Question

Let $S$ be the area bounded by the curve $y=\sin x,0\le x\le \pi$ and the $x-$axis and $T$ be the area bounded by the curves $y=\sin x,0\le x\le \frac{\pi }{2},$ $y=a\cos x,0\le x\le \frac{\pi }{2}$ and the $x-$axis, where $a\in R^{+}$. If $S:T=1:\frac{1}{3},$ then which of the following is(are) CORRECT ?

• A. The value of $a$ is $\frac{4}{3}$
• B. The value of $S$ is equal to $1$.
• C. The value of $T$ is equal to $\frac{2}{3}$
• D. The value of $a$ is $\frac{2}{3}$

Answer 1

Answer: (A), (C)

The value of $T$ is equal to $\frac{2}{3}$

Area bounded by curve $y=\sin xx\in [0,\pi ]$
$S={\int }_0^{\pi }\sin xdx$
$\Rightarrow S=[-\cos x{]}_0^{\pi }$
$\Rightarrow S=-\cos \pi +\cos 0$
$\Rightarrow S=-(-1)+1=2$
As $S=2$ and $S:T=1:\frac{1}{3}$ (given )
We get, $T=\frac{2}{3}$
Calculating the value of $a$
Intersection point: $\sin x=a\cos x\Rightarrow x={\tan }^{-1}a$

$T={\int }_0^{{\tan }^{-1}a}\sin xdx+{\int }_{{\tan }^{-1}a}^{\pi /2}a\cos xdx=\frac{2}{3}$
$\Rightarrow [-\cos x{]}_0^{{\tan }^{-1}a}+a[\sin x{]}_{{\tan }^{-1}a}^{\pi /2}=\frac{2}{3}$
$\Rightarrow -\cos \left({\tan }^{-1}a\right)+1+a(1-\sin ({\tan }^{-1}a\left)\right)=\frac{2}{3}$
$\Rightarrow -\frac{1}{\sqrt{1+a^2}}+1+a-\frac{a^2}{\sqrt{1+a^2}}=\frac{2}{3}$
$\Rightarrow 1+a-[\frac{1}{\sqrt{1+a^2}}+\frac{a^2}{\sqrt{1+a^2}}]=\frac{2}{3}$
$\Rightarrow 1+a-\left[\frac{1+a^2}{\sqrt{1+a^2}}\right]=\frac{2}{3}$
$\Rightarrow 1+a-\sqrt{1+a^2}=\frac{2}{3}$
$\Rightarrow (1+a)-\sqrt{1+a^2}=\frac{2}{3}$
$\Rightarrow (1+a)-\frac{2}{3}=\sqrt{1+a^2}$
$\Rightarrow a+\frac{1}{3}=\sqrt{1+a^2}$
Squaring on both sides, we get
${(a+\frac{1}{3})}^2={\left(\sqrt{1+a^2}\right)}^2$
$\Rightarrow a^2+\frac{1}{9}+\frac{2a}{3}=1+a^2$
$\Rightarrow \frac{2a}{3}=1-\frac{1}{9}$
$\Rightarrow \frac{2a}{3}=\frac{8}{9}$
$\Rightarrow a=\frac{4}{3}$
and $T=1+\frac{4}{3}-\sqrt{1+14/9}$
$=\frac{7}{3}-\frac{5}{3}$
$\therefore T=\frac{2}{3}$