Practicing ML Aggarwal Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.4 is the ultimate need for students who intend to score good marks in examinations.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.4

Question 1.

If a function is derivable at a point, is it necessary that it must be continuous at that point?

Solution:

Yes, every differentiable functions is continuous.

Let f(x) be differentiable at x = a.

∴ \(\underset{x \rightarrow a}{\mathrm{Lt}}\) \(\frac{f(x)-f(a)}{x-a}\) = f'(a) ……….(1)

We want to prove that f(x) is continuous at x = a

i.e. \(\underset{x \rightarrow a}{\mathrm{Lt}}\) f(x) = f(a)

Now \(\underset{x \rightarrow a}{\mathrm{Lt}}\) [f(x) – f(a)] = \(\underset{x \rightarrow a}{\mathrm{Lt}}\) \(\) (x – a)

= \(\) . \(\underset{x \rightarrow a}{\mathrm{Lt}}\) (x – a)

= f'(a) × (a – a) [using (1)]

= f'(a) × 0 = 0

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

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

Question 2.

If a function is continuous at a point, is it necessary that it must be derivable at that point?

Solution:

No, every continuous function is not differentiable.

e.g. f(x) = |x|

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

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

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

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

and f(0) = |0| = 0

∴ L.H.L. = R.H.L. = f(0) = 0

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

differentiability at x = 0 :

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

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

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

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

and f(0) = |0| = 0

∴ Lf'(0) ≠ Rf'(0)

Thus f(x) is not differentiable at x = 0

Hence every continuous function is not differentiable.

Question 3.

Is the function f(x) = |x| derivable at x = 0?

Solution:

Lf’ (0) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{f(x)-f(0)}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{|x|-|0|}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{-x}{x}\)

[as x → 0^{–}

⇒ x < 0

∴ |x| = – x]

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

Rf’ (0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{|x|-0}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x-0}{x}\) = 1

[∵ as x → 0^{+}

⇒ x > 0

⇒ |x| = + x]

∴ Lf'(0) = R f'(0)

Thus f(x) = |x| is not derivable at x = 0.

Question 4.

Is the function f(x) = cot x derivable at x = 0?

Solution:

Let f(x) = cot x,

D_{f} = R – {multiple of π}

Since x → 0^{+}, cot x → ∞

and if x → 0^{–}, cot x → – ∞,

∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) does not exist and hence f is discontinuous at x = 0

and hence f is not derivable at x = 0.

[since if a function is discontinuous then it never be differentiable]

Question 5.

Is the function f(x) = sec x derivable at x = \(\frac{\pi}{2}\)?

Solution:

Let f(x) = sec x,

D_{f} = R – {odd multiple of \(\frac{\pi}{2}\)}

as x → \(\frac{\pi^{-}}{2}\)

⇒ sec x → + ∞ and as x^{2} → \(\frac{\pi^{+}}{2}\)

⇒ sec x → – ∞

∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) f(x) does not exist.

Thus f is not continuous at x = \(\frac{\pi}{2}\) and hence f(x) is not derivable at x = \(\frac{\pi}{2}\).

[since if a function is discontinuous then it never be differentiable].

Question 6.

When a function is called derivable?

Solution:

A function f(x) is said to be differentiable at x = a

if \(\underset{x \rightarrow a}{\mathrm{Lt}}\) \(\frac{f(x)-f(a)}{x-a}\) exists finitely.

Question 7.

If a function is derivable at x = c, then write the value of \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x).

Solution:

Since f(x) is derivable at x = c

∴ f'(c) = \(\ {Lt}_{x \rightarrow c} \frac{f(x)-f(c)}{x-c}\)

Now \(\underset{x \rightarrow c}{\mathrm{Lt}}\) [f(x) – f(c)]

= \(\ {Lt}_{x \rightarrow c} \frac{f(x)-f(c)}{x-c}\) × (x – c)

= \(\ {Lt}_{x \rightarrow c} \frac{f(x)-f(t)}{x-c}\) . \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (x – c)

= f'(c) (c – c)

= f'(c) × 0 = 0

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

Question 8.

Which of the following functions are derivable?

