Question

If $x=a{\cos }^3\theta {\sin }^2\theta ,y=a{\sin }^3\theta {\cos }^2\theta$ and $\frac{(x^2+y^2{)}^p}{(xy{)}^q}(p,qϵN)$ is independent of $\theta$, then

• A. $p=4$
• B. $p=5$
• C. $q=4$
• D. $q=-5$

Answer 1

Answer: (B), (C)

The correct options are
B $p=5$
C $q=4$

$x^2+y^2={\left(a{\cos }^3\theta {\sin }^2\theta \right)}^2+{\left(a{\sin }^3\theta {\cos }^2\theta \right)}^2=a^2{\sin }^4\theta {\cos }^4\theta$

$xy=\left(a{\cos }^3\theta {\sin }^2\theta \right)\left(a{\sin }^3\theta {\cos }^2\theta \right)=a^2{\sin }^5\theta {\cos }^5\theta$

Therefore,

$\frac{(x^2+y^2{)}^p}{(xy{)}^q}=\frac{a^{2p}(\sin \theta \cos \theta {)}^{4p}}{a^{2q}(\sin \theta \cos \theta {)}^{5q}}$

Is Independent of $\theta$ when $4p=5q$

i.e. $p=5,q=4$

Hence, option ${}^{\prime }{B}^{\prime }$ and ${}^{\prime }{C}^{\prime }$ are correct.