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,
Df = 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,
Df = R – {odd multiple of \(\frac{\pi}{2}\)}
as x → \(\frac{\pi^{-}}{2}\)
⇒ sec x → + ∞ and as x2 → \(\frac{\pi^{+}}{2}\)
⇒ 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) ;
Df = R
Let c ∈ Df 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 Df.
∴ f is derivable at every point of its domain.
Hence f is a derivable function.
(ii) Let f(x) = cos x;
Df = R
Let c ∈ Df 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 ;
Df = R – {odd multiple of \(\frac{\pi}{2}\)}
Let c ∈ Df 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 ;
Df = R – {even multiple of \(\frac{\pi}{2}\)}
Let c ∈ Df be any arbitrary element.
which exists finitely.
Thus f is differentiable at x = c but c be any arbitrary element of Df.
∴ f is differentiable at every point of its domain.
Thus f(x) is a derivable function.
(v) Let f(x) = sec x;
Df = R – {odd multiple of \(\frac{\pi}{2}\)}
Let c ∈ Df 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 Df = R – {even multiple of \(\frac{\pi}{2}\)}
Let c ∈ Df 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, Df = 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) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) ax2
= 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 g1 (h) = h
and g2(h) = cos \(\frac{1}{h}\)
\(\underset{h \rightarrow 0}{\mathrm{Lt}}\) g1 (h) = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h = 0
Here |g2 (h)| = |cos \(\frac{1}{h}\)| ≤ 1 ∀ h ∈ R – {0}
⇒ g2 (h) is bounded in the detected ngd of 0.
∴ \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) g1 (h) g2 (h) = 0
⇒ \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h cos \(\frac{1}{h}\) = 0
Thus f(x) is differentiable at x = c.