RD Sharma Class 12 Ex 10.1 Solutions Chapter 10 Differentiability

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter10
Exercise10.1
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 10.1 Solutions Chapter 10 Differentiability

Question 1. Show that f(x) = |x – 3| is continuous but not differentiable at x = 3.

Solution:

f(x) = |x - 3| = \begin{cases}-(x - 3)\ \ \ , x<3\\x - 3\ \ \ \ \ \ \ \   , x\ge 3\end{cases}

f(3) = 3 – 3 = 0

LHL = lim_{x\to3^-}f(x)
= lim_{h\to0}f(3-h)

= lim_{h\to0}(h)

= 0

RHL = lim_{x\to3^+}f(x)
= lim_{h\to0}f(3+h)
=  lim_{h\to0}(h)

= 0

Since LHL = RHL, f(x) is continuous at x = 3.

Now, LHD at (x = 3) = lim_{x\to3^-}\frac{f(x)-f(3)}{x-3}

= lim_{h\to0}\frac{h}{-h}

= –1

RHD at (x = 3) = lim_{x\to3^+}\frac{f(x)-f(3)}{x-3}
= lim_{h\to0}\frac{h}{h}

= 1

Since (LHD at x = 3) ≠ (RHD at x = 3)

f(x) is continuous but not differentiable at x =3.

Question 2. Show that f (x) = x1/3 is not differentiable at x = 0.

Solution:

(LHD at x = 0) = lim_{x\to0^-}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{f(0-h)-f(0)}{0-h-0}
= lim_{h\to0}(-1)^{-2/3}h^{-2/3}

= Undefined

(RHD at x = 0) = lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{f(0+h)-f(0)}{0+h-0}
= lim_{h\to0}h^{-2/3}

= Undefined

Clearly LHD and RHD do not exist at 0.

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

Question 3. Show that f(x) = \begin{cases}12x-13,x\le3\\2x^2+5,x>3\end{cases}” height=”79″ width=”281″><strong> is differentiable at x = 3.</strong><span class=

Solution:

(LHD at x = 3) = lim_{x\to3^-}\frac{f(x)-f(3)}{x-3}

= lim_{h\to0}\frac{(-12h)}{-h}

= 12

RHD at x = 3 = lim_{x\to3^+}\frac{f(x)-f(3)}{x-3}

= lim_{h\to0}\frac{h(2h+12)}{h}

= 12

Since LHL = RHL

f(x) is differentiable at x = 3.

Question 4. Show that the function f is defined as follows is continuous at x = 2, but not differentiable thereat:

f(x) = \begin{cases}3x-2\ \ \ \ \ \ \ \ \ \ ,0<x\le1\\2x^2-2\ \ \ \ \ \ \ \ ,1<x\le2\\5x-4\ \ \ \  \ \ \ \  \ \  ,x>2\end{cases}

Solution:

f(2) = 2(2)2 – 2 = 6

LHL = lim_{x\to2^-}f(2-h)
= lim_{h\to0}[2(2-h)^2-(2-h)]

= 8 – 2 

= 6

RHL = lim_{x\to2^+}f(x)
= lim_{h\to0}f(2+h)
= lim_{h\to0}5(2+h) - 4

= 6

Clearly LHL = RHL at x = 2

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

Question 5. Discuss the continuity and differentiability of the function f(x) = |x| + |x -1| in the interval of (-1, 2).

Solution:

f(x) = \begin{cases}x+x+1\ \ \ ,-1<x<0\\1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,0\lex\le1\\-x-x+1,1<x<2\end{cases}
f(x) = \begin{cases}2x+1,-1<x<0\\1,0\lex\le1\\-2x+1,1<x<2\end{cases}

(LHD at x = 0) = lim_{x\to0^-}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{2x}{x}

= 2

(RHD at x = 0) = lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{0}{x}

= 0

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

Question 6. Find whether the following function is differentiable at x = 1 and x = 2 or not.

f(x) = \begin{cases}x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ ,x\le1\\2-x\ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ ,1\lex\le2\\-2+3x-x^2\ \ ,x>2\end{cases}.

Solution:

(LHD at x = 1) = lim_{x\to1^-}\frac{f(x)-f(1)}{x-1}

= lim_{x\to1^-}\frac{x-1}{x-1}

= 1

(RHD at x = 1) = lim_{x\to1^+}\frac{f(x)-f(1)}{x-1}

= lim_{x\to1^+}\frac{1-x}{x-1}

= –1

Clearly LHD ≠ RHD at x = 1

So f(x) is not differentiable at x = 1.

(LHD at x = 2) = lim_{x\to2^-}\frac{f(x)-f(2)}{x-2}

= lim_{x\to2^-}\frac{2-x}{x-2}

= –1

(RHD at x = 2) = lim_{x\to2^+}\frac{f(x)-f(2)}{x-2}

= lim_{x\to2^+}(1-x)

= –1

Clearly LHL = RHL at x = 2

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

Question 7(i). Show that f(x) = \begin{cases}x^msin[\frac{1}{x}],x≠0\\0\ \ \ \ \ \  \ \ \ \ \ \  \ \ ,x=0\end{cases}  is differentiable at x = 0, if m>1.

