Students can track their progress and improvement through regular use of ISC Maths Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.2

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.2

Question 1.

Is the function f defined by f(x) = |x| continuous at

(i) x = 0

(ii) x = 1

(iii) x = \(\frac {1}{2}\) ?

Solution:

(i) Here, \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) [x]

= 0 – 1

= – 1

and \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) [x] = 0

∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\)

Thus, f is discontinuous at x = 0.

(ii) Given, f(x) = [x]

\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) [x]

= 1 – 1 = 0

∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)

Thus, f is discontinuous at x = 1.

(iii) \(\underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}}\) [x] = 0

[as x → \(\frac{1}{2}\)^{–}

⇒ x < \(\frac{1}{2}\)

⇒ [x] = 0]

\(\underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}}\) [x] = 0

[as x → \(\frac{1}{2}\)^{+}

⇒ x > latex]\frac{1}{2}[/latex]

∴ [x] = 0]

Aliter:

\(\mathrm{Lt}_{x \rightarrow \frac{1}{2}}\) f(x) = \(\mathrm{Lt}_{x \rightarrow \frac{1}{2}}\)[x]

= [latex]\frac{1}{2}[/latex] = 0

∴ \(\underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}}\) f(x)

= 0

= f(\(\frac{1}{2}\))

= [latex]\frac{1}{2}[/latex]

= 0

Thus f is continuous at x = \(\frac{1}{2}\).

Examine the following (2 to 7) functions for continuity :

Question 2.

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

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

x^2 & \text {, if } x<0 \end{array}\right.\) (NCERT) (ii) f(x) = \(\left\{\begin{array}{lll} x+5 & \text { if } & x \leq 1 \\ x-5 & \text { if } & x>1

\end{array}\right.\) (NCERT)

Solution:

(i) Clearly D_{f} = R, so we examine the continuity of f at all x ∈ R.

Let c ∈ R be arbitrary real number. Then three cases arises.

Case – I:

When c > 0, f(c) = c

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x = c = f(c)

∴ f is continuous for all c > 0

Case – II:

When c < 0, f(c) = c^{2}

∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x^{2}

= c^{2}

= f(c)

∴ f is continuous for all c < 0

Case – III:

When c = 0 then f(c) = f(0) = 0

\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x^{2} = 0 = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0

∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = 0

Thus f is continuous at x = 0

Thus on combining all three cases, f is continuous ∀ c ∈ R i.e. at all x ∈ R.

Hence, f is a continuous function.

(ii) Clearly D_{f} = R, So we examine f for continuity at all x ∈ R. Let c ∈ R be any arbitrary real number. Three cases arises.

Case – I:

When c < 1, f(c) = c + 5

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x + 5

= c + 5

= f(c)

∴ f is continuous for all c < 1. Case – II: When c > 1, f(c) = c – 5

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x – 5

= c – 5

= f(c)

∴ f is continuous for all c > 1

Case – III:

When c = 1, then f(c) = f(1)

= 1 + 5 = 6

L.H.L. = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x + 5

= 1 + 5

= 6

R.H.L. = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) x – 5

= 1 – 5

= – 4

∴ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x)

∴ f is discontinuous at x = 1.

Hence on combining all threee cases, f is continuous for all c ∈ R – {1} i.e. ∀ x ∈ R – {1}

Question 3.

(i) f(x) = \(\left\{\begin{array}{cc}

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

\frac{3}{2} x & \text {, if } x \geq 2

\end{array}\right.\) (NCERT)

(ii) f(x) = \(\left\{\begin{array}{cc}

\sin x-\cos x & , x \neq 0 \\

-1 & , x=0

\end{array}\right.\) (NCERT)

Solution:

(i) When x < 2,

f(x) = 2x – 1, which is polynomial in x and hence continuous everywhere.

∴ f(x) is continuous for each x < 2. Where x > 2 ;

f(x) = \(\frac{3 x}{2}\) which is polynomial in x and hence continuous for each x > 2.

at x = 2 :

L.H.L = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) 2x – 1

= 4 – 1

= 3

and R.H.L = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) \(\frac{3 x}{2}\)

= \(\frac{3 \times 2}{2}\) = 3

also f(2) = 3

∴ L.H.L = R.H.L = f(2)

Hence f(x) is continuous at x = 2.

