science – Oxford Solutions
• Class 6 Science – Oxford Solutions

NEET BIOLOGY

• Evolution
• Breathing and Exchange of Gases
• Anatomy of Flowering Plants
• Body Fluids and Circulation
• Human Health and Disease
• Microbes in Human Welfare
• Cell Cycle and Cell Division
• Biotechnology and Its Applications
• Biodiversity and Conservation
• Morphology of Flowering Plants

For additional reference and to access the official NCERT Class 10 Maths textbook, visit the NCERT website. This will help you understand the concepts covered in Class 12  more effectively.

Visit NCERT Official Website: NCERT Class 12 Maths Book