Question

The equation $\sin x (\sin\, x + \cos x) = k$ has real solutions, where $k$ is a real number. Then

• A. $0 \le k\le\frac{1+\sqrt{2}}{2}$
• B. $2-\sqrt{3}\le k\le2+\sqrt{3}$
• C. $0\le k\le2-\sqrt{3}$
• D. $\frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$

Answer 1

Answer: (D)

Let, $f(x) =\sin x(\sin x+\cos x) $
$=\sin ^{2} x+\sin x \cos x$
$=\frac{1-\cos 2 x}{2}+\frac{2 \sin x \cos x}{2} $
$=\frac{1}{2}-\frac{1}{2} \cos .2 x+\frac{1}{2} \sin 2 x $
$=\frac{1}{2}+\frac{1}{2}(\sin 2 x-\cos 2 x)$
Now, $-\sqrt{(1)^{2}+(1)^{2}} \leq \sin 2 x-\cos 2 x$
$\leq \sqrt{(1)^{2}+(1)^{2}}$
$\left[\because \sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\right] $
$ \Rightarrow -\sqrt{2} \leq \sin 2 x-\cos 2 x \leq \sqrt{2} $
$ \Rightarrow -\frac{\sqrt{2}}{2} \leq \frac{\sin 2 x-\