Thus f(x) is continuous everywhere.

(iii) When x ≠ 0 i.e. x < 0 and x > 0

f (x) = sin x – cos x,

since sin x and cos x.

both are continuous everywhere

∴ f(x) = sin x – cos x is also continuous for each x ≠ 0.

since difference of two continuous functions is continuous.

at x = 0 :

\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) f(- h)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [sin (- h) – cos (- h)]

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [- sin h – cos h]

= 0 – 1

= – 1

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) f(h)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [sin h – cos h]

= 0 – 1

= – 1

and f(0) = – 1

∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(0) = – 1

Thus f is continuous at x = 0.

Hence f(x) is continuous everywhere and nowhere discontinuous.

Question 4.

(i) f(x) = |x| (NCERT)

(ii) f(x) = |x – 5| (NCERT)

Solution:

(i) Modulus function is given by

|x| = \(\left[\begin{array}{rl}

-x & ; \quad x<0 \\

x & ; \quad x \geq 0

\end{array}\right.\)

Here D_{f} = R

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) (0 + h) = 0

\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x

= – \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) (0 – h) = 0

∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)

Thus, modulus function is continuous at x = 0

f(x) = |x| is continuous at x = 0.

(ii) f(x) = \(\left\{\begin{array}{cc}

\frac{x}{x}=1 ; & x>0 \\

\frac{x}{x}=-1 ; & x<0 \\

0 & ; x=0

\end{array}\right.\)

When x < 0 :

f(x) = – 1, which is a constant function and hence continuous everywhere.

∴ f(x) is continuous for all x < 0.

When x > 0 ;

f(x) = 1, which is a constant function and hence continuous for all each x > 0.

at x = 0;

L.H.L = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) – 1

= – 1

and R.H.L = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 1 = 1

∴ L.H.L. ≠ R.H.L.

Thus, f(x) is not continuous at x = 0.

Hencef(x) is continuous at each point except x = 0.

Question 5.

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

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

Solution:

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

∴ D_{f} = R – {- 5}

Let c be any arbitrary point of its domain.

∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow c} \frac{x^2-25}{x+5}\)

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

= c – 5

= f(c)

Thus f is continuous at x = c, but c is arbitrary.

Hence f is continuous at every point of its domain.

(ii) Given, f(x) = \(\frac{1}{x-5}\),

Now f is not defined at x = 5.

∴ D_{f} = R – {- 5}.

Let c be any arbitrary point of its domain.

∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow c} \frac{1}{x-5}\)

= \(\frac{1}{c-5}\)

= f(c)

Therefore f is continuous at x = c, but c is arbitrary.

Hence f is continuous at every point of its domain.

Question 6.

(i) f(x) = \(\left\{\begin{array}{cc}

\frac{\sin 2 x}{x}, & \text { if } x>0 \\

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

\end{array}\right.\) (NCERT)

(ii) f(x) = \(\left\{\begin{array}{cc}

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

0, & x=0

\end{array}\right.\) (NCERT)

Solution:

(i) f(x) = \(\left\{\begin{array}{cc}

\frac{\sin 2 x}{x}, & \text { if } x>0 \\

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

\end{array}\right.\)

Here D_{f} = R, so we examine f for continuity at all x ∈ R.

Let c ∈ Rbe any arbitrary real number. Three cases arises.

Case – I :

When c < 0 ; f(c) = 2

and \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2

= 2 = f(c)

∴ f is continuous for all c < 0. Case – II : When c > 0 ;

f(c) = \(\frac{\sin 2 c}{c}\)

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) \(\frac{\sin 2 x}{x}\)

= \(\frac{\sin 2 c}{c}\)

= f(c)

∴ f is continuous for all c > 0.

