Question

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

• A. $p+q=6$
• B. $4p=5q$
• C. $p=q$
• D. $pq=16$

Answer 1

The correct option is B $4p=5q$

$x=a{\cos }^3\theta {\sin }^2\theta$

$y=a{\sin }^3\theta {\cos }^2\theta$

Then, ${(x^2+y^2)}^p={(a^2{\cos }^6\theta {\sin }^4\theta +a^2{\sin }^6\theta {\cos }^4\theta )}^p$

$={\left[a^2{\sin }^4\theta {\cos }^4\theta \right({\cos }^2\theta +{\sin }^2\theta \left)\right]}^p$

$=(a^2{\sin }^4\theta {\cos }^4\theta {)}^p$

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

Now,

$\frac{{(x^2+y^2)}^p}{{\left(xy\right)}^q}$

$=\frac{{\left(a^2{\sin }^4\theta {\cos }^4\theta \right)}^p}{(a^2{\cos }^5\theta {\sin }^5\theta {)}^q}$

$=a^{2p-2q}\times (\sin \theta \cos \theta {)}^{4p-5q}$

Now, $\frac{{(x^2+y^2)}^p}{{\left(xy\right)}^q}$ is independent of $\theta$ if,

$4p-5q=0$

$\therefore 4p=5q$

Hence, option 'B' is correct.