Solution:

(LHD at x = 0) = lim_{x\to0^-}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{f(0-h)-f(0)}{0-h-0}
= lim_{h\to0}\frac{(-h)^msin[\frac{-1}{h}]-0}{-h}

= 0 × k

= 0

(RHD at x = 0) = lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{(0+h)-f(0)}{0+h-0}
= lim_{h\to0}\frac{(h)^msin[\frac{-1}{h}]-0}{-h}

=  0 × k

= 0

Clearly LHL = RHL at x = 0

Hence f(x) is differentiable at x = 0.

Question 7(ii) Show that f(x) = \begin{cases}x^msin[\frac{1}{x}],x≠0\\0\ \ \ \ \ \  \ \ \ \ \ \  \ \ ,x=0\end{cases}  is not differentiable at x = 0, if 0<m<1.

Solution:

(LHD at x = 0) = lim_{x\to0^-}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{f(0-h)-f(0)}{0-h-0}
= lim_{h\to0}\frac{(-h)^msin[\frac{-1}{h}]-0}{-h}

= Not defined

(RHD at x = 0) = lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{(0+h)-f(0)}{0+h-0}
= lim_{h\to0}\frac{(h)^msin[\frac{-1}{h}]-0}{-h}

= Not defined

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

Question 7(iii) Show that f(x) = \begin{cases}x^msin[\frac{1}{x}],x≠0\\0\ \ \ \ \ \  \ \ \ \ \ \  \ \ ,x=0\end{cases}  is not differentiable at x = 0if m≤0.

Solution:

(LHD at x = 0) = lim_{x\to0^-}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{f(0-h)-f(0)}{0-h-0}
= lim_{h\to0}\frac{(-h)^msin[\frac{-1}{h}]-0}{-h}

= Not defined

(RHD at x = 0) = lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{(0+h)-f(0)}{0+h-0}
= lim_{h\to0}\frac{(h)^msin[\frac{-1}{h}]-0}{-h}

= Not defined

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

Question 8. Find the value of a and b so that the function f(x)= \begin{cases}x^2+3x+a,x\le1\\bx+2,x>1\end{cases} ” height=”79″ width=”313″><strong> is differentiable at each real value of x.</strong><span class=

Solution:

(LHD at x = 1) = lim_{x\to1^-}\frac{f(x)-f(1)}{x-1}

= lim_{h\to0}\frac{h^2-5h}{-h}

= 5

(RHD at x = 2) = lim_{x\to1^+}\frac{f(x)-f(1)}{x-1}

= lim_{h\to0}\frac{b+bh+2-b-2}{h}

= b

Since f(x) is differentiable at x = 1,so

b = 5

Hence, 4 + a = b + 2

or, a = 7 – 4 = 3

Hence, a = 3 and b = 5.

Question 9. Show that the function f(x) = \begin{cases}|2x-3|[x]\ \ \ \ \ \ \ \ ,x\ge 1\\sin[\frac{πx}{2}]\ \ \ \ \ \ \ \ \ \ \ \  \ \ ,x<1\end{cases}  is notdifferentiable at x =1.

Solution:

(LHD at x = 1) = lim_{x\to1^-}\frac{f(x)-f(1)}{x-1}

= lim_{h\to0}\frac{f(1-h)-1}{1-h-1}
= lim_{h\to0}\frac{cos[πh/2]-1}{-h/2}

= 0

(RHD at x =1) =  lim_{x\to1^+}\frac{f(x)-f(1)}{x-1}

= lim_{h\to0}\frac{f(1+h)-f(1)}{1+h-1}
= lim_{h\to0}\frac{-2-2h+3-1}{h}

= –2

Since (LHD at x = 1) ≠ (RHD at x = 1)

f(x) is continuous but not differentiable at x =1.

Question 10. If f(x) = \begin{cases}ax^2-b\ \ \ \ \ \  \ ,|x|<1\\\frac{1}{|x|}\ \ \  \ \ \ \ \ \ \ \ \ \ \ \ ,|x|\ge 1\end{cases}  is differentiable at x = 1, find a and b.

Solution:

We know f(x) is continuous at x = 1.

So, a – b = 1                        …..(1)

(LHD at x = 1) = lim_{x\to1^-}\frac{f(x)-f(1)}{x-1}

= lim_{h\to0}\frac{a(1-h)^2-(a-1)-1}{-h}

Using (1), we get

= lim_{h\to0}(2a-ah)

= 2a

(RHD at x =1) = lim_{x\to1^+}\frac{f(x)-f(1)}{x-1}

= lim_{h\to0}\frac{-1}{1+h}

= –1

Since f(x) is differentiable, LHL = RHL

or, 2a = –1

a = –1/2

Substituting a = –1/2 in (1), we get,

b = –1/2 – 1

b = –3/2

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