Question

If $\beta +{\cos }^2,\alpha ,\beta +{\sin }^2\alpha$ are the roots of $x^2+2bx+c=0$ and $\gamma +{\cos }^4\alpha ,\gamma +{\sin }^4\alpha$ are the roots of $x^2+2Bx+C=0$, then :

• A. $b-B=c-C$
• B. $b^2-B^2=c-C$
• C. $b^2-B^2=4(c-C)$
• D. $4(b^2-B^2)=c-C$

Answer 1

Answer: (B)

$b^2-B^2=c-C$

Given

$\beta +co{s}^2\alpha ,\beta +si{n}^2\alpha$ are the roots of equation $x^2+2bx+c=0$

Also

$\gamma +co{s}^4\alpha ,\gamma +si{n}^4\alpha$ are the roots of equation $x^2+2Bx+C=0$

As we know that difference in roots of equation $a{x}^2+bx+c=0$ is given by,

$=\frac{\sqrt{b^2-4ac}}{a}$

For first equation

$\beta +co{s}^2\alpha -\beta -si{n}^2\alpha =\frac{\sqrt{(2b{)}^2-4c}}{1}=\sqrt{4b^2-4c}$

$co{s}^2\alpha -si{n}^2\alpha =\sqrt{4b^2-4c}-\left(1\right)$

For second equation

$\gamma +co{s}^4\alpha -\gamma -si{n}^4\alpha =\frac{\sqrt{(2B{)}^2-4C}}{1}=\sqrt{4B^2-4C}$

$(co{s}^2\alpha +si{n}^2\alpha )(co{s}^2\alpha -si{n}^2\alpha )=\sqrt{4B^2-4C}$

$\sqrt{4b^2-4c}=\sqrt{4B^2-4C}$ ( from eqs.(1))

$⟹b^2-c=B^2-C⟹b^2-B^2=c-C$