Question

If $\sin \alpha ,\cos \alpha$ be the roots of the equation $x^2-bx+c=0$. Then, which of the following statements is/are correct?

• A. $c\le \frac{1}{2}$
• B. $b\le \sqrt{2}$
• C. $c>\frac{1}{2}$
• D. $b>\sqrt{2}$

Answer 1

Answer: (A), (B)

The correct options are
A $b\le \sqrt{2}$
B $c\le \frac{1}{2}$

Given equation is $x^2-bx+c=\mathrm{0.....}\left(i\right)$
and roots are $\sin \alpha$ and $\cos \alpha =b.....\left(ii\right)$
and $\sin \alpha \cdot \cos \alpha =c...\left(iii\right)$

$\because {\sin }^2\alpha +{\cos }^2\alpha ={(\sin \alpha +\cos \alpha )}^2-2\sin \alpha .\cos \alpha$

$\Rightarrow 1=b^2-2c$ [from eqs $\left(ii\right)$ and $\left(iii\right)$]

$\because -\sqrt{1+1}\le (\sin \alpha +\cos \alpha )\le \sqrt{1+1}$

$\Rightarrow -\sqrt{2}\le b\le \sqrt{2}$ [from eq $\left(ii\right)$]

$2\sin \alpha .\cos \alpha =2c$ [from eq. $\left(iii\right)$]

$\Rightarrow \sin 2\alpha =2c$

$\Rightarrow 2c\le 1$

$\Rightarrow c\le \frac{1}{2}$