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If $\sin \alpha, \cos \alpha$ be the roots of the equation $x^{2}-b x+c=0 .$ Then, which of the following statements is/are correct?
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Verified Answer
The correct answers are:
$c \leq \frac{1}{2}$, $b \leq \sqrt{2}$
Given equation is $x^{2}-b x+c=0$
and roots are $\sin \alpha$ and $\cos \alpha$
Also $\sin \alpha+\cos \alpha=b$
and $\sin \alpha \cdot \cos \alpha=c$
$$
\begin{array}{r}
\because \sin ^{2} \alpha+\cos ^{2} \alpha=(\sin \alpha+\cos \alpha)^{2} \\
-2 \sin \alpha \cdot \cos \alpha
\end{array}
$$
$\Rightarrow \quad 1=b^{2}-2 c$
[from Eqs. (ii) and (iii)]
$-\sqrt{1+1} \leq(\sin \alpha+\cos \alpha) \leq \sqrt{1+1}$
$\Rightarrow \quad-\sqrt{2} \leq b \leq \sqrt{2}$
and $2 \sin \alpha ; \cos \alpha=x 2 c$
[from Eq. (iii)] $\begin{array}{ll}\Rightarrow & \sin 2 \alpha=2 c(\because-1 \leq \sin 2 \alpha \leq 1) \\ \Rightarrow & 2 c \leq 1 \Rightarrow c \leq \frac{1}{2}\end{array}$
and roots are $\sin \alpha$ and $\cos \alpha$
Also $\sin \alpha+\cos \alpha=b$
and $\sin \alpha \cdot \cos \alpha=c$
$$
\begin{array}{r}
\because \sin ^{2} \alpha+\cos ^{2} \alpha=(\sin \alpha+\cos \alpha)^{2} \\
-2 \sin \alpha \cdot \cos \alpha
\end{array}
$$
$\Rightarrow \quad 1=b^{2}-2 c$
[from Eqs. (ii) and (iii)]
$-\sqrt{1+1} \leq(\sin \alpha+\cos \alpha) \leq \sqrt{1+1}$
$\Rightarrow \quad-\sqrt{2} \leq b \leq \sqrt{2}$
and $2 \sin \alpha ; \cos \alpha=x 2 c$
[from Eq. (iii)] $\begin{array}{ll}\Rightarrow & \sin 2 \alpha=2 c(\because-1 \leq \sin 2 \alpha \leq 1) \\ \Rightarrow & 2 c \leq 1 \Rightarrow c \leq \frac{1}{2}\end{array}$
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