Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1 | Set 2

Chapter 5 on Continuity and Differentiability is a crucial part of calculus in Class 12 mathematics. It builds upon the concept of limits and introduces students to the fundamental ideas of continuous functions and differentiation. This chapter lays the groundwork for understanding rates of change, optimization problems, and many advanced mathematical concepts.

Content of this article has been merged with Chapter 5 Continuity And Differentiability- Exercise 5.1 as per the revised syllabus of NCERT.

Question 18. For what value of λ is the function defined by

f(x)= \begin{cases} \lambda(x^2-2x), \hspace{0.2cm}x \leq0\\ 4x+1,\hspace{0.2cm}x>0 \end{cases}

continuous at x = 0? What about continuity at x = 1?

Solution:

To be continuous function, f(x) should satisfy the following at x = 0:

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)

Continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2-2x)

= λ(0²- 2(0)) = 0

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x+1)

= λ4(0) + 1 = 1

Function value at x = 0, f(0) = \lambda(0^2-2(0)) = 0

As, 0 = 1 cannot be possible

Hence, for no value of λ, f(x) is continuous.

But here, \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)

Continuity at x = 1,

Left limit = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (4x+1)

= (4(1) + 1) = 5

Right limit = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x+1)

= 4(1) + 1 = 5

Function value at x = 1, f(1) = 4(1) + 1 = 5

As, \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 5

Hence, the function is continuous at x = 1 for any value of λ.

Question 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. 

Solution:

[x] is greatest integer function which is defined in all integral points, e.g.

[2.5] = 2

[-1.96] = -2

x-[x] gives the fractional part of x.

e.g: 2.5 - 2 = 0.5

c be an integer

Let's check the continuity at x = c,

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])

= (c - (c - 1)) = 1

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])

= (c - c) = 0

Function value at x = c, f(c) = c - [c]= c - c = 0

As, \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Hence, the function is discontinuous at integral.

c be not an integer

Let's check the continuity at x = c,

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])

= (c - (c - 1)) = 1

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])

= (c - (c - 1)) = 1

Function value at x = c, f(c) = c - [c] = c - (c - 1) = 1

As, \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)=f(1)=1

Hence, the function is continuous at non-integrals part.

Question 20. Is the function defined by f(x) = x² – sin x + 5 continuous at x = π?

Solution:

Let's check the continuity at x = π,

f(x) = x² – sin x + 5

Let's substitute, x = π+h

When x⇢π, Continuity at x = π

Left limit = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (x^2 – sin \hspace{0.1cm}x + 5)

= (π² – sinπ + 5) = π² + 5

Right limit = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+}(x^2 – sin \hspace{0.1cm}x + 5)

= (π² – sinπ + 5) = π² + 5

Function value at x = π, f(π) = π² – sin π + 5 = π² + 5

As, \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi)

Hence, the function is continuous at x = π .

Question 21. Discuss the continuity of the following functions:

(a) f(x) = sin x + cos x 

Solution:

Here, 

f(x) = sin x + cos x

Let's take, x = c + h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So, 

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (sin(c + h) + cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B - sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}  ((sinc cosh + cosc sinh) + (cosc cosh − sinc sinh))

\lim_{h \to 0} f(c+h)  = ((sinc cos0 + cosc sin0) + (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h)  = (sinc + cosc) = f(c)

Function value at x = c, f(c) = sinc + cosc

As, \lim_{x \to c} f(x)  = f(c) = sinc + cosc

Hence, the function is continuous at x = c.

(b) f(x) = sin x – cos x

Solution:

Here,

f(x) = sin x - cos x

Let's take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}  (sin(c + h) − cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B - sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} ((sinc cosh + cosc sinh) − (cosc cosh − sinc sinh))

\lim_{h \to 0} f(c+h)  = ((sinc cos0 + cosc sin0) − (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h)  = (sinc − cosc) = f(c)

Function value at x = c, f(c) = sinc − cosc

As, \lim_{x \to c} f(x)  = f(c) = sinc − cosc

Hence, the function is continuous at x = c.

(c) f(x) = sin x . cos x

Solution:

Here,

f(x) = sin x + cos x

Let's take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} sin(c + h) × cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B - sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0}  ((sinc cosh + cosc sinh) × (cosc cosh − sinc sinh))

\lim_{h \to 0} f(c+h)  = ((sinc cos0 + cosc sin0) × (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h)  = (sinc × cosc) = f(c)

Function value at x = c, f(c) = sinc × cosc

As, \lim_{x \to c} f(x)  = f(c) = sinc × cosc

Hence, the function is continuous at x = c.

