Class 12 RD Sharma Solutions - Chapter 10 Differentiability - Exercise 10.1
Last Updated :
23 Aug, 2024
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≥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≤3\\2x^2+5,x>3\end{cases} is differentiable at x = 3.
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≤1\\2x^2-2\ \ \ \ \ \ \ \ ,1<x≤2\\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≤x≤1\\-x-x+1,1<x<2\end{cases}
f(x) = \begin{cases}2x+1,-1<x<0\\1,0≤x≤1\\-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≤1\\2-x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,1≤x≤2\\-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≤1\\bx+2,x>1\end{cases} is differentiable at each real value of x.
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≥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|≥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
Summary
Exercise 10.1 focuses on the concept of differentiability of functions. It covers topics such as determining if a function is differentiable at a point or in an interval, finding the value of constants to make a function differentiable, and understanding the relationship between continuity and differentiability. Students are expected to apply the definition of differentiability, use limit concepts, and analyze function behavior at critical points.
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