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Question: Let \[a,b \in R\]and \(f:R \to R\) be defined by \(f(x) = a\cos \left( {\left| {{x^3} - x} \right|} ...

Let a,b∈Ra,b \in Rand f:R→Rf:R \to R be defined by f(x)=acos⁡(∣x3−x∣)+b∣x∣sin⁡(∣x3+x∣)f(x) = a\cos \left( {\left| {{x^3} - x} \right|} \right) + b\left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right). Then ff is
(A) Differentiate at x=0x = 0 if a=0a = 0 and b=1b = 1
(B) Differentiate at x=1x = 1 if a=1a = 1 and b=0b = 0
(C) Not Differentiate at x=0x = 0 if a=1a = 1 and b=0b = 0
(D) Not Differentiate at x=1x = 1 if a=1a = 1 and b=1b = 1

Explanation

Solution

To solve this problem we find the right and left hand derivatives of a given function at x=0x = 0 and x=1x = 1 by putting value of a,ba,b as given in options.
The left hand derivative of f(x)f(x) at x=ax = a is f′(a−)=lim⁡h→0f(a−h)−f(a)−hf{'}({a^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a - h) - f(a)}}{{ - h}}
And right hand derivative of f(x)f(x) at x=ax = a is f′(a+)=lim⁡h→0f(a+h)−f(a)hf{'}({a^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}

