Question

Let f be a real valued function defined on the interval (-1, 1) such that ${e^{ - x}}f\left( x \right) = 2 + \int_0^x {\sqrt {{t^4} + 1} dt}$, for all $x \in \left( { - 1,1} \right)$ and let ${f^{ - 1}}$ be the inverse function of $f$. Then ${\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right)$ is equal to
$\left( a \right)$ 1
$\left( b \right)$ 1/3
$\left( c \right)$ 1/2
$\left( d \right)$ 1/e

Answer 1

In this particular question use the concept that inverse function is a function that reverses another function so if ${f^{ - 1}}$ be the inverse function of $f$, then $f\left( {{f^{ - 1}}\left( x \right)} \right)$ = x, then differentiate both sides w.r.t. x to calculate the value of f’ (x) in terms of h’ (f (x)), so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given data:
Let f be a real valued function defined on the interval (-1, 1), such that ${e^{ - x}}f\left( x \right) = 2 + \int_0^x {\sqrt {{t^4} + 1} dt}$, for all $x \in \left( { - 1,1} \right)$ and let${f^{ - 1}}$ be the inverse function of $f$.
So if ${f^{ - 1}}$ be the inverse function of $f$, then $f\left( {{f^{ - 1}}\left( x \right)} \right)$ = x............... (1)
Now differentiate the above equation w.r.t x we have,
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( {{f^{ - 1}}\left( x \right)} \right)} \right] = \dfrac{d}{{dx}}x$
Now as we know that $\dfrac{d}{{dx}}u\left( {g\left( x \right)} \right) = u'g\left( x \right)\dfrac{d}{{dx}}g\left( x \right),\dfrac{d}{{dx}}x = 1$, so use this property in the above equation we have,
$\Rightarrow f'\left( {{f^{ - 1}}\left( x \right)} \right)\dfrac{d}{{dx}}{f^{ - 1}}\left( x \right) = 1$
$\Rightarrow f'\left( {{f^{ - 1}}\left( x \right)} \right){\left( {{f^{ - 1}}\left( x \right)} \right)^\prime } = 1$, $\left[ {\because \dfrac{d}{{dx}}{f^{ - 1}}\left( x \right) = {{\left( {{f^{ - 1}}\left( x \right)} \right)}^\prime }} \right]$
Now substitute in place of x, 2 we have,

$\Rightarrow f'\left( {{f^{ - 1}}\left( 2 \right)} \right){\left( {{f^{ - 1}}\left( 2 \right)} \right)^\prime } = 1$
$\Rightarrow {\left( {{f^{ - 1}}\left( 2 \right)} \right)^\prime } = \dfrac{1}{{f'\left( {{f^{ - 1}}\left( 2 \right)} \right)}}$.................. (2)
Now it is given that,
${e^{ - x}}f\left( x \right) = 2 + \int_0^x {\sqrt {{t^4} + 1} dt}$................. (3)
Now putting x = 0 in the above equation we have,
$\Rightarrow {e^{ - 0}}f\left( 0 \right) = 2 + \int_0^0 {\sqrt {{t^4} + 1} dt}$
Now as we know that $\int_0^0 {f\left( t \right)dt = 0,{e^{ - 0}} = 1}$
$\Rightarrow f\left( 0 \right) = 2$................ (4)
Now from equation (1) and (4) we have,
$f\left( {{f^{ - 1}}\left( x \right)} \right)$= x, $f\left( 0 \right) = 2$
So on comparing, ${f^{ - 1}}\left( 2 \right) = 0$
Now substitute this value in equation (2) we have,
$\Rightarrow {\left( {{f^{ - 1}}\left( 2 \right)} \right)^\prime } = \dfrac{1}{{f'\left( 0 \right)}}$............. (5)
Now differentiate equation (3) w.r.t x we have,
$\Rightarrow \dfrac{d}{{dx}}\left[ {{e^{ - x}}f\left( x \right)} \right] = \dfrac{d}{{dx}}\left( {2 + \int_0^x {\sqrt {{t^4} + 1} dt} } \right)$
Now as we know that $\dfrac{d}{{dx}}\left( {\int_0^x {\sqrt {{t^4} + 1} dt} } \right) = {\left( {\sqrt {{t^4} + 1} } \right)_{t = x}} - {\left( {\sqrt {{t^4} + 1} } \right)_{t = 0}}$, $\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}b + b\dfrac{d}{{dx}}a,\dfrac{d}{{dx}}{e^{ - x}} = - {e^{ - x}}$ and the differentiation of constant term is zero, so use these properties in the above equation we have,
$\Rightarrow \left[ {{e^{ - x}}f'\left( x \right) + f\left( x \right)\left( { - {e^{ - x}}} \right)} \right] = \left( {0 + {{\left( {\sqrt {{t^4} + 1} } \right)}_{t = x}} - {{\left( {\sqrt {{t^4} + 1} } \right)}_{t = 0}}} \right)$
$\Rightarrow {e^{ - x}}f'\left( x \right) - f\left( x \right){e^{ - x}} = \sqrt {{x^4} + 1} - 0$
$\Rightarrow {e^{ - x}}f'\left( x \right) - f\left( x \right){e^{ - x}} = \sqrt {{x^4} + 1}$
Now substitute in place of x, 0 we have,
$\Rightarrow {e^{ - 0}}f'\left( 0 \right) - f\left( 0 \right){e^{ - 0}} = \sqrt {{0^4} + 1}$
$\Rightarrow f'\left( 0 \right) - f\left( 0 \right) = 1$
Now from equation (4), f (0) = 2, so we have,
$\Rightarrow f'\left( 0 \right) - 2 = 1$
$\Rightarrow f'\left( 0 \right) = 1 + 2 = 3$
Now substitute this value in equation (5) we have,
$\Rightarrow {\left( {{f^{ - 1}}\left( 2 \right)} \right)^\prime } = \dfrac{1}{3}$
$\Rightarrow {\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right) = \dfrac{1}{3}$
So this is the required answer.
Hence option (b) is the correct answer.

Note :Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation property which is given as $\dfrac{d}{{dx}}u\left( {g\left( x \right)} \right) = u'g\left( x \right)\dfrac{d}{{dx}}g\left( x \right),\dfrac{d}{{dx}}x = 1$, $\dfrac{d}{{dx}}\left( {\int_0^x {\sqrt {{t^4} + 1} dt} } \right) = {\left( {\sqrt {{t^4} + 1} } \right)_{t = x}} - {\left( {\sqrt {{t^4} + 1} } \right)_{t = 0}}$,$\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}b + b\dfrac{d}{{dx}}a,\dfrac{d}{{dx}}{e^{ - x}} = - {e^{ - x}}$so differentiate the equation (1) by using first property and equation (3) by using second properties and then substitute the values as above we will get the required value of ${\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right)$.