Question

If $\frac{dy}{dx}-y{\log }_{e}2=2^{\sin x}(\cos x-1){\log }_{e}2$ and $y\left(\frac{\pi }{2}\right)=2,$ then

• A. $y\left(0\right)=1$
• B. $y\left(0\right)=0$
• C. Area bounded by the curve $y=f\left(x\right),$ $x=0,x=1$ and $x$-axis lies in $(1,2)$
• D. Area bounded by the curve $y=f\left(x\right),$ $x=0,x=1$ and $x$-axis lies in $(2,4)$

Answer 1

Answer: (A), (C)

The correct options are
A $y\left(0\right)=1$
C Area bounded by the curve $y=f\left(x\right),$ $x=0,x=1$ and $x$-axis lies in $(1,2)$

I.F. $=e^{-\int {\log }_{e}2dx}=e^{-x\ln 2}=2^{-x}$
$\therefore$ Solution is $y\cdot 2^{-x}=\int 2^{\sin x-x}(\cos x-1)\ln 2dx$

Let $\sin x-x=t\Rightarrow (\cos x-1)dx=dt$
Then, $y\cdot 2^{-x}=\int 2^t\ln 2dt$
$\therefore y\cdot 2^{-x}=2^{\sin x-x}+c\Rightarrow y=2^{\sin x}+c\cdot 2^x$

Since $y\left(\frac{\pi }{2}\right)=2,\therefore c=0$
$\therefore y=2^{\sin x}$
$\Rightarrow y\left(0\right)=1$

As $0<x<1$
$\Rightarrow 0<\sin x<1$
$\Rightarrow 1<2^{\sin x}<2$
$\Rightarrow 1<\underset{0}{\overset{1}{\int }}{2}^{\sin x}<2$
$\therefore$ Area bounded $\in (1,2)$