(i) sin x

(ii) cos x

(iii) tan x

(iv) cot x

(v) sec x

(vi) cosec x

Solution:

(i) Let sin x = f(x) ;

D_{f} = R

Let c ∈ D_{f} be any arbitrary element

Now f'(c) = \(\ {Lt}_{n \rightarrow, n} \frac{f(c+h)-f(c)}{n}\)

= \(\ {Lt}_{h \rightarrow 0} \frac{\sin (c+h)-\sin c}{h}\)

= \(\ {Lt}_{h \rightarrow 0} \frac{2 \cos \left(\frac{c+h+c}{2}\right) \sin \left(\frac{c+h-c}{2}\right)}{h}\)

= \(\ {Lt}_{h \rightarrow 0} \frac{2 \cos \left(c+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}\)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) cos (c + \(\frac{h}{2}\)) 0 . \(\ {Lt}_{x \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}\)

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

= cos c . 1

which exists finitely.

Thus f is differentiable or derivable at x = c but c be any arbitrary element of D_{f}.

∴ f is derivable at every point of its domain.

Hence f is a derivable function.

(ii) Let f(x) = cos x;

D_{f} = R

Let c ∈ D_{f} be any arbitrary element.

f'(c) = \(\ {Lt}_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\)

= \(\ {Lt}_{h \rightarrow 0} \frac{\cos (c+h)-\cos c}{h}\)

= \(\ {Lt}_{h \rightarrow 0} \frac{-2 \sin \left(\frac{c+c+h}{2}\right) \sin \left(\frac{c+h-c}{2}\right)}{h}\)

= – 2 \(\ {Lt}_{h \rightarrow 0} \sin \left(c+\frac{h}{2}\right) \cdot \ {Lt}_{h \rightarrow 0} \frac{\sin \frac{h}{2}}{2 \times \frac{h}{2}}\)

= – sin c, which exists finitely

∴ f is derivable at x = c

but c be any arbitrary element of domain of f.

∴ f is differentiable at every point of its domain and hence f be a derivable function.

(iii) Let f(x) = tan x ;

D_{f} = R – {odd multiple of \(\frac{\pi}{2}\)}

Let c ∈ D_{f} be any arbitrary element.

Thus f is differentiable at x = c

but c be any arbitrary element of domain of f.

∴ f is differentiable at every point of its domain.

Hence f be a derivable function.

(iv) Let f(x) = cot x ;

D_{f} = R – {even multiple of \(\frac{\pi}{2}\)}

Let c ∈ D_{f} be any arbitrary element.

which exists finitely.

Thus f is differentiable at x = c but c be any arbitrary element of D_{f}.

∴ f is differentiable at every point of its domain.

Thus f(x) is a derivable function.

(v) Let f(x) = sec x;

D_{f} = R – {odd multiple of \(\frac{\pi}{2}\)}

Let c ∈ D_{f} be any arbitrary element.

∴ f is differentiable at x = c

but c be any arbitrary element of domain of f.

∴ f be differentiable at every point of its domain.

∴ f be a derivable function.

(vi) Let f(x) = cosec x

Then D_{f} = R – {even multiple of \(\frac{\pi}{2}\)}

Let c ∈ D_{f} be any arbitrary element.

Thus f is differentiable at x = c

but c be any arbitrary element of domain of f.

∴ f is differentiable at every point of its domain.

Hence f is derivable function.

Question 9.

Examine the function f(x) = \(\begin{cases}1+x & , \text { if } x \leq 2 \\ 5-x & \text { if } x>2\end{cases}\) for differentiability at x = 2.

Solution:

at x = 2

Lf'(2) = \(\ {Lt}_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}\)

= \(\ {Lt}_{x \rightarrow 2^{-}} \frac{1+x-(1+2)}{x-2}\)

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

= 1

Rf'(2) = \(\ {Lt}_{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}\)

= \(\ {Lt}_{x \rightarrow 2^{+}} \frac{5-x-(1+2)}{x-2}\)

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

= – 1

∴ Lf'(2) ≠ Rf'(2)

Thus, f(x) is not differentiable at x = 2.

Question 10.

If f(2) = 4 and f'(2) = 4, then evaluate \(\ {Lt}_{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\).

