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)

$\Theta \frac{a+0}{2}=x \Rightarrow a=2 x$
$\text { and } \frac{0+ b }{2}= y \Rightarrow b =2 y $
$\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 $
$\therefore \text { Area bounded }=\int\limits_1^{\sqrt{ e }} \frac{2}{ x } dx =2(\ln x )_1^{\sqrt{e}}=1$