Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1

Question 1.

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

Solution:

(i) At x = 0, \(\lim _{x \rightarrow 0}\)f(x) = \(\lim _{x \rightarrow 0}\) (5x – 3) = – 3

and f(0) = -3.

∴ f is continuous at x = 0.

(ii) At x = – 3, \(\lim _{x \rightarrow -3}\)f(x) = \(\lim _{x \rightarrow -3}\) (5x – 3) = -18

and f(- 3) = – 18.

∴ f is continuous at x = – 3.

(iii) At x = 5, \(\lim _{x \rightarrow 5}\) f(x) = \(\lim _{x \rightarrow 5}\) (5x – 3) = 22

and f(5) = 22

∴ f is continuous at x = 5.

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Question 2.

Examine the continuity of the function f(x) = 2x² – 1 at x = 3.

Solution:

\(\lim _{x \rightarrow 3}\) f(x) = \(\lim _{x \rightarrow 3}\) (2x² – 1) = 17

and f(3) = 17.

∴ f is continuous at x = 3.

Question 3.

Examine the following functions for continuity:

(a) f(x) = x – 5

(b) f(x) = \(\frac { 1 }{ x-5 }\)

(c) f(x) = \(\frac{x^{2}-25}{x+5}\)

(d) f(x) = |x-5|

Solution:

(a) f(x) = x – 5

x – 5 is a polynomial.

Therefore, it is continuous at each x ∈ R.

(b) f(x) = \(\frac { 1 }{ x-5 }\)

At x = 5, f(x) is not defined.

∴ f is not continuous at x = 5.

When x ≠ 5, \(\lim _{x \rightarrow c}\) = \(\frac { 1 }{ c-5 }\)

Also, f(c) = \(\frac { 1 }{ c-5 }\)

∴ f is continuous at x ∈ R – {5}.

(c) f(x) = \(\frac{x^{2}-25}{x+5}\)

At x = – 5, function/is not defined.

∴ f is discontinuous at x = – 5.

At x = c ≠ – 5,

\(\lim _{x \rightarrow c}\) f(x) = \(\lim _{x \rightarrow c}\) \(\frac{x^{2}-25}{x+5}\) = c – 5

and f(c) = c – 5.

∴ f is continuous for all x ∈ R – {- 5).

(d) f(x) = |x – 5|

At x = 5, f(5) = |5 – 5| = 0

\(\lim _{x \rightarrow c}\) |x – 5| = 0

∴ f is continuous at x = 5.

At x = c > 5,

\(\lim _{x \rightarrow c}\) |x – 5| = c – 5 [c > 5]

Also, f(c) = c – 5.

∴ f is continuous at x = c > 5.

Similarly, at x = c < 5,

\(\lim _{x \rightarrow c}\) |x – 5| = 5 – c and f(c) = 5 – c.

∴ f is continuous at x = c < 5.

Thus, f is continuous for all x ∈ R.

Question 4.

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

Solution:

f(x) = x^{n} is a polynomial which is continuous for all x ∈ R.

Hence, f is continuous at x = n, n ∈ N.

Question 5.

Is the function f defined by

f(x) = \(\left\{\begin{array}{l}

x, \text { if } x \leq 1 \\

5, \text { if } x>1

\end{array}\right.\)

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

Solution:

(i) At x = 0,

\(\lim _{x \rightarrow 0^{-}}\) f(x) = \(\lim _{x \rightarrow 0^{-}}\) x = 0.

\(\lim _{x \rightarrow 0^{+}}\) f(x) = \(\lim _{x \rightarrow 0^{+}}\) x = 0.

Also, f(0) = 0.

∴ f is continuous at x = 0.

(ii) At x =1,

\(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{-}}\) (x) = 1.

\(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) (x) = 5.

\(\lim _{x \rightarrow 1^{-}}\) f(x) ≠ \(\lim _{x \rightarrow 1^{+}}\) (x)

∴ f is discontinuous at x = 1.

(iii) At x = 2,

\(\lim _{x \rightarrow 2}\) f(x) = 5 and f(2) = 5.