Case – III :

When c = 0

Then f(c) = f(0) = 2

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\sin 2 x}{x}\)

= \(\ {Lt}_{x \rightarrow 0} 2 \times \frac{\sin 2 x}{2 x}\)

= 2 \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\sin 2 x}{x}\)

= 2 × 1 = 2

= f(0)

[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]

∴ f is continuous for all c = 0.

So on combining all three cases, f is continous for all c ∈ R i.e. for all x ∈ R.

(ii) f(x) = \(\left\{\begin{array}{cc}

\frac{x}{x}=1 ; & x>0 \\

\frac{x}{x}=-1 & ; x<0 \\

0 & ; x=0

\end{array}\right.\)

When x < 0

f(x) = – 1, which is a constant function and hence continuous everywhere.

∴ f(x) is continuous for all x < 0. When x > 0 ;

f(x) = 1. which is a constant function and hence continuous for all each x > 0.

at x = 0;

L.H.L = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) – 1 = – 1

and R.H.L. = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 1 = 1

∴ L.H.L ≠ R.H.L

Thus, f(x) is not continuous at x = 0.

Hence f(x) is continuous at each point except x = 0.

Question 7.

(i) f(x) = \(\left\{\begin{array}{ccc}

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

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

\end{array}\right.\) (NCERT)

(ii) f(x) = tan x, 0 ≤ x ≤ \(\frac{\pi}{4}\).

Solution:

(i) When x < 1,

f(x) = x^{10} – 1,

which is polynomial in x and hence continuous for each x < 1. When x > 1,

f(x) = x^{2},

which is polynomial in x and hence continuous for each x > 1.

at x = 1

\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) x^{10}

= (1)^{10} – 1

= 1 – 1 = 0

\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) x^{2}

= 1^{2} = 1

and f(1) = 1^{10} – 1 = 0

∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)

Hence f(x) is discontinuous at x = 1.

Thus, the function f(x) is continuous everywhere except x = 1.

(ii) f(x) = tan x

given D_{f} = [0, \(\frac{\pi}{4}\)]

Let c ∈ D_{f} be any arbitrary point

Here, \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) tan x

= tan c

= f(c)

∴ f is continuous at x = c but c is any arbitrary point.

Hence f is continuous at every point of its domain.

Locate the points discontinuity (if any) of the following (8 to 10) functions:

Question 8.

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

2 x+3, \text { if } x \leq 2 \\

2 x-3, \text { if } x>2

\end{array}\right.\) (NCERT)

(ii) f(x) = \(\begin{cases}x^3-3, & \text { if } x \leq 2 \\ x^2+1, & \text { if } x>2\end{cases}\) (NCERT)

Solution:

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

2 x+3, \text { if } x \leq 2 \\

2 x-3, \text { if } x>2

\end{array}\right.\)

Here D_{f} = R

Since examine the continuity of f at all x ∈ R.

Let c ∈ R be any arbitrary point.

Case – I:

When c < 2,

f(x) = 2c + 3

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (2x + 3)

= 2c + 3

= f(c)

∴ f is continuous for all c < 2.

Case – II:

When c > 2,

f(c) = 2c + 3

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2x – 3

= 2c – 3

= f(c)

∴ f is continuous for all c > 2.

Case – III:

When c = 2,

f(c) = 2c – 3

\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) (2x – 3)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [2 (2 + h) – 3]

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [2h + 1]

= 1

\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) (2x + 3)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [2 (2 – h) + 3]

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [7 – 2h]

= 7

∴ \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x)

Thus f is discontinuous at x = 2.

Thus on combining all three cases, f is continuous for all x = c except x = 2 but c is arbitrary.

Hence f is continuous everywhere on R – {2} and has one point of discontinuity at x = 2.

(ii) Given f(x) = \(\begin{cases}x^2-3 & ; \quad x \leq 2 \\ x^2+1 & ; \quad x>2\end{cases}\)

Here D_{f} = R, so we examine the continuity of function f at all x ∈ R.

Let c ∈ R be any arbitrary point.

Three cases arises.

Case – I :

When c < 2, f(c) = c^{2} – 3

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x^{2} – 3

= c^{2} – 3

= f(c)

∴ f is continuous for all c < 2. Case – II : When c > 2,

f(c) = c^{2} + 1

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x^{2} + 1)

= c^{2} + 1

= f(c)

∴ f is continuous for all x = c > 2.

