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.
$$x + 1 + \frac{x}{{\sqrt {1 + {x^2}} }}$$
B.
$$x + \frac{x}{{\sqrt {1 + {x^2}} }}$$
C.
$$1 + \frac{x}{{\sqrt {1 + {x^2}} }}$$
D.
$$\frac{x}{{\sqrt {1 + {x^2}} }}$$
Answer :
$$x + 1 + \frac{x}{{\sqrt {1 + {x^2}} }}$$
Solution :
It is given that, $$f\left( x \right) > x,$$ for all $$x > 1.$$ So, area bounded by $$y = f\left( x \right),\,y = x$$ and the lines $$x = 1,\,x =
t$$ is given by $$\int_1^t {\left\{ {f\left( x \right) - x} \right\}dx} $$
But this area is given equal to $$\left( {t + \sqrt {1 + {t^2}} - \sqrt 2 - 1} \right)$$ sq unit.
Therefore, $$\int_1^t {\left\{ {f\left( x \right) - x} \right\}dx = t + \sqrt {1 + {t^2}} - \sqrt 2 - 1,{\text{ for all }}t > 1} $$
On differentiating both sides w.r.t. $$t,$$ we get
$$\eqalign{
& f\left( t \right) - t = 1 + \frac{t}{{\sqrt {1 + {t^2}} }}{\text{ for all }}t > 1 \cr
& \Rightarrow f\left( t \right) = t + 1 + \frac{t}{{\sqrt {1 + {t^2}} }}{\text{ for all }}t > 1 \cr
& {\text{Hence, }}f\left( x \right) = x + 1 + \frac{x}{{\sqrt {1 + {x^2}} }}{\text{ for all }}x > 1 \cr} $$