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{align}
  & f\left( x \right)=\dfrac{1}{2\sqrt{{{x}^{2}}+1}}\times 2x \\
 & f\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}} \\
\end{align} $
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.