Question 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution:

Continuity of cosine

Here,

f(x) = cos x

Let's take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (cos\hspace{0.1cm} (c+h))

Using the trigonometric identities, we get

cos(A + B) =  cos A cos B - sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (cosc cosh − sinc sinh)

\lim_{h \to 0} f(c+h)  = (cosc cos0 − sinc sin0)

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h)  = (cosc) = f(c)

Function value at x = c, f(c) = (cosc)

As, \lim_{x \to c} f(x)  = f(c) = (cosc)

Hence, the cosine function is continuous at x = c.

Continuity of cosecant

Here,

f(x) = cosec x = \frac{1}{sin \hspace{0.1cm}x}

Domain of cosec is R - {nπ}, n ∈ Integer

Let's take, x = c + h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin \hspace{0.1cm}(c+h)})

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})\\ \lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c})

Function value at x = c, f(c) = \frac{1}{sin\hspace{0.1cm} c}

As, \lim_{x \to c} f(x) = f(c) = \frac{1}{sin\hspace{0.1cm} c}

Hence, the cosecant function is continuous at x = c.

Continuity of secant

Here,

f(x) = sec x = \frac{1}{cos \hspace{0.1cm}x}

Let's take, x = c + h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos \hspace{0.1cm}(c+h)})

Using the trigonometric identities, we get

cos(A + B) =  cos A cos B - sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c})

Function value at x = c, f(c) = \frac{1}{cos\hspace{0.1cm} c}

As, \lim_{x \to c} f(x) = f(c) = \frac{1}{cos\hspace{0.1cm} c}

Hence, the secant function is continuous at x = c.

Continuity of cotangent

Here,

f(x) = cot x = \frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}

Let's take, x = c+h

When x⇢c then h⇢0

\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)

So,

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos \hspace{0.1cm}(c+h)}{sin \hspace{0.1cm}(c+h)})

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B - sin A sin B

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})

cos 0 = 1 and sin 0 = 0

\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})

\lim_{h \to 0} f(c+h) = (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})

Function value at x = c, f(c) = \frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}

As, \lim_{x \to c} f(x) = f(c) = \frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}

Hence, the cotangent function is continuous at x = c.

Question 23. Find all points of discontinuity of f, where

f(x)= \begin{cases} \frac{sin \hspace{0.1cm}x}{x}, \hspace{0.2cm}x <0\\ x+1,\hspace{0.2cm}x\geq0 \end{cases}

Solution:

Here,

From the two continuous functions g and h, we get

\frac{g(x)}{h(x)}     = continuous when h(x) ≠ 0

For x < 0, f(x) = \frac{sin \hspace{0.1cm}x}{x}     , is continuous

Hence, f(x) is continuous x ∈ (-∞, 0)

Now, For x ≥ 0, f(x) = x + 1, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (0, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 0) U (0, ∞)= R - {0}

Let's check the continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\frac{sin \hspace{0.1cm}x}{x})\\= 1

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x+1)\\= (1+0)\\= 1

Function value at x = 0, f(0) = 0 + 1 = 1

As, \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1

Hence, the function is continuous at x = 0.

Hence, the function is continuous for any real number.

Question 24. Determine if f defined by

f(x)= \begin{cases} x^2sin\frac{1}{x}, \hspace{0.2cm}x \neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}

is a continuous function?

Solution:

Here, as it is given that

For x = 0, f(x) = 0, which is a constant

As constant are continuous, hence f(x) is continuous x ∈ = R - {0}

Let's check the continuity at x = 0,

As, we know range of sin function is [-1,1]. So, -1 ≤ sin(\frac{1}{x})     ≤ 1 which is a finite number.

Limit = \lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 sin(\frac{1}{x}))

= (0² ×(finite number)) = 0

Function value at x = 0, f(0) = 0

As, \lim_{x \to 0} f(x) = f(0).

Hence, the function is continuous for any real number.

Question 25. Examine the continuity of f, where f is defined by

f(x)= \begin{cases} sin\hspace{0.1cm}x-cos\hspace{0.1cm}x, \hspace{0.2cm}x\neq0\\ -1,\hspace{0.2cm}x=0 \end{cases}

Solution:

Continuity at x = 0,

Left limit = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sin0 − cos0) = 0 − 1 = −1

Right limit = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sin0 − cos0) = 0 − 1 = −1

Function value at x = 0, f(0) = sin 0 - cos 0 = 0 - 1 = -1

As, \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1

Hence, the function is continuous at x = 0.

Continuity at x = c (real number c≠0),

Left limit = \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sinc − cosc)

Right limit = \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)

= (sinc − cosc)

Function value at x = c, f(c) = sin c - cos c

As, \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) = (sin\hspace{0.1cm}c-cos\hspace{0.1cm}c)

So concluding the results, we get

The function f(x) is continuous at any real number.

