Let \[a,b \in R\]and $f:R \to R$ be defined by $f(x) = a\cos \left( {\left| {{x^3} - x} \right|} \right) + b\left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$. Then $f$ is
(A) Differentiate at $x = 0$ if $a = 0$ and $b = 1$
(B) Differentiate at $x = 1$ if $a = 1$ and $b = 0$
(C) Not Differentiate at $x = 0$ if $a = 1$ and $b = 0$
(D) Not Differentiate at $x = 1$ if $a = 1$ and $b = 1$
Answer
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Hint: To solve this problem we find the right and left hand derivatives of a given function at $x = 0$ and $x = 1$ by putting value of $a,b$ as given in options.
The left hand derivative of $f(x)$ at $x = a$ is $f{'}({a^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a - h) - f(a)}}{{ - h}}$
And right hand derivative of $f(x)$ at $x = a$ is $f{'}({a^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$
Complete step by step answer:
Given function: $f(x) = a\cos \left( {\left| {{x^3} - x} \right|} \right) + b\left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
Now we go through first option:
1. Differentiate at $x = 0$ if $a = 0$ and $b = 1$
Now we put value of a and b in given function
From this we can say $f(x) = \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
Now find derivative:
Left hand derivative at $x = 0$
$f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}$
So $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| { - h} \right|\sin \left( {\left| { - {h^3} - h} \right|} \right) - 0}}{{ - h}}$
$\because f(0) = 0 \times \sin 0 = 0$
$\because \left| { - h} \right| = h$
So $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\sin \left( {\left| { - {h^3} - h} \right|} \right)}}{{ - h}}$
Now $f{'}({0^ - }) = \dfrac{{\sin \left( 0 \right)}}{{ - 1}}$
$\because \sin 0 = 0$
So $f{'}({0^ - }) = 0$
So left hand derivative at $x = 0$ is $0$
Similarly we find left hand derivative
For right hand derivative of $f(x)$ at $x = 0$ is $f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
$ \Rightarrow \;$
$ \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 $\left| h \right| = h$
And after putting limit value we get
$ \Rightarrow f{'}({0^ + }) = \sin 0$
$ \Rightarrow f{'}({0^ + }) = 0$
So left hand derivative and right hand derivative are equal
$f{'}({0^ + }) = f{'}({0^ - }) = 0$ so function is derivable at $x = 0$
2. Differentiate at $x = 1$ if $a = 1$ and $b = 0$
So put value of $a,b$ we get $f(x) = 1 \times \cos \left( {\left| {{x^3} - x} \right|} \right) + 0 \times \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
$f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)$
So
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 - h) - f(1)}}{{ - h}}$
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(1 - h)}^3} - (1 - h)} \right|} \right) - \cos (1 - 1)}}{{ - h}}$
Now we put limit
So $f{'}({1^ - }) = \dfrac{{\cos \left( {1 - 1} \right) - \cos (0)}}{{ - 1}}$ and $\cos 0 = 1$
So $f{'}({1^ - }) = \dfrac{{1 - 1}}{1} = 0$
Similarly right hand limit $f{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 + h) - f(1)}}{h}$
$f{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(1 + h)}^3} - (1 + h)} \right|} \right) - \cos (1 - 1)}}{h}$
Now putting limit value
$f{'}({1^ + }) = \dfrac{{\cos \left( {1 - 1} \right) - \cos (0)}}{1}$
So $f{'}({1^ + }) = \dfrac{{1 - 1}}{1} = 0$
So the left hand derivative is equal to right hand derivative
So we can say function is Differentiate at $x = 1$ if $a = 1$ and $b = 0$
So answer is option A and B is correct .