∴ f is continuous at x = 2.

Question 6.

f(x) = \(\left\{\begin{array}{l}

2 x+3, x \leq 2 \\

2 x-3, x>2

\end{array}\right.\)

Solution:

f(x) = \(\left\{\begin{array}{l}

2 x+3, x \leq 2 \\

2 x-3, x>2

\end{array}\right.\)

At x – 2, L.H.L. = \(\lim _{x \rightarrow 2^{-}}\) (2x + 3) = 7

f(2) = 2 x 2 + 3 = 7

R.H.L. = \(\lim _{x \rightarrow 2^{+}}\) (2x – 3) = 2 x 2 – 3 = 1

⇒ L.H.L. ≠ R.H.L.

∴ f is discontinuous at x = 2.

At x = c < 2, \(\lim _{x \rightarrow c}\) (2x + 3) = 2c + 3 = f(c).

f is continuous at x = c < 2. At x = c > 2, \(\lim _{x \rightarrow c}\) (2x – 3) = 2c – 3 = f(c).

∴ f is continuous at x = c > 2.

⇒ Point of discontinuity is x = 2.

Question 7.

f(x) = \(\left\{\begin{array}{l}

|x|+3, \text { if } x \leq-3 \\

-2 x, \quad \text { if }-3<x<3 \\ 6 x+2, \text { if } x>3

\end{array}\right.\)

Solution:

f(x) = \(\left\{\begin{array}{l}

|x|+3, \text { if } x \leq-3 \\

-2 x, \quad \text { if }-3<x<3 \\ 6 x+2, \text { if } x>3

\end{array}\right.\)

At x = – 3, L.H.L. = \(\lim _{x \rightarrow 3^{-}}\) (|x|+3)

= \(\lim _{x \rightarrow -3^{-}}\) (-x + 3) = 3 + 3 = 6.

f(- 3) = |- 3| + 3 = 6.

R.H.L. = \(\lim _{x \rightarrow -3^{+}}\) f(x) = \(\lim _{x \rightarrow -3^{+}}\) (- 2x) = 6.

L.H.L. = R.H.L. = f(- 3).

⇒ f is continuous at x = – 3.

At x = 3, L.H.L. = \(\lim _{x \rightarrow 3^{-}}\) f(x) = \(\lim _{x \rightarrow 3^{-}}\) (- 2x) = – 6

R.H.L. = \(\lim _{x \rightarrow 3^{+}}\) f(x) = \(\lim _{x \rightarrow 3^{+}}\) (6x + 2) = 20

f(3) is not defined.

∴ L.H.L. ≠ R.H.L. ≠ f(3)

∴ f is discontinuous at x = 3.

At x = c < – 3,

\(\lim _{x \rightarrow -c}\) (|x| + 3) = – c + 3 = f(c)

⇒ \(\lim _{x \rightarrow -c}\) f(x) = f(c)

⇒ f is continuous at x = c < – 3.

At x = c, when – 3 < x < 3,

\(\lim _{x \rightarrow 1^{-}}\) (- 2x) =-2c =f(c) ⇒ \(\lim _{x \rightarrow 1^{-}}\) f(x) = f(c).

∴ f is continuous at x = c, where -3 < c < 3.

At x = c, \(\lim _{x \rightarrow c}\)(6x + 2) = 6c + 2 = f(c).

⇒ \(\lim _{x \rightarrow c}\)f(x) = f(c)

⇒ f is continuous at x = c > 3.

Question 8.

f(x) = \(\left\{\begin{array}{ll}

\frac{|x|}{x}, & \text { if } x \neq 0 \\

0, & \text { if } x=0

\end{array}\right.\)

Solution:

Question 9.

f(x) = \(\left\{\begin{array}{ll}

\frac{x}{|x|}, & \text { if } x<0 \\

-1, & \text { if } x \geq 0

\end{array}\right.\)

Solution:

∴ \(\lim _{x \rightarrow c}\)f(x) = f(c) ⇒ f is continuous at x = c < 0. At x = c > 0, \(\lim _{x \rightarrow c}\) f(x) = – 1

Also, f(c) = – 1.

