RD Sharma Class 12 Ex 10.1 Solutions Chapter 10 Differentiability

Leave a Comment / Class 12 Maths / By

Here we provide RD Sharma Class 12 Ex 10.1 Solutions Chapter 10 Differentiability for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 10.1 Solutions Chapter 10 Differentiability book pdf download. Now you will get step-by-step solutions to each question.

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 = 0, if 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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *