Question

If ${\sin }^2\alpha ,3{\cos }^2\alpha$ are the roots of the equation $4x^2-3px+k=0$, then-

• A. $p=4-\frac{8}{3}{\sin }^2\alpha$
• B. $p=\frac{4}{3}$
• C. $k=3{\sin }^{2}2\alpha$
• D. $k=3\sin 2\alpha$

Answer 1

Answer: (A), (C)

The correct options are
A $p=4-\frac{8}{3}{\sin }^2\alpha$
C $k=3{\sin }^{2}2\alpha$

Hence
Sum of roots is
${\sin }^2\alpha +3{\cos }^2\alpha =\frac{3p}{4}$ ...(i)
And
$3{\sin }^2\alpha .{\cos }^2\alpha =\frac{k}{4}$
Or
$3.[4{\sin }^2\alpha .{\cos }^2\alpha ]=k$
Or
$3{\sin }^{2}2\alpha =k$
Now
${\sin }^2\alpha +3{\cos }^2\alpha =\frac{3p}{4}$
Or
${\sin }^2\alpha +3-3{\sin }^2\alpha =\frac{3p}{4}$
Or
$3-2{\sin }^2\alpha =\frac{3p}{4}$
Or
$4-\frac{8}{3}{\sin }^2\alpha =p$.