Question

Let $S$ be the area bounded by the curve $y=\sin x(0 \leq x \leq \pi)$ and the $x$-axis and $T$ be the area bounded by the curves $y=\sin x\left(0 \leq x \leq \frac{\pi}{2}\right), y=a \cos x\left(0 \leq x \leq \frac{\pi}{2}\right)$ 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 $S =\int\limits_0^\pi \sin xdx =2$, so $T =\frac{2}{3}$ where
Now $T=\int\limits_0^{\tan ^{-1} a} \sin x d x+\int\limits_{\tan ^{-1} a}^{\pi / 2} a \cos x d x=\frac{2}{3}$ 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}$
$\Rightarrow a =\frac{4}{3}$