Solution:

Given f(2) = 4 and f'(2) = 4 …………(1)

\(\ {Lt}_{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}=\ {Lt}_{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2}\)

\(\ {Lt}_{x \rightarrow 2} \frac{(x-2) f(2)-2(f(x)-f(2))}{x-2}=\ {Lt}_{x \rightarrow 2} \frac{(x-2) f(2)}{x-2}-2 \ {Lt}_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}\)

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

[Since f'(a) = \(\ {Lt}_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\)]

= f(2) – 2 f'(2)

= 4 – 2 × 4

= – 4

Question 11.

Examine the following function for continuity at x = I and differentiability at x = 2. f(x) = \(\left\{\begin{array}{cc}

5 x-4, & 0<x<1 \\

4 x^2-3 x & , 1<x<2 \\

3 x+4 & , \quad x \geq 2

\end{array}\right.\)

Solution:

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

5 x-4, & 0<x<1 \\

4 x^2-3 x & , 1<x<2 \\

3 x+4 & , \quad x \geq 2

\end{array}\right.\)

Clearly f(x) is not defined at x = 1.

∴ f(x) is discontinuous at x = 1

and hence f(x) is not differentiable at x = 1.

at x = 2 ;

∴ Lf'(2) ≠ Rf'(2)

Thus f(x) is not differentiable at x = 2.

Question 12.

Show that the function f(x) = 2x – |x| is continuous at x = 0 but not differentiable at x = 0.

Solution:

Given f(x) = 2x – |x|

Continuity at x = 0:

Since, D_{f} = R

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

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

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

[∵ as x → 0^{+}

⇒ x > 0

∴ |x| = x]

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

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

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

[∵ as x → 0^{–}

⇒ x < 0

∴ |x| = – x]

Also f(0) = 2 × 0 – |0|

= 0 – 0

= 0

∴ f(x) is continuous at x = 0.

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

Differentiability at x = 0:

Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x}\)

= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{\{2 x-(-x)\}-0}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{3 x}{x}\)

= 3

Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{(2 x-x)-(0-0)}{x-0}\)

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

= 1

∴ Lf'(0) ≠ Rf'(0)

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

Hence f(x) is continuous at x = 0 but not differentiable at x = 0.

Question 12 (old).

Show that f(x) = |x – 5| is continuous but not differentiable at x = 5.

Solution:

Given f(x) = |x – 5|

= \(\left\{\begin{array}{cc}

x-5 & ; \quad x \geq 5 \\

-(x-5) & ; x<5 \end{array}\right.\) Continuity

at x = 5 :

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

= – (5 – 5)

= 0

and \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (x – 5)

= (5 – 5)

= 0.

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

Hence f(x) is continuous at x = 5.

Differentiability at x = 5 Lf'(5) = \(\ {Lt}_{h \rightarrow 0^{-}} \frac{f(5+h)-f(5)}{h}\)

= \(\ {Lt}_{h \rightarrow 0^{-}}-\frac{(5+h-5)-(5-5)}{h}\)

= \(\ {Lt}_{h \rightarrow 0^{-}} \frac{-h}{h}\)

= – 1 Rf'(5)

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{f(5+h)-f(5)}{h}\)

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{(5+h-5)-(5-5)}{h}\)

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{h}{h}\)

= 1

∴ Rf'(5) ≠ Lf'(5)

Thus, f(x) is not differentiable at x = 5

Hence f (x) is continuous but not differentiable at x = 5.

Question 13.

Show that f(x) = |x – 4| is continuous but not differentiable at x = 4. (ISC 2019) Solution: Given f(x) = |x – 4| = \(\left\{\begin{array}{cc} x-4 & ; x>4 \\

-(x-4) & ; x<4 \\

0 & ; x=4

\end{array}\right.\)

L.H.L. = \(\underset{x \rightarrow 4^{-}}{\mathrm{Lt}}\) f(x)

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

R.H.L. = \(\underset{x \rightarrow 4^{+}}{\mathrm{Lt}}\) f(x)

= \(\underset{x \rightarrow 4^{+}}{\mathrm{Lt}}\) x – 4 = 0

∴ L.H.L. = R.H.L.

⇒ \(\underset{x \rightarrow 4}{\mathrm{Lt}}\) f(x) = 0

and f(x) = 0 at x = 4

⇒ f(4) = 0

Thus, \(\underset{x \rightarrow 4}{\mathrm{Lt}}\) f(x) = f(4)

∴ f(x) is continuous at x = 4.

