Q72 / 200 · Manipal (MET) 2014 · Mathematics
If $a \cos ^3 \alpha+3 a \cos \alpha \sin ^2 \alpha=m$ and $a \sin ^3 \alpha+3 a \cos ^2 \alpha \sin \alpha=n$, then $(m+n)^{2 / 3}+(m-n)^{2 / 3}$ is equal to
- A$2 a^3$
- B$2 a^{1 / 3}$
- Correct answerC$2 a^{2 / 3}$
- D$2 a^3$
Solution
Given relations are
$\begin{aligned}
& a \cos ^3 \alpha+3 a \cos \alpha \sin ^2 \alpha=m ...(i)\\
& a \sin ^3 \alpha+3 a \cos ^2 \alpha \sin \alpha=n...(ii)
\end{aligned}$
On adding Eq. (i) and (ii) we get $\begin{aligned} (m+n)= & a \cos ^3 \alpha+3 a \cos \alpha \sin ^2 \alpha \\ & +3 a \cos ^2 \alpha \sin \alpha+a \sin ^3 \alpha \\ = & a(\cos \alpha+\sin \alpha)^3 \end{aligned}$
Similarly on subtracting Eq. (ii) from Eq. (i), we get $\begin{aligned} & m-n=a(\cos \alpha-\sin \alpha)^3 \\ & \text { Now, }(m+n)^{2 / 3}+(m-n)^{2 / 3} \\ & =a^{2 / 3}\left\{(\cos \alpha+\sin \alpha)^2+(\cos \alpha-\sin \alpha)^2\right\} \\ & =a^{2 / 3}\left\{2\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\right\}=2 a^{2 / 3} \end{aligned}$
On adding Eq. (i) and (ii) we get $\begin{aligned} (m+n)= & a \cos ^3 \alpha+3 a \cos \alpha \sin ^2 \alpha \\ & +3 a \cos ^2 \alpha \sin \alpha+a \sin ^3 \alpha \\ = & a(\cos \alpha+\sin \alpha)^3 \end{aligned}$
Similarly on subtracting Eq. (ii) from Eq. (i), we get $\begin{aligned} & m-n=a(\cos \alpha-\sin \alpha)^3 \\ & \text { Now, }(m+n)^{2 / 3}+(m-n)^{2 / 3} \\ & =a^{2 / 3}\left\{(\cos \alpha+\sin \alpha)^2+(\cos \alpha-\sin \alpha)^2\right\} \\ & =a^{2 / 3}\left\{2\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\right\}=2 a^{2 / 3} \end{aligned}$
Hence, the correct answer is option: 'C'.