Question

If the area bounded by the x-axis, curve y = f (x) and the lines x = 1, x = b is equal to $\sqrt{{{b}^{2}}+1}-\sqrt{2}$ for all b > 1, then f (x) is
a. $\sqrt{x-1}$
b. $\sqrt{x+1}$
c. $\sqrt{{{x}^{2}}+1}$
d. $\dfrac{x}{\sqrt{1+{{x}^{2}}}}$

Answer 1

Hint : We will first find the value of f (x) by generalizing the given relation, $\int\limits_{1}^{b}{f\left( x \right)dx}=\sqrt{{{b}^{2}}+1}-\sqrt{2}$ and then we will differentiate the obtained general term to get the value of f (x).

Complete step-by-step answer :
It is given in the question that the area bounded by the x-axis, curve y = f (x) and the lines x = 1, x = b is equal to $\sqrt{{{b}^{2}}+1}-\sqrt{2}$ for all b > 1 and we have been asked to find the value of f (x).
We can represent the given data as follows.

We know that the area under the curve y = (x) and x-axis is given by the integration of y = f (x) from 1 to b, as we have the lines x = 1 and x = b. So, we get,
Area = $\int\limits_{1}^{b}{f\left( x \right)}$
Now, we have been given that $\int\limits_{1}^{b}{f\left( x \right)dx}$ is equal to $\sqrt{{{b}^{2}}+1}-\sqrt{2}$ , so we can write,
$\int\limits_{1}^{b}{f\left( x \right)dx}=\sqrt{{{b}^{2}}+1}-\sqrt{2}$
We can also write $\sqrt{2}$ as $\sqrt{1+1}$ . So, we get,
$\int\limits_{1}^{b}{f\left( x \right)dx}=\sqrt{{{b}^{2}}+1}-\sqrt{1+1}$
Now, if we observe $\sqrt{{{b}^{2}}+1}-\sqrt{1+1}$ , we get to know that we can generalize it using the general function $\left[ \sqrt{{{x}^{2}}+1} \right]_{1}^{b}$ , so we get,
$\int\limits_{1}^{b}{f\left( x \right)dx}=\left[ \sqrt{{{x}^{2}}+1} \right]_{1}^{b}$
We know that integration is the reverse of differentiation, so we can write it as,
$f\left( x \right)=\dfrac{d\left( \sqrt{{{x}^{2}}+1} \right)}{dx}$
Now, we will differentiate $\left( \sqrt{{{x}^{2}}+1} \right)$ with respect to x. We know that the derivative of $\sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$ . So, we can write the derivative of $\left( \sqrt{{{x}^{2}}+1} \right)$ as,
$\begin{aligned} & f\left( x \right)=\dfrac{1}{2\sqrt{{{x}^{2}}+1}}\times 2x \\\ & f\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}} \\\ \end{aligned}$
So, the correct answer is “Option D”.

Note : Many time, the students take the opposite limit after generalizing the function, f (x), they may take the function, $\int{f\left( x \right)dx}=\left[ \sqrt{{{x}^{2}}+1} \right]_{b}^{1}$ which gives the opposite result. They might get a negative sign in the area of the region found out, but it is also correct as we will neglect the negative sign because the area cannot be negative.