Question

The curve $C_1:y=e^{-x}$ and $C_2:y=e^{-x}\sin x(x>0)$ touches each other at infinitely many points. Let $0<x_1<x_2<\cdots <x_n<\cdots$ be the abscissa of these points of contact. If $A_n$ denotes the area bounded by two curves & ordinates $x=x_n$ and $x=x_{n+1}$, then

• A. $A_1=\frac{1}{2}\frac{(e^{2\pi }-1)}{e^{5\pi /2}}$
• B. $A_2=\frac{1}{2}\frac{(e^{2\pi }-1)}{e^{9\pi /2}}$
• C. $A_1,A_2,A_3\text{are in A.P.}$
• D. $A_1,A_2,A_3\text{are in G.P.}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $A_1=\frac{1}{2}\frac{(e^{2\pi }-1)}{e^{5\pi /2}}$
B $A_2=\frac{1}{2}\frac{(e^{2\pi }-1)}{e^{9\pi /2}}$
D $A_1,A_2,A_3\text{are in G.P.}$

$e^{-x}=e^{-x}\sin x\Rightarrow \sin x=1$
$\Rightarrow x_n=2(n-1)\pi +\frac{\pi }{2};(n\in N)$
$\int e^{-x}(1-\sin x)dx=\frac{e^{-x}}{2}(\sin x+\cos x)-e^{-x}+c$

$\text{So, Area bounded}:$
$A_n={\int }_{x_n}^{x_{n+1}}{e}^{-x}(1-\sin x)dx=[\frac{e^{-x}}{2}(\sin x+\cos x)-e^{-x}{]}_{x_n}^{x_{n+1}}=\frac{1}{2}(e^{2\pi }-1).e^{-\frac{(4n+1)\pi }{2}}$
$\text{So,}{A}_1=\frac{1}{2}\frac{(e^{2\pi }-1)}{e^{5\pi /2}},A_2=\frac{1}{2}\frac{(e^{2\pi }-1)}{e^{9\pi /2}},A_3=\frac{1}{2}\frac{(e^{2\pi }-1)}{e^{13\pi /2}}$
$\text{Clearly,}{A}_1,A_2,A_3\text{are in G.P with}r=\frac{1}{e^{2\pi }}$