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Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.1 | Set 2
  • Last Updated : 30 Apr, 2021

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 = 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 – 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 ∈ Interger

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.

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