∴ \(\lim _{x \rightarrow c}\) f(x) = f(c) ⇒ f is continuous at x = c > 0.

There is no point of discontinuity for this function in its domain.

Question 10.

f(x) = \(\left\{\begin{array}{ll}

x+1, & \text { if } x \geq 1 \\

x^{2}+1, & \text { if } x<1 \end{array}\right.\)

Solution:

f(x) = \(\left\{\begin{array}{ll}

x+1, & \text { if } x \geq 1 \\

x^{2}+1, & \text { if } x<1 \end{array}\right.\)

At x = 1, L.H.L. = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{-}}\) (x² + 1) = 2,

R.H.L. = \(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) (x + 1) = 2

and f(1) = 1 + 1 = 2.

⇒ f is continuous at x = 1.

At x = c > 1, \(\lim _{x \rightarrow c}\) f(x) = \(\lim _{x \rightarrow c}\) (x + 1) = c + 1 = f(c).

⇒ f is continuous at x = c > 1.

At x = c < 1, \(\lim _{x \rightarrow c}\) f(x) = \(\lim _{x \rightarrow c}\) (x² + 1) = c² + 1 = f(c).

⇒ f is continuous at x = c < 1.

There is no point of discontinuity at any noint x ∈R.

Question 11.

f(x) = \(\left\{\begin{array}{l} x^{3}-3, \text { if } x \leq 2 \\ x^{2}+1, \text { if } x>2 \end{array}\right.\)

Solution:

\(\left\{\begin{array}{l} x^{3}-3, \text { if } x \leq 2 \\ x^{2}+1, \text { if } x>2 \end{array}\right.\)

At x = 2,

L.H.L. = \(\lim _{x \rightarrow 2^{-}}\) (x³ – 3) = 8 – 3 = 5,

R.H.L. = \(\lim _{x \rightarrow 2^{+}}\) (x² + 1) = 4 + 1 = 5

and f(2) = 2³ – 3 = 8 – 3 = 5.

⇒ f is continuous at x = 2.

At x = c < 2, \(\lim _{x \rightarrow c}\) (x³ – 3) = c³ – 3 = f(c).

⇒ f is continuous at x < 2.

At x = c > 2, \(\lim _{x \rightarrow c}\) (x² + 1) = c² + 1 = f(c).

⇒ f is continuous at x > 2.

Hence, f is continuous for all x ∈ R.

∴ There is no point of discontinuity.

Question 12.

f(x) = \(\left\{\begin{array}{l}

x^{10}-1, \text { if } x \leq 1 \\

x^{2}, \quad \text { if } x>1

\end{array}\right.\)

Solution:

⇒ f is continuous at x = 1.

At x = c < 1, \(\lim _{x \rightarrow c}\) (x<sup>10</sup> – 1) = c<sup>10</sup> – 1 = f(c).

⇒ f is continuous at x < 1.

At x = c > 1, \(\lim _{x \rightarrow c}\) (x²) = c² = f(c).

⇒ f is continuous at x > 1.

Hence, f is continuous at all points x ∈ R – [1]

∴ Point of discontinuity is x = 1.

Question 13.

Is the function defined by

f(x) = \(\left\{\begin{array}{l}

x+5, \text { if } x \leq 1 \\

x-5, \text { if } x>1

\end{array}\right.\)

a continuous function?

Solution:

At x = 1,

L.H.L. = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{-}}\) (x + 5) = 6,

R.H.L. = \(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) (x – 5) = – 4,

and

f(1) = 1 + 5 = 6

f(1) = L.H.L ≠ R.H.L

⇒ f is continuous at x = 1.

At x = c < 1, \(\lim _{x \rightarrow c}\) (x + 5) = c + 5 = f(c).

⇒ f is continuous at x < 1.

At x = c > 1, \(\lim _{x \rightarrow c}\) (x – 5) = c – 5 = f(c).

⇒ f is continuous at x > 1.