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

Question 26. f(x)= \begin{cases} \frac{k\hspace{0.1cm}cos\hspace{0.1cm}x}{\pi-2x}, \hspace{0.2cm}x\neq\frac{\pi}{2}\\ 3,x=\frac{\pi}{2} \end{cases} \hspace{0.1cm}\hspace{0.1cm}  at x = π/2.

Solution:

Continuity at x = π/2

Let's take x = \frac{\pi}{2}+h

When x⇢π/2 then h⇢0

Substituting x = \frac{\pi}{2}   +h, we get

cos(A + B) = cos A cos B - sin A sin B

Limit = \lim_{h \to 0} f(\frac{\pi}{2}+h) = \lim_{h \to 0} (\frac{k\hspace{0.1cm}cos(\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)}\\= \lim_{h \to 0} (\frac{k(cos(\frac{\pi}{2})cos h-sin(\frac{\pi}{2})sinh)}{\pi-\pi-2h)}\\= \lim_{h \to 0} (\frac{k(0 \times cos h-1\times sinh)}{-2h)}\\= \lim_{h \to 0} (\frac{k(-sinh)}{-2h)}\\ = \frac{k}{2} \lim_{h \to 0} (\frac{(sinh)}{h)}\\ = \frac{k}{2}

Function value at x = \frac{\pi}{2}, f(\frac{\pi}{2})   = 3

As, \lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})   should satisfy, for f(x) being continuous

k/2 = 3

k = 6

Question 27. f(x)= \begin{cases} kx^2,x\leq2\\ 3,x>2 \end{cases} \hspace{0.1cm}\hspace{0.1cm}  at x = 2

Solution:

Continuity at x = 2

Left limit = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2)

= k(2)² = 4k

Right limit = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3)\\= 3

Function value at x = 2, f(2) = k(2)² = 4k

As, \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2)    should satisfy, for f(x) being continuous

4k = 3

k = 3/4

Question 28. f(x)= \begin{cases} kx+1,x\leq\pi\\ cos \hspace{0.2cm}x,x>\pi \end{cases} \hspace{0.1cm}\hspace{0.1cm}  at x = π

Solution:

Continuity at x = π

Left limit = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx+1)

= k(π) + 1

Right limit = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (cos x)

= cos(π) = -1

Function value at x = π, f(π) = k(π) + 1

As, \lim_{x \to \pi^-}f(x) = \lim_{x \to \pi^+} f(x)= f(\pi)    should satisfy, for f(x) being continuous

kπ + 1 = -1

k = -2/π

Question 29. f(x)= \begin{cases} kx+1,x\leq5\\ 3x-5,x>5 \end{cases} \hspace{0.1cm}\hspace{0.1cm}  at x = 5

Solution:

Continuity at x = 5

Left limit = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx+1)

= k(5) + 1 = 5k + 1

Right limit = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x-5)

= 3(5) - 5 = 10

Function value at x = 5, f(5) = k(5) + 1 = 5k + 1

As, \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)= f(5)   should satisfy, for f(x) being continuous

5k + 1 = 10

k = 9/5 

Question 30. Find the values of a and b such that the function defined by

f(x)= \begin{cases} 5,x\leq2\\ ax+b,2<x<10\\ 21,x\geq10 \end{cases}

is a continuous function

Solution:

Continuity at x = 2

Left limit = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5)\\= 5

Right limit = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax+b)\\= 2a+b

Function value at x = 2, f(2) = 5

As, \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2)   should satisfy, for f(x) being continuous at x = 2

2a + b = 5 ........................(1)

Continuity at x = 10

Left limit = \lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax+b)

= 10a + b

Right limit = \lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} (21)

= 21

Function value at x = 10, f(10) = 21

As, \lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x)= f(10)   should satisfy, for f(x) being continuous at x = 10

10a + b = 21 ........................(2)

Solving the eq(1) and eq(2), we get

a = 2

b = 1

Question 31. Show that the function defined by f(x) = cos (x²) is a continuous function

Solution:

Let's take

g(x) = cos x

h(x) = x²

g(h(x)) = cos (x²)

To prove g(h(x)) continuous, g(x) and h(x) should be continuous.

Continuity of g(x) = cos x

Let's check the continuity at x = c

x = c + h

g(c + h) = cos (c + h)

When x⇢c then h⇢0

cos(A + B) =  cos A cos B - sin A sin B

Limit = \lim_{h \to 0} g(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0}  (cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

Function value at x = c, g(c) = cos c

As, \lim_{x \to c} g(x) = g(c) = cos\hspace{0.1cm} c

The function g(x) is continuous at any real number.