3. Not Differentiate at $x = 0$ if $a = 1$ and $b = 0$
After putting value of $a,b$
$f(x) = 1 \times \cos \left( {\left| {{x^3} - x} \right|} \right) + 0 \times \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
$f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)$
Now we have to find left and right hand derivative
Right hand derivative $f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
$f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(0 + h)}^3} - (0 + h)} \right|} \right) - \cos (0)}}{h}$
$f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {({h^3} - h)} \right|} \right) - \cos (0)}}{h}$
Now if we put limit we see that numerator is exact equal to zero $\because \cos 0 - \cos 0 = 0$
So overall limit is equal to zero it does not matter what is in numerator
So $f{'}({0^ + }) = 0$
Now left hand derivative $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}$
$f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(0 - h)}^3} - (0 - h)} \right|} \right) - \cos (0)}}{{ - h}}$
Now we put limit
So $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {( - {h^3} + h)} \right|} \right) - \cos (0)}}{{ - h}}$
Now if we put limit we see that numerator is exact equal to zero $\because \cos 0 - \cos 0 = 0$
So overall limit is equal to zero it does not matter what is in numerator
So $f{'}({0^ - }) = 0$
So from here we see that left hand and right hand derivatives are equal so option C is wrong statement.
4. Not Differentiate at $x = 1$ if $a = 1$ and $b = 1$
Now put value of $a,b$
$f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right) + \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
Now we have to find that given above function is derivable or not at $x = 1$
As we see earlier that $f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)$ is differentiable at $x = 1$
(refer prove of option B)
Now we prove that $f(x) = \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$ is derivable or not at $x = 1$
So left hand derivative
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 - h) - f(1)}}{{ - h}}$
So $f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| {1 - h} \right|\sin \left( {\left| {{{(1 - h)}^3} - (1 - h)} \right|} \right) - \sin 2}}{{ - h}}$
If we put limit
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( 0 \right) - \sin 2}}{0}$
This goes $\infty $ because denominator is zero
$f{'}({1^ - }) = \infty $
Now right hand derivative $f{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 + h) - f(1)}}{h}$
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| {1 + h} \right|\sin \left( {\left| {{{(1 + h)}^3} - (1 + h)} \right|} \right) - \sin 2}}{h}$
Now we put limit
$f{'}({1^ - }) = \dfrac{{\sin 0 - \sin 2}}{0}$
As we it goes to $\infty $ because numerator is exact zero
So $f{'}({1^ - }) = \infty $
So given function at $x = 1$ and $a = b = 1$ 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.
The left hand derivative of $f(x)$ at $x = a$ is $f{'}({a^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a - h) - f(a)}}{{ - h}}$
And right hand derivative of $f(x)$ at $x = a$ is $f{'}({a^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$
Complete step by step answer:
Given function: $f(x) = a\cos \left( {\left| {{x^3} - x} \right|} \right) + b\left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
Now we go through first option:
1. Differentiate at $x = 0$ if $a = 0$ and $b = 1$
Now we put value of a and b in given function
From this we can say $f(x) = \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
Now find derivative:
Left hand derivative at $x = 0$
$f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}$
So $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| { - h} \right|\sin \left( {\left| { - {h^3} - h} \right|} \right) - 0}}{{ - h}}$
$\because f(0) = 0 \times \sin 0 = 0$
$\because \left| { - h} \right| = h$
So $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\sin \left( {\left| { - {h^3} - h} \right|} \right)}}{{ - h}}$
Now $f{'}({0^ - }) = \dfrac{{\sin \left( 0 \right)}}{{ - 1}}$
$\because \sin 0 = 0$
So $f{'}({0^ - }) = 0$
So left hand derivative at $x = 0$ is $0$
Similarly we find left hand derivative
For right hand derivative of $f(x)$ at $x = 0$ is $f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
$ \Rightarrow \;$
$ \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 $\left| h \right| = h$
And after putting limit value we get