Complete step by step answer:
Given function: f(x)=acos⁡(∣x3−x∣)+b∣x∣sin⁡(∣x3+x∣)f(x) = a\cos \left( {\left| {{x^3} - x} \right|} \right) + b\left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)
Now we go through first option:
1. Differentiate at x=0x = 0 if a=0a = 0 and b=1b = 1
Now we put value of a and b in given function
From this we can say f(x)=∣x∣sin⁡(∣x3+x∣)f(x) = \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)
Now find derivative:
Left hand derivative at x=0x = 0
f′(0−)=lim⁡h→0f(0−h)−f(0)−hf{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}
So f′(0−)=lim⁡h→0∣−h∣sin⁡(∣−h3−h∣)−0−hf{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| { - h} \right|\sin \left( {\left| { - {h^3} - h} \right|} \right) - 0}}{{ - h}}
∵f(0)=0×sin⁡0=0\because f(0) = 0 \times \sin 0 = 0
∵∣−h∣=h\because \left| { - h} \right| = h
So f′(0−)=lim⁡h→0hsin⁡(∣−h3−h∣)−hf{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\sin \left( {\left| { - {h^3} - h} \right|} \right)}}{{ - h}}
Now f′(0−)=sin⁡(0)−1f{'}({0^ - }) = \dfrac{{\sin \left( 0 \right)}}{{ - 1}}
∵sin⁡0=0\because \sin 0 = 0
So f′(0−)=0f{'}({0^ - }) = 0
So left hand derivative at x=0x = 0 is 00
Similarly we find left hand derivative
For right hand derivative of f(x)f(x) at x=0x = 0 is f′(0+)=lim⁡h→0f(0+h)−f(0)hf{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}
⇒  \Rightarrow \;
⇒f′(0+)=lim⁡h→0∣h∣sin⁡(∣h3+h∣)−0h\Rightarrow f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| h \right|\sin \left( {\left| {{h^3} + h} \right|} \right) - 0}}{h}
Now we know ∣h∣=h\left| h \right| = h
And after putting limit value we get
⇒f′(0+)=sin⁡0\Rightarrow f{'}({0^ + }) = \sin 0
⇒f′(0+)=0\Rightarrow f{'}({0^ + }) = 0
So left hand derivative and right hand derivative are equal
f′(0+)=f′(0−)=0f{'}({0^ + }) = f{'}({0^ - }) = 0 so function is derivable at x=0x = 0
2. Differentiate at x=1x = 1 if a=1a = 1 and b=0b = 0
So put value of a,ba,b we get f(x)=1×cos⁡(∣x3−x∣)+0×∣x∣sin⁡(∣x3+x∣)f(x) = 1 \times \cos \left( {\left| {{x^3} - x} \right|} \right) + 0 \times \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)
f(x)=cos⁡(∣x3−x∣)f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)
So
f′(1−)=lim⁡h→0f(1−h)−f(1)−hf{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 - h) - f(1)}}{{ - h}}
f′(1−)=lim⁡h→0cos⁡(∣(1−h)3−(1−h)∣)−cos⁡(1−1)−hf{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(1 - h)}^3} - (1 - h)} \right|} \right) - \cos (1 - 1)}}{{ - h}}
Now we put limit
So f′(1−)=cos⁡(1−1)−cos⁡(0)−1f{'}({1^ - }) = \dfrac{{\cos \left( {1 - 1} \right) - \cos (0)}}{{ - 1}} and cos⁡0=1\cos 0 = 1
So f′(1−)=1−11=0f{'}({1^ - }) = \dfrac{{1 - 1}}{1} = 0
Similarly right hand limit f′(1+)=lim⁡h→0f(1+h)−f(1)hf{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 + h) - f(1)}}{h}
f′(1+)=lim⁡h→0cos⁡(∣(1+h)3−(1+h)∣)−cos⁡(1−1)hf{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(1 + h)}^3} - (1 + h)} \right|} \right) - \cos (1 - 1)}}{h}
Now putting limit value
f′(1+)=cos⁡(1−1)−cos⁡(0)1f{'}({1^ + }) = \dfrac{{\cos \left( {1 - 1} \right) - \cos (0)}}{1}
So f′(1+)=1−11=0f{'}({1^ + }) = \dfrac{{1 - 1}}{1} = 0
So the left hand derivative is equal to right hand derivative
So we can say function is Differentiate at x=1x = 1 if a=1a = 1 and b=0b = 0
So answer is option A and B is correct .
3. Not Differentiate at x=0x = 0 if a=1a = 1 and b=0b = 0
After putting value of a,ba,b
f(x)=1×cos⁡(∣x3−x∣)+0×∣x∣sin⁡(∣x3+x∣)f(x) = 1 \times \cos \left( {\left| {{x^3} - x} \right|} \right) + 0 \times \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)
f(x)=cos⁡(∣x3−x∣)f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)
Now we have to find left and right hand derivative
Right hand derivative f′(0+)=lim⁡h→0f(0+h)−f(0)hf{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}
f′(0+)=lim⁡h→0cos⁡(∣(0+h)3−(0+h)∣)−cos⁡(0)hf{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(0 + h)}^3} - (0 + h)} \right|} \right) - \cos (0)}}{h}
f′(0+)=lim⁡h→0cos⁡(∣(h3−h)∣)−cos⁡(0)hf{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {({h^3} - h)} \right|} \right) - \cos (0)}}{h}
Now if we put limit we see that numerator is exact equal to zero ∵cos⁡0−cos⁡0=0\because \cos 0 - \cos 0 = 0
So overall limit is equal to zero it does not matter what is in numerator
So f′(0+)=0f{'}({0^ + }) = 0
Now left hand derivative f′(0−)=lim⁡h→0f(0−h)−f(0)−hf{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}
f′(0−)=lim⁡h→0cos⁡(∣(0−h)3−(0−h)∣)−cos⁡(0)−hf{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(0 - h)}^3} - (0 - h)} \right|} \right) - \cos (0)}}{{ - h}}
Now we put limit
So f′(0−)=lim⁡h→0cos⁡(∣(−h3+h)∣)−cos⁡(0)−hf{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {( - {h^3} + h)} \right|} \right) - \cos (0)}}{{ - h}}
Now if we put limit we see that numerator is exact equal to zero ∵cos⁡0−cos⁡0=0\because \cos 0 - \cos 0 = 0
So overall limit is equal to zero it does not matter what is in numerator
So f′(0−)=0f{'}({0^ - }) = 0
So from here we see that left hand and right hand derivatives are equal so option C is wrong statement.
4. Not Differentiate at x=1x = 1 if a=1a = 1 and b=1b = 1
Now put value of a,ba,b
f(x)=cos⁡(∣x3−x∣)+∣x∣sin⁡(∣x3+x∣)f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right) + \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)
Now we have to find that given above function is derivable or not at x=1x = 1
As we see earlier that f(x)=cos⁡(∣x3−x∣)f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right) is differentiable at x=1x = 1
(refer prove of option B)
Now we prove that f(x)=∣x∣sin⁡(∣x3+x∣)f(x) = \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right) is derivable or not at x=1x = 1
So left hand derivative
f′(1−)=lim⁡h→0f(1−h)−f(1)−hf{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 - h) - f(1)}}{{ - h}}
So f′(1−)=lim⁡h→0∣1−h∣sin⁡(∣(1−h)3−(1−h)∣)−sin⁡2−hf{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| {1 - h} \right|\sin \left( {\left| {{{(1 - h)}^3} - (1 - h)} \right|} \right) - \sin 2}}{{ - h}}
If we put limit
f′(1−)=lim⁡h→0sin⁡(0)−sin⁡20f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( 0 \right) - \sin 2}}{0}
This goes ∞\infty because denominator is zero
f′(1−)=∞f{'}({1^ - }) = \infty
Now right hand derivative f′(1+)=lim⁡h→0f(1+h)−f(1)hf{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 + h) - f(1)}}{h}
f′(1−)=lim⁡h→0∣1+h∣sin⁡(∣(1+h)3−(1+h)∣)−sin⁡2hf{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| {1 + h} \right|\sin \left( {\left| {{{(1 + h)}^3} - (1 + h)} \right|} \right) - \sin 2}}{h}
Now we put limit
f′(1−)=sin⁡0−sin⁡20f{'}({1^ - }) = \dfrac{{\sin 0 - \sin 2}}{0}
As we it goes to ∞\infty because numerator is exact zero
So f′(1−)=∞f{'}({1^ - }) = \infty
So given function at x=1x = 1 and a=b=1a = b = 1 is derivable
So option D is wrong

Therefore, only the option (A) and (B) are correct.

Note:
Analyse the given information and go step by step while proceeding through the solution. Notice that the use of the chain rule of differentiation is a crucial part of the solution to this problem. Be careful with the use of braces while solving to avoid any confusion. In questions like these, there{{'}}s no choice other than checking for each option one by one.