Differentiability at x = 4 :

Lf'(4) = \(\underset{x \rightarrow 4^{-}}{\mathrm{Lt}}\) \(\frac{f(x)-f(4)}{x-4}\)

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{f(4-h)-f(y)}{4-h-4}\)

[put x = 4 – h

as x → 4^{–}

⇒ x < 4

⇒ 4 – h < 4

⇒ – h < 0

⇒ h → 0^{+}]

= \(\underset{h \rightarrow 0^{+}}{\ {Lt}} \frac{|-h|-0}{-h}\)

= \(\underset{h \rightarrow 0}{\ {Lt}} \frac{h}{-h}\)

= – 1

and Rf'(4) = \(\ {Lt}_{h \rightarrow 0^{+}} \frac{f(x)-f(4)}{x-4}\)

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{f(4+h)-f(4)}{4+h-4}\)

[put x = 4 + h

as x → 4^{–}

⇒ x < 4

⇒ 4 – h < 4

⇒ – h < 0

⇒ h → 0^{+}]

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{|4+h-4|-0}{h}\)

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{|h|}{h}\)

= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{h}{h}\)

= 1

∴ Lf'(4) ≠ Rf'(4)

Thus f(x) is not differentiable at x = 4.

Hence f(x) is continuous but not differentiable at x = 4.

Question 14.

If the functionf (x) = |x – 3 | + |x – 4|, then show that f is not differentiable at x = 3 and x = 4.

Solution:

Given, f(x) = |x – 3 | + |x – 4|

= \(\left\{\begin{array}{ccc}

-(x-3)-(x-4) & ; & x<3 \\

+(x-3)-(x-4) & ; & 3 \leq x<4 \\

(x-3)+(x-4) & ; & x \geq 4

\end{array}\right.\)

∴ f(x) = \(\left\{\begin{array}{ccc}

-2 x+7 & ; & x<3 \\

1 & ; & 3 \leq x<4 \\

2 x-7 & ; & x \geq 4

\end{array}\right.\)

differentiability at x = 3:

Lf'(3) = \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) \(\frac{f(x)-f(3)}{x-3}\)

= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{-2 x+7-1}{x-3}\)

= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{-2(x-3)}{x-3}\)

= – 2

Rf'(3) = \(\ {Lt}_{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}\)

= \(\ {Lt}_{x \rightarrow 3^{+}} \frac{1-1}{x-3}\)

= 0

∴ Lf'(3) ≠ Rf'(3)

Differentiability at x = 4 :

Lf'(4) = \(\ {Lt}_{x \rightarrow 4^{-}} \frac{f(x)-f(4)}{x-4}\)

= \(\ {Lt}_{x \rightarrow 4^{-}} \frac{1-(8-7)}{x-4}\)

= 0

Rf'(4) = \(\ {Lt}_{x \rightarrow 4^{+}} \frac{f(x)-f(4)}{x-4}\)

= \(\ {Lt}_{x \rightarrow 4^{+}} \frac{2 x-7-(8-7)}{x-4}\)

= \(\ {Lt}_{x \rightarrow 4^{+}} \frac{2 x-8}{x-4}\)

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

= 2

∴ Lf'(4) ≠ Rf'(4)

Thus, f(x) is not differentiable at x = 4.

Hence f is not differentiable at x = 3 and x = 4.

Question 15.

Show that the function f is continuous at x = 1 for all values of a where f(x) = \(\left\{\begin{array}{cc}

a x^2+1, & x \geq 1 \\

x+a & , x<1

\end{array}\right.\).

Find its right and left hand derivatives at x = 1.

Hence, find the condition for the existence of the derivative at x = 1.

Solution:

Continuity at x = 1 :

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

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

= a + 1

also f(1) = a + 1

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

= f(1)

= a + 1

Thus f is continuous at x = 1, for all values of a.

Differentiability at x = 1 :

Now f(x) is differentiable at x = 1

∴ Lf'(1) = Rf'(1)

⇒ 1 = 2a

⇒ a = \(\frac{1}{2}\).

Question 16.

Prove that the function f(x) = \(\left\{\begin{array}{cc}

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

1, & x=0

\end{array}\right.\) is not differentiable at x = 0.