Case – III:

When c = 2

\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\)

= \(\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}\) [(2 + h)^{2} + 1] = 5

\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) (x^{3} – 3)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [(2 – h)^{3} – 3]

= 8 – 3 = 5

∴ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = f(2)

Hence f is continuous everywhere on R.

∴ f is continuous everywhere on R.

Hence it has no point of discontinuity.

Question 9.

(i) f(x) = \(\left\{\begin{array}{cc}

\frac{x^4-16}{x-2}, & x \neq 2 \\

16, & x=2

\end{array}\right.\)

(ii) f(x) = \(\left\{\begin{array}{cc}

\frac{\sin 3 x}{x}+\cos x & , x>0 \\

4-3 x & , x \leq 0

\end{array}\right.\)

Solution:

(i) We note that D_{f} = R, then we have to examine f for continuity for all x ∈ R and c ∈ R be any arbitrary element.

\(\ {Lt}_{x \rightarrow 2} \frac{x^4-16}{x-2}\)

= \(\ {Lt}_{x \rightarrow 2} \frac{\left(x^2-4\right)\left(x^2+4\right)}{x-2}\)

= \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) (x + 2) (x^{2} + 4)

= 4 . 8

= 32

also f(2) = 16

∴ \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) f(x) ≠ f(2).

Thus f is discontinuous at x = 2.

at c ≠ 2 be any arbitrary point, then f(c) = \(\frac{c^4-16}{c-2}\)

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{x^4-16}{x-2}\)

= \(\frac{c^4-16}{c-2}\)

= f(c)

∴ f is continuous at x = c ≠ 2

f(x) has only one point of discontinuity at x = 2.

(ii) Here D_{f} = R, so we examine f for continuity at all x ∈ R

Let c ∈ R be any real number. Three cases arise.

Case – I:

When c < 0, f(c) = 4 – 3c

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 4 – 3x

= 4 – 3c

= f(c)

∴ f is continuous for all c < 0

Case – II:

When c > 0,

f(c) = \(\frac{\sin 3 c}{c}\) + cos c

∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) \(\frac{\sin 3 x}{x}\) + cos x

= \(\frac{\sin 3 c}{c}\) + cos c

= f(c)

∴ f is continuous for all c > 0

Case – III:

When c = 0

Then f(c) = f(0)

= 4 – 3 × 0

= 4

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (\(\frac{\sin 3 x}{x}\) + cos x)

= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) 3 (\(\frac{\sin 3 x}{x}\)) + \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) cos x

= 3 × 1 + 1 = 4

[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]

\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) 4 – 3x

= 4 – 3 × 0 = 4

also f(0) = 4

∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)

Thus f is continuous at x = 0

Hence on combining all three cases, f is continuous for all c ∈ R i.e. for all x ∈ R.

Thus f be a continuous function.

Hence there is no point for which f is discontinuous.

Question 10.

(i) f(x) = \(\left\{\begin{array}{ccc}

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

\end{array}\right.\) (NCERT)

(ii) f(x) = \(\left\{\begin{array}{cc}

x+2, & x \leq 1 \\

x-2, & 1<x<2 \\

0, & x \geq 2

\end{array}\right.\)

Solution:

(i) When x < 0 and x > 1

f(x) = 2x and 4x, which is polynomial in x and hence continuous everywhere.

∴ f(x) is continuous for each x < 0 and x > 1.

When 0 < x < 1 ; f(x) = 0, which is a constant function.

and hence continuous everywhere.

∴ f(x) is continuous for each x ∈ (0, 1).

at x = 0 ;

L.H.L. = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 2x = 0

R.H.L. = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = 0

and f(0) = 0

∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(0)

Thus f(x) is continuous at x = 0.

at x = 1 ;

L.H.L. = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) 0 = 0

R.H.L. = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\)

= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) 4x

= 4 × 1 = 4

∴ L.H.L. ≠ R.H.L.

Thus f(x) is continuous at x = 1.