$ \Rightarrow f{'}({0^ + }) = \sin 0$
$ \Rightarrow f{'}({0^ + }) = 0$
So left hand derivative and right hand derivative are equal
$f{'}({0^ + }) = f{'}({0^ - }) = 0$ so function is derivable at $x = 0$
2. Differentiate at $x = 1$ if $a = 1$ and $b = 0$
So put value of $a,b$ we get $f(x) = 1 \times \cos \left( {\left| {{x^3} - x} \right|} \right) + 0 \times \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
$f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)$
So
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 - h) - f(1)}}{{ - h}}$
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(1 - h)}^3} - (1 - h)} \right|} \right) - \cos (1 - 1)}}{{ - h}}$
Now we put limit
So $f{'}({1^ - }) = \dfrac{{\cos \left( {1 - 1} \right) - \cos (0)}}{{ - 1}}$ and $\cos 0 = 1$
So $f{'}({1^ - }) = \dfrac{{1 - 1}}{1} = 0$
Similarly right hand limit $f{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 + h) - f(1)}}{h}$
$f{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(1 + h)}^3} - (1 + h)} \right|} \right) - \cos (1 - 1)}}{h}$
Now putting limit value
$f{'}({1^ + }) = \dfrac{{\cos \left( {1 - 1} \right) - \cos (0)}}{1}$
So $f{'}({1^ + }) = \dfrac{{1 - 1}}{1} = 0$
So the left hand derivative is equal to right hand derivative
So we can say function is Differentiate at $x = 1$ if $a = 1$ and $b = 0$
So answer is option A and B is correct .
3. Not Differentiate at $x = 0$ if $a = 1$ and $b = 0$
After putting value of $a,b$
$f(x) = 1 \times \cos \left( {\left| {{x^3} - x} \right|} \right) + 0 \times \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
$f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)$
Now we have to find left and right hand derivative
Right hand derivative $f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
$f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(0 + h)}^3} - (0 + h)} \right|} \right) - \cos (0)}}{h}$
$f{'}({0^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {({h^3} - h)} \right|} \right) - \cos (0)}}{h}$
Now if we put limit we see that numerator is exact equal to zero $\because \cos 0 - \cos 0 = 0$
So overall limit is equal to zero it does not matter what is in numerator
So $f{'}({0^ + }) = 0$
Now left hand derivative $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}$
$f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {{{(0 - h)}^3} - (0 - h)} \right|} \right) - \cos (0)}}{{ - h}}$
Now we put limit
So $f{'}({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left| {( - {h^3} + h)} \right|} \right) - \cos (0)}}{{ - h}}$
Now if we put limit we see that numerator is exact equal to zero $\because \cos 0 - \cos 0 = 0$
So overall limit is equal to zero it does not matter what is in numerator
So $f{'}({0^ - }) = 0$
So from here we see that left hand and right hand derivatives are equal so option C is wrong statement.
4. Not Differentiate at $x = 1$ if $a = 1$ and $b = 1$
Now put value of $a,b$
$f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right) + \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$
Now we have to find that given above function is derivable or not at $x = 1$
As we see earlier that $f(x) = \cos \left( {\left| {{x^3} - x} \right|} \right)$ is differentiable at $x = 1$
(refer prove of option B)
Now we prove that $f(x) = \left| x \right|\sin \left( {\left| {{x^3} + x} \right|} \right)$ is derivable or not at $x = 1$
So left hand derivative
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 - h) - f(1)}}{{ - h}}$
So $f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| {1 - h} \right|\sin \left( {\left| {{{(1 - h)}^3} - (1 - h)} \right|} \right) - \sin 2}}{{ - h}}$
If we put limit
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( 0 \right) - \sin 2}}{0}$
This goes $\infty $ because denominator is zero
$f{'}({1^ - }) = \infty $
Now right hand derivative $f{'}({1^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(1 + h) - f(1)}}{h}$
$f{'}({1^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left| {1 + h} \right|\sin \left( {\left| {{{(1 + h)}^3} - (1 + h)} \right|} \right) - \sin 2}}{h}$
Now we put limit
$f{'}({1^ - }) = \dfrac{{\sin 0 - \sin 2}}{0}$
As we it goes to $\infty $ because numerator is exact zero
So $f{'}({1^ - }) = \infty $
So given function at $x = 1$ and $a = b = 1$ 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.
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