Solution:

Continuity at x = 0

\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\ {Lt}_{x \rightarrow 0^{-}} \frac{x}{|x|}\)

= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{x}{-x}\)

= – 1

[as x → 0^{–}

⇒ x < 0

∴ |x| = – x]

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

= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x}{x}\)

[as x → 0^{+}

⇒ x > 0

∴ |x| = x]

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

Thus f is discontinuous at x = 0

∴ f is not differentiable at x = 0

[since every differentiable function is continuous. So if a function is discontinuous at a point. Then it can’t be differentiable at that point.

Question 17.

Examine for continuity and differentiability, each of the following functions:

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

x \sin \frac{1}{x}, & x<0 \\ 0, & x \geq 0 \end{array} \text { at } x=0\right.\)

(ii) f(x) = \(\left\{\begin{array}{cc} |x| \sin \frac{1}{x}, & x>0 \\

0, & x \leq 0

\end{array} \text { at } x=0\right.\)

(iii) f(x) = \(\left\{\begin{array}{c}

(x-c)^2 \cos \frac{1}{x-c}, x \neq c \\

0, x=c

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

Solution:

(i) For continuity:

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

= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x sin \(\frac{1}{x}\)

= 0

[Let g(x) = x ;

h(x) = sin \(\frac{1}{x}\)

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

and | h(x) |= |sin \(\frac{1}{x}\)| ≤ 1 ∀ x ∈ R – {0}

Thus h(x) is bounded in the detected ngd of 0]

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

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

= 0

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

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

differentiability at x = 0 :

f’_{–} (0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{x \sin \frac{1}{x}-0}{x}\)

= \(\underset{x \rightarrow 0^{-}}{\ e{Lt}} \sin \frac{1}{x}\)

which does not exists

[∵ sin \(\frac{1}{x}\) oscillates between – 1 and 1]

f’_{+} (0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{0-0}{x}\)

= 0

f’_{–} (0) = f’_{+} (0)

Thus f(x) is not differentiable at x = 0.

(ii) Continuity at x = 0:

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

= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) |x| sin \(\frac{1}{x}\)

= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x sin \(\frac{1}{x}\) = 0

[let g(x) = x and h(x) = sin \(\frac{1}{x}\)

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

and | h(x) | = |sin \(\frac{1}{x}\)| ≤ 1 ∀ x ∈ x – {0}

∴ h(x) is bounded in the detected ngd of 0.

∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g (x) h (x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x sin \(\frac{1}{x}\)

= 0

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

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

∴ L.H.L. = R.H.L. = f(0) = 0

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

Differentiability at x = 0:

f’_{+} (0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{|x| \sin \frac{1}{x}-0}{x}\)

= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x \sin \frac{1}{x}}{x}\)

= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) sin \(\frac{1}{x}\)

which does not exists.

[∵ sin \(\frac{1}{x}\) is a number oscillates between – 1 and 1]

f’_{–} (0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)

= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{0-0}{x}\)

= 0

∴ f’_{+} (0) ≠ f’_{–} (0)

Therefore f(x) is not differentiable at x = 0.

(iii) Continuity at x = c:

\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x – c)^{2} cos \(\frac{1}{x-c}\)

= 0

Let g(x) = (x – c)^{2}

and h(x) = cos \(\frac{1}{x-c}\)

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

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

= 0

and |h(x)| = |cos \(\frac{1}{x-c}\)| ≤ 1 ∀ x ∈ R – {c}

Thus h(x) is bounded in the detected ngd of c

∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) g(x) h(x) = 0

⇒ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x – c)^{2} cos \(\frac{1}{x-c}\) = 0

and f(c) = 0

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

Differentiability at x = c:

f'(c) = \(\ {Lt}_{x \rightarrow c} \frac{f(x)-f(c)}{x-c}\)

= \(\ {Lt}_{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}\)

= \(\ {Lt}_{h \rightarrow 0} \frac{(c+h-c)^2 \cos \frac{1}{c+h-c}-0}{h}\)

= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h cos \(\frac{1}{h}\)

= 0

Let g_{1} (h) = h

and g_{2}(h) = cos \(\frac{1}{h}\)

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

Here |g_{2} (h)| = |cos \(\frac{1}{h}\)| ≤ 1 ∀ h ∈ R – {0}

⇒ g_{2} (h) is bounded in the detected ngd of 0.

∴ \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) g_{1} (h) g_{2} (h) = 0

⇒ \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h cos \(\frac{1}{h}\) = 0

Thus f(x) is differentiable at x = c.