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 $y=\sin x\left(0\le x\le \frac{\pi }{2}\right),y=a\cos x\left(0\le x\le \frac{\pi }{2}\right)$ 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 $S$ is equal to 1 .
• B. The value of T is equal to $\frac{2}{3}$.
• C. The value of a is $\frac{2}{3}$.
• D. The value of $a$ is $\frac{4}{3}$.

Answer 1

Answer: (B), (D)

We have $S=\underset{0}{\overset{\pi }{\int }}\sin xdx=2$, so $T=\frac{2}{3}$ where
Now $T=\underset{0}{\overset{{\tan }^{-1}a}{\int }}\sin xdx+\underset{{\tan }^{-1}a}{\overset{\pi /2}{\int }}a\cos xdx=\frac{2}{3}$ i.e. $-\cos \left({\tan }^{-1}a\right)+1+a\left(1-\sin \left({\tan }^{-1}a\right)\right)=\frac{2}{3}$,
i.e. $-\frac{1}{\sqrt{1+a^2}}+1+a-\frac{a^2}{\sqrt{1+a^2}}=\frac{2}{3}\Rightarrow (a+1)-\sqrt{a^2+1}=\frac{2}{3}$
$\Rightarrow a+\frac{1}{3}=\sqrt{a^2+1}$
$\Rightarrow a=\frac{4}{3}$