Continuity of h(x) = x²

Let's check the continuity at x = c

Limit = \lim_{x \to c} h(x) = \lim_{x \to c} (x^2)

= c²

Function value at x = c, h(c) = c²

As, \lim_{x \to c} h(x) = h(c) = c^2

The function h(x) is continuous at any real number.

As, g(x) and h(x) is continuous then g(h(x)) = cos(x²) is also continuous.

Question 32. Show that the function defined by f(x) = | cos x | is a continuous function. 

Solution:

Let's take

g(x) = |x|

m(x) = cos x

g(m(x)) = |cos x|

To prove g(m(x)) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x - 0|, |x| = x when x ≥ 0 and |x| = -x when x < 0

Let's check the continuity at x = c

When c < 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c

Function value at x = c, g(c) = |c| = -c

As, \lim_{x \to c} g(x) = g(c) = -c

When c ≥ 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c

Function value at x = c, g(c) = |c| = c

As, \lim_{x \to c} g(x) = g(c) = c

The function g(x) is continuous at any real number.

Continuity of m(x) = cos x

Let's check the continuity at x = c

x = c + h

m(c + h) = cos (c + h)

When x⇢c then h⇢0

cos(A + B) =  cos A cos B - sin A sin B

Limit = \lim_{h \to 0} m(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0} (cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

Function value at x = c, m(c) = cos c

As, \lim_{x \to c} m(x) = m(c) = cos \hspace{0.1cm}c

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then g(m(x)) = |cos x| is also continuous.

Question 33. Examine that sin | x | is a continuous function.

Solution:

Let's take

g(x) = |x|

m(x) = sin x

m(g(x)) = sin |x|

To prove m(g(x)) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x-0|, |x|=x when x≥0 and |x|=-x when x<0

Let's check the continuity at x = c

When c < 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c

Function value at x = c, g(c) = |c| = -c

As, \lim_{x \to c} g(x) = g(c) = -c

When c ≥ 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c

Function value at x = c, g(c) = |c| = c

As, \lim_{x \to c} g(x) = g(c) = c

The function g(x) is continuous at any real number.

Continuity of m(x) = sin x

Let's check the continuity at x = c

x = c + h

m(c + h) = sin (c + h)

When x⇢c then h⇢0

sin(A + B) = sin A cos B + cos A sin B

Limit = \lim_{h \to 0} m(c+h) = \lim_{h \to 0} (sin(c+h))\\ = \lim_{h \to 0}  (sinc cosh + cosc sinh) 

= sinc cos0 + cos csin0 = sinc

Function value at x = c, m(c) = sin c

As, \lim_{x \to c} m(x) = m(c) = sin c

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then m(g(x)) = sin |x| is also continuous.

Question 34. Find all the points of discontinuity of f defined by f(x) = | x | – | x + 1 |

Solution:

Let's take

g(x) = |x|

m(x) = |x + 1|

g(x) - m(x) = | x | – | x + 1 |

To prove g(x) - m(x) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x - 0|, |x| = x when x≥0 and |x| = -x when x < 0

Let's check the continuity at x = c

When c < 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c

Function value at x = c, g(c) = |c| = -c

As, \lim_{x \to c} g(x) = g(c) = -c

When c ≥ 0

Limit = \lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c

Function value at x = c, g(c) = |c| = c

As, \lim_{x \to c} g(x) = g(c) = c

The function g(x) is continuous at any real number.

Continuity of m(x) = |x + 1|

As, we know that modulus function works differently.

In |x + 1|, |x + 1| = x + 1 when x ≥ -1 and |x + 1| = -(x + 1) when x < -1

Let's check the continuity at x = c

When c < -1

Limit = \lim_{x \to c} m(x) = \lim_{x \to c} (|x+1|)\\= \lim_{x \to c} -(x+1)

= -(c + 1)

Function value at x = c, m(c) = |c + 1| = -(c + 1)

As, \lim_{x \to c} m(x) = m(c) = -(c+1)

When c ≥ -1

Limit = \lim_{x \to c} m(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x+1)

= c + 1

Function value at x = c, m(c) = |c| = c + 1

As, \lim_{x \to c} m(x)  = m(c) = c + 1

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then g(x) - m(x) = |x| – |x + 1| is also continuous.

Summary

Exercise 5.1 focuses on the concept of continuity of functions. It covers various types of functions including piecewise functions, absolute value functions, and trigonometric functions. Students learn to analyze continuity at specific points, find points of discontinuity, and determine conditions for functions to be continuous. This exercise provides a solid foundation for understanding differentiability, which is covered later in the chapter.