⇒ f is continuous at all points x ∈ R, except x = 1.

Question 14.

f(x) = \(\left\{\begin{array}{l}

3, \text { if } 0 \leq x \leq 1 \\

4, \text { if } 1<x<3 \\

5, \text { if } 3 \leq x \leq 10

\end{array}\right.\)

Solution:

f(x) = \(\left\{\begin{array}{l}

3, \text { if } 0 \leq x \leq 1 \\

4, \text { if } 1<x<3 \\

5, \text { if } 3 \leq x \leq 10

\end{array}\right.\)

In the interval 0 ≤ x < 1, f(x) = 3.

f is continuous in this interval.

At x = 1,

L.H.L. = \(\lim _{x \rightarrow 1^{-}}\) f(x) = 3

R.H.L. = \(\lim _{x \rightarrow 1^{+}}\) f(x) = 4

⇒ f is discontinous at x = 1 because L.H.L. ≠ RH.L

In the interval 1 < x < 3, f(x) = 4.

∴ f is continuous in this interval.

At x = 3,

L.H.L. = \(\lim _{x \rightarrow 3^{-}}\) f(x) = 4

R.H.L. = \(\lim _{x \rightarrow 3^{+}}\) f(x) = 5

⇒ f is discontinous at x = 1 because L.H.L. ≠ RH.L

In the interval 3 ≤ x ≤ 10, f(x) = 5.

∴ f is continuous in this interval except for x = 3

⇒ f is not continous at x = 1 and x = 3.

Question 15.

f(x) = \(\left\{\begin{array}{l}

2 x, \text { if } x<0 \\ 0, \text { if } 0 \leq x \leq 1 \\ 4 x, \text { if } x>1

\end{array}\right.\)

Solution:

At x = 0,

L.H.L. = \(\lim _{x \rightarrow 0^{-}}\) (2x) = 0

R.H.L. = \(\lim _{x \rightarrow 0^{+}}\) (0) = 0

and f(0) = 0.

⇒ f is continous at x = 0.

At x = 1,

L.H.L. = \(\lim _{x \rightarrow 1^{+}}\) (0) = 0

R.H.L. = \(\lim _{x \rightarrow 1^{+}}\) (4x) = 4

and f(1) = 0.

∴ f(1) = L.H.L. ≠ R.H.L.

∴ f is continuous at x = 1.

When x < 0, f(x) = 2x, being a polynomial, is continuous at all points x < 0. When x > 1, f(x) = 4x, being a polynomial, is continuous at all points x > 1.

When 0 ≤ x ≤ 1, f(x) = 0 is a continuous function, except for the point x = 1.

∴ The point of discontinuity is x =1.

Question 16.

f(x) = \(\left\{\begin{array}{ll}

-2, & \text { if } x<-1 \\

2 x, & \text { if }-1<x<1 \\ 2, & \text { if } x>1

\end{array}\right.\)

Solution:

At x = – 1,

L.H.L. = \(\lim _{x \rightarrow 1^{-}}\) f(x) = -2,

f(-1) = – 2

and R.H.L. = \(\lim _{x \rightarrow 1^{+}}\) f(x) = – 2.

⇒ f is continous at x = – 1.

At x = 1,

L.H.L. = \(\lim _{x \rightarrow 1^{-}}\) f(x) = 2

f(1) = 2

R.H.L. = \(\lim _{x \rightarrow 1^{+}}\) f(x) = 2

∴ f is continuous at x = 1.

Hence, f is continuous function.

Question 17.

Find the values of a and b so that the function defined by

f(x) = \(\left\{\begin{array}{ll}

a x+1, & \text { if } x \leq 3 \\

b x+3, & \text { if } x>3

\end{array}\right.\)

is continuous at x = 3.

Solution:

For any arbitary value of b, we can find the value of a corresponding to the value of b. Thus, there are infinitely many values of a and b.

Question 18.

For what value of λ, is the function

f(x) = \(\left\{\begin{array}{l}

\lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\

4 x+1, \quad \text { if } x>0

\end{array}\right.\)

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

Solution:

So, f is continuous at x = 1.