(ii) D_{f} = R;

So we examine for continuity for all values of x ∈ R.

Let c ∈ R be any real number.

Five cases arise.

Case – I :

When c < 1

then f(c) = c + 2

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x + 2

= c + 2

= f(c)

∴ f is continuous for all c < 1.

Casae – II :

When 1 < c < 2

then f(c) = c – 2

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x – 2)

= c – 2

= f(c)

∴ f is continuous for all c where 1 < c < 2.

Case – III:

When c > 2 then f(c) = 0

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 0

= 0 = f(c)

∴ f is continuous for all c where c < 2.

Case – IV:

at c = 1 then

f(c) = f(1)

= 1 + 2 = 3

\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) x – 2

= 1 – 2 = – 1

\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x + 2

= 1 + 2 = 3

∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) Thus f is discontinuous at x = c = 1.

Case – V:

at x = c = 2 then f(c) = f(2) = 0 \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) x – 2

= 2 – 2 = 0

\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 0 = 0

∴ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x)

= f(0) = 0

Thus f is continuous at x = c = 2.

On combining all five cases, f is discontinuous at x = 1.

Question 11.

Find the value of the constant k so that the function f defined by f(x) = \(\begin{cases}k x^2 & , x \leq 2 \\ x-3, & x>2\end{cases}\) may be continuous.Solution:

at x = 2:

\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) x – 3

= 2 – 3

= – 1

and \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) kx^{2}

= \(\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}\) k (2 – h)^{2}

= 4k

As f(x) be a continuous function.

So f(x) is also continuous at x = 2.

∴ \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(2)

⇒ – 1 = 4k

⇒ k = \(\frac{1}{4}\).

Question 12.

For what choice a and b are the following functions continuous:

(i) f(x) = \(\left\{\begin{array}{ccc}

-5 & \text {, if } & x \leq-1 \\

a x-b & \text {, if } & -1<x<3 \\ 7 & \text {, if } & x \geq 3 \end{array}\right.\) (ii) f(x) = \(\left\{\begin{array}{cc} x^2 & , x \leq 0 \\ a x+b & , x>0

\end{array}\right.\)

Solution:

(i) Given f(x) = \(\left\{\begin{array}{ccc}

-5 & \text {, if } & x \leq-1 \\

a x-b & \text {, if } & -1<x<3 \\ 7 & \text {, if } & x \geq 3 \end{array}\right.

\) Since f be a continuous function.

Thus f must be continuous at x = – 1 and x = 3.

at x = – 1 :

\(\underset{x \rightarrow-1^{-}}{\ {Lt}}\) f(x) = \(\underset{x \rightarrow-1^{-}}{\ {Lt}}\) – 5

= – 5 \(\underset{x \rightarrow-1^{+}}{\ {Lt}}\) f(x)

= \(\underset{x \rightarrow-1^{+}}{\ {Lt}}\) ax – b

= – a – b

also f (- 1) = – a – b

Since f is continuous at x = – 1

∴ \(\underset{x \rightarrow-1^{-}}{\ {Lt}}\) f(x) = \(\underset{x \rightarrow-1^{+}}{\ {Lt}}\)

f(x) = f(- 1)

∴ – 5 = – a – b ⇒ a + b = 5 …………(1)

at x = 3 :

\(\underset{x \rightarrow 3^{+}}{\ {Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\ {Lt}}\) 7

= 7 \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) f(x)

= \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) ax – b

= 3a – b

Also f(3) = 7

Since f is continuous at x = 3.

∴ \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) f(x)

= \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) f(x) = f(3)

∴ 3a – b = 7 ……….(2)

On solving (1) and (2)

simultaneously, we get a = 3 and b = 2.

(ii) Given, f(x) = \(\left\{\begin{array}{cc} x^2 & , x \leq 0 \\ a x+b & , x>0

\end{array}\right.\)

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (ax + b)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) ah + b = b

\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x^{2}

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) (- h)^{2} = 0

Also f(0) = 0

Since the function f(x) is continuous everywhere.

∴ function is continous at x = 0

Thus \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)

∴ b = 0 and a can have any value.