Question

Let a curve $y=f(x)$ be such that the segment of tangent contained between coordinate axes is bisected by the point of contact. If $f(1)=2$, then the area bounded by $y=f(x)$, lines $x=1,x=\sqrt{e}$ and $x$-axis is given by

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$\Theta \frac{a+0}{2}=x\Rightarrow a=2x$
$\text{ and }\frac{0+b}{2}=y\Rightarrow b=2y$
$\therefore \text{ Slope of tangent }=\frac{dy}{dx}=\frac{-b}{a}=\frac{-y}{x}$
$\Rightarrow \int \frac{dy}{y}+\int \frac{dx}{x}=\int 0$
$\Rightarrow \ln x+\ln y=\ln c\Rightarrow xy=c$
$\Theta (1,2)\text{ is in it }$
$\therefore xy=2$
$\therefore \text{ Area bounded }=\underset{1}{\overset{\sqrt{e}}{\int }}\frac{2}{x}dx=2(\ln x{)}_1^{\sqrt{e}}=1$