⇒ f is not continuous at x = 0 for any value of λ, but f is continuous at x = 1 for all values of λ.

Question 19.

Show that 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:

Let c be an integer.

⇒ f is not continuous at all integral points.

Note:

To find :

Question 20.

Is the function defined by x² – sin x + 5 continuous at x = π?

Solution:

Let f(x) = x² – sin x + 5.

Hence, f is continuous at x = π.

Alternatively:

g(x) = x² + 5 is a polynomial and h(x) = sin x

∴ g is continuous for all x ∈ R.

h(x) = sin x, which is continuous for all x ∈ R

∴ f = g – h is also continuous for x ∈ R.

Question 21.

Discuss the continuity of the following functions:

(a) f(x) = sin x + cos x

(b) f(x) = sin x – cos x

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

Solution:

As above, f is continuous for all x ∈ R.

(c)

f(x) = sin x cos x = \(\frac { 1 }{ 2 }\)(2 sin x cos x)

= \(\frac { 1 }{ 2 }\) sin 2x.

Again, f is continuous for all x ∈ R.

Question 22.

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

Solution:

Let f(x) = cos x.

At x = c, c ∈ R,

\(\lim _{x \rightarrow c}\) cos x = cos c = f(c).

∴ f is continuous for all values of x ∈ R.

(b) Let f(x) = sec x.

sec x is not defined at x = (2n + 1) \(\frac { π }{ 2 }\), n ∈Z.

Also, at x = \(\frac { π }{ 2 }\),

∴ f is not continuous at x = \(\frac { π }{ 2 }\) or at x = (2n + 1) \(\frac { π }{ 2 }\).

At x = c ≠ (2n + 1)\(\frac { π }{ 2 }\),

\(\lim _{x \rightarrow c}\) sec x = sec c = f(c).

Hence, f is continuous at x ∈ R except at x = (2n + 1) \(\frac { π }{ 2 }\), where n ∈ Z.

(c) f(x) = cosec x

f is not defined at x = nπ.

⇒ f is not continuous at x = nπ.

(d) f(x) = cot x

f is not defined at x = nπ.

Thus, f is not continuous at x = π or at x = nπ.

At x = c ≠ nπ,

\(\lim _{x \rightarrow c}\) cot x = cot c = f(c).

f is continuous at all points x ∈ R except x = nπ, where n ∈ Z.

Question 23.

Find all the points of discontinuity of f, where

f(x) = \(\left\{\begin{array}{l}

\frac{\sin x}{x}, \text { if } x<0 \\

x+1, \text { if } x \geq 0

\end{array}\right.\)

Solution:

∴ f is continuous at x = 0.

When x < 0, sin x and x both are continuous.

∴ \(\frac { sin x }{ x }\) is also continuous.

When x > 0, f(x) = x + 1 is a polynomial.

∴ f is continuous.

⇒ f is not discontinuous at any point.

Question 24.

Determine if f defined by f(x) = \(\left\{\begin{array}{l}

x^{2} \sin \frac{1}{x}, \text { if } x \neq 0 \\

0, \quad \text { if } x=0

\end{array}\right.\) is a continuous function?

Solution:

sin \(\frac { 1}{ h }\) lies between -1 and 1, a finite quantity.

∴ h² sin \(\frac { 1}{ h }\) → 0 as h → 0.

∴ L.H.L. = 0

Similarly, \(\lim _{x \rightarrow 0^{+}}\)(x² sin \(\frac { 1}{ h }\)) = 0

Also, f(0) = 0 [Gievn]

∴ L.H.L = R.H.L = f(c)

∴ f is continuous for all x ∈ R.

Question 25.

Examine the continuity of f, where f is defined by f(x) = \(\left\{\begin{array}{r}

\sin x-\cos x, \text { if } x \neq 0 \\

-1, \quad \text { if } x=0

\end{array}\right.\)

Solution:

∴ f is continuous at x = 0.

Alternatively :

sin x and cos x both are continuous for all x ∈ R.

