Question

Let \[f:\left[ {0:2} \right] \to R\] be a function which is continuous on \[\left[ {0,2} \right]\]and is differentiable on \[\left( {0,2} \right)\] with \[f\left( 0 \right) = 1\].
Let \[F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} \] for \[x \in \left[ {0,2} \right]\]. If \[F'\left( x \right) = f'\left( x \right)\] for all \[x \in \left( {0,2} \right)\], then \[F\left( 2 \right)\] equals:
A. \[{e^2} - 1\]
B. \[{e^4} - 1\]
C. \[e - 1\]
D. \[{e^4}\]

Answer 1

Hint: In this problem we first use the given data to find the expression of \[f\left( x \right)\] by integration and using the data \[F'\left( x \right) = f'\left( x \right)\] and then we find the value of \[c\]. Then find the expression of \[F\left( x \right)\] and substitute \[x = 2\] to get the required solution.

Complete step-by-step answer:
Given that \[F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} \] for \[x \in \left[ {0,2} \right]\]
Since the function \[f\left( x \right)\] is continuous on \[\left[ {0,2} \right]\] and is differentiable on\[\left( {0,2} \right)\] differentiating the above expression w.r.t \[x\] we have
\[
   \Rightarrow \dfrac{d}{{dx}}\left( {F\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {\int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} } \right) \\
   \Rightarrow F'\left( x \right) = f\left( {\sqrt {{x^2}} } \right)\dfrac{d}{{dx}}\left( {{x^2}} \right) - f\left( 0 \right)\dfrac{d}{{dx}}\left( 0 \right) \\
   \Rightarrow F'\left( x \right) = f\left( {\left| x \right|} \right)\left( {2x} \right) - f\left( 0 \right) \times 0 \\
\]
Since \[x \in \left[ {0,2} \right]\],\[\left| x \right|\]=\[x\] then
\[
   \Rightarrow F'\left( x \right) = f\left( x \right)\left( {2x} \right) - f\left( 0 \right) \times 0 \\
   \Rightarrow F'\left( x \right) = 2f\left( x \right)x \\
\]
We are provided that \[F'\left( x \right) = f'\left( x \right)\] then
\[
   \Rightarrow F'\left( x \right) = f'\left( x \right) = 2f\left( x \right)x \\
   \Rightarrow \dfrac{{f'\left( x \right)}}{{f\left( x \right)}} = 2x \\
\]
On integrating the above expression, we get
\[
   \Rightarrow \int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \int {2x} dx \\
   \Rightarrow \log \left( {f\left( x \right)} \right) = 2\left( {\dfrac{{{x^2}}}{2}} \right) + c \\
   \Rightarrow \log \left( {f\left( x \right)} \right) = {x^2} + c \\
\]
We have \[f\left( 0 \right) = 1\] so we will proceed by putting \[x = 0\] then above expression becomes
\[
   \Rightarrow \log \left( {f\left( 0 \right)} \right) = {\left( 0 \right)^2} + c \\
   \Rightarrow \log \left( 1 \right) = c \\
  \therefore c = 0 \\
\]
Then the expression becomes
\[
   \Rightarrow \log \left( {f\left( x \right)} \right) = {x^2} \\
  \therefore f\left( x \right) = {e^{{x^2}}} \\
\]
Then, \[F\left( x \right) = \int_0^{{x^2}} {{e^x}} dx\]
\[ \Rightarrow F\left( x \right) = {e^{{x^2}}} - 1\]
Then putting \[x = 2\]
\[
   \Rightarrow F\left( 2 \right) = {e^2}^{^2} - 1 \\
   \Rightarrow F\left( 2 \right) = {e^4} - 1 \\
\]
\[\therefore \]\[F\left( 2 \right) = {e^4} - 1\]
Thus, the correct option is B. \[{e^4} - 1\].

So, the correct answer is “Option B”.

Note: To solve these kinds of problems, always remember the formula \[\dfrac{d}{{dx}}\left( {\int_{l.l}^{u.l} {f\left( x \right)dx} } \right) = f\left( {u.l} \right)\dfrac{d}{{dx}}\left( {u.l} \right) - f\left( {l.l} \right)\dfrac{d}{{dx}}\left( {l.l} \right)\] and \[\log 1 = 0\]. Here \[f'\left( x \right)\] represents that it is the first derivative of \[f\left( x \right)\] i.e., \[\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)\].