Question

The area bounded by the curve $y = f\left( x \right)$, $y = x$ and the lines $x = 1$, $x = t$ is $\left( {t + \sqrt {1 + {t^2}} } \right) - \sqrt 2 - 1$ sq. unit, for all $t > 1$. If $f\left( x \right)$ satisfying $f\left( x \right) > x$ for all $x > 1$, then $f\left( x \right)$ is equal to
A. $x + 1 + \dfrac{x}{{\sqrt {1 + {x^2}} }}$
B. $x + \dfrac{x}{{\sqrt {1 + {x^2}} }}$
C. $1 + \dfrac{x}{{\sqrt {1 + {x^2}} }}$
D. $\dfrac{x}{{\sqrt {1 + {x^2}} }}$

Answer 1

Hint: In this question, we are given the equation of two curves and two lines. Firstly, calculate the area of the curves by using the function $f\left( x \right) - x$ integrate it with respect to $dx$ from $x = 1$to $x = t$ and equate it to $\left( {t + \sqrt {1 + {t^2}} } \right) - \sqrt 2 - 1$. Here, we have subtracted $x$ from $f\left( x \right)$ because $f\left( x \right) > x$. Then, apply the formulas of differentiation and in the last put $t = x$.

Formula Used: Differentiation formula –
$\dfrac{d}{{dx}}c = 0$, where $c$ is the constant
$\dfrac{d}{{dx}}\left( x \right) = 1$
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
Relation between integration and differentiation –
$\dfrac{{dy}}{{dx}} \rightleftharpoons \int {ydx} $

Complete step by step Solution:
Given that,
The area bounded by the curves $y = f\left( x \right)$, $y = x$ and the lines $x = 1$, $x = t$ is $\left( {t + \sqrt {1 + {t^2}} } \right) - \sqrt 2 - 1$
Now, $f\left( x \right) > x$ is also given. Therefore, the function which we’ll take to calculate the area will be $f\left( x \right) - x$
Now, according to the equations of the lines, the values of $x$ are $1$ and $t$.
Thus, the area of the given curves and the lines will be;
$A = \int\limits_1^t {\left( {f\left( x \right) - x} \right)} dx$
But according to the question area is $\left( {t + \sqrt {1 + {t^2}} } \right) - \sqrt 2 - 1$
It implies that,
$\int\limits_1^t {\left( {f\left( x \right) - x} \right)} dx = \left( {t + \sqrt {1 + {t^2}} } \right) - \sqrt 2 - 1$
As we know that, $\dfrac{d}{{dx}}c = 0$, where $c$ is the constant and $\dfrac{d}{{dx}}\left( x \right) = 1$ , $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
Differentiate the above equation with respect to $t$, and we get
$f\left( t \right) - t = 1 + \dfrac{t}{{\sqrt {1 + {t^2}} }}$
Also, written as $f\left( t \right) = 1 + t + \dfrac{t}{{\sqrt {1 + {t^2}} }}$
Now, at $t = x$
We get, $f\left( x \right) = 1 + x + \dfrac{t}{{\sqrt {1 + {x^2}} }}$ for all $x > 1$

Therefore, the correct option is (A).

Note:Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.