∴ sin x – cos x is continuous for all x ∈ R.

⇒ f is continuous for all x ∈ R for all x ∈ R.

Question 26.

f(x) = \(\left\{\begin{array}{r}

\frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\

3, \text { if } x=\frac{\pi}{2}

\end{array}\right. \text { at } x=\frac{\pi}{2}\)

Solution:

Question 27.

The function is defined by

f(x) = \(\left\{\begin{array}{l}

k x^{2}, \text { if } x \leq 2 \\

3, \text { if } x>2

\end{array}\right. \text { at } x=2\)

Solution:

Question 28.

The function is defined by

f(x) = \(\left\{\begin{array}{ll}

k x+1, & \text { if } x \leq \pi \\

\cos x, & \text { if } x>\pi

\end{array}\right. \text { at } x=\pi\)

Solution:

Question 29.

The function is defined by

f(x) = \(\left\{\begin{array}{l}

k x+1, \text { if } x \leq 5 \\

3 x-5, \text { if } x>5

\end{array}\right. \text { at } x=5\)

Solution:

L.H.L. = \(\lim _{x \rightarrow 5^{-}}\) (kx + 1) = 5k +1,

f(5) = k . 5 + 1 – 5k + 1

and

R.H.L. = \(\lim _{x \rightarrow 5^{+}}\) (3x – 5) = 10.

f is continuous, if L.H.L. = R.H.L. = f(5).

∴ 5k + 1 = 10

∴ k = \(\frac { 9 }{ 5 }\).

Question 30.

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

f(x) = \(\left\{\begin{array}{cl}

5, & \text { if } x \leq 2 \\

a x+b, & \text { if } 2<x<10 \\

21, & \text { if } x \geq 10

\end{array}\right.\)

is a continous function.

Solution:

At x = 2, L.H.L. = \(\lim _{x \rightarrow 2^{-}}\) (5) = 5,

f(2) = 5

and R.H.L. = \(\lim _{x \rightarrow 2^{+}}\) (ax + b) = 2a + b.

f is continuous at x = 2, if 2a + b = 5 … (1)

At x = 10, L.H.L. = \(\lim _{x \rightarrow 10^{-}}\) f(x) = \(\lim _{x \rightarrow 10^{-}}\) (ax + b)

= 10a + b

and R.H.L. = \(\lim _{x \rightarrow 10^{+}}\) f(x) = \(\lim _{x \rightarrow 10^{+}}\) (21) = 21.

Also, f(10) = 21.

∴ f is continuous at x = 10, if 10a + b = 21. … (2)

Subtracting (1) from (2),

8a = 21 – 5 = 16. ∴ a = 2.

From (1), 2 x 2 + b – 5. ∴ b = 1.

Hence, a = 2 and b = 1.

Question 31.

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

Solution:

Now, f(x) = cos x².

Let g(x) = cos x and h(x) = x².

∴ (goh) (x) = g(h(x)) = cos x².

Now, g and h both are continuous for all x ∈ R.

∴ f(x) = (goh)(x) = cos x² is also continuous at all x ∈ R.

Question 32.

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

Solution:

Let g(x) = | x | and f(x) – cos x.

∴ f(x) = (goh)(x) = g(h(x))

= g(cos x) = | cos x|.

Now, g{x) = | x | and h(x) = cos x

both are continuous for all values of x ∈ R.

∴ (goh)(x) is also continuous.

Hence, f(x) = (goh)(x)

= | cos x | is continuous for all values of x ∈ R.

Question 33.

Examine if sin | x | is a continuous function.

Solution:

Let g(x) = sin x and h(x) = | x |.

∴ (goh)(x) = g(h(x)) = g(| X |) = sin | x | = f(x).

Now, g(x) = sin x and h(x) = | x |

both are continuous for all x ∈ R.

∴ f(x) = (goh)(x)

= sin | x | is continuous at all x ∈ R.

Question 34.

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

Solution:

∴ f is continous at x = 0

⇒ There is no point of discontinuity.

Hence, f is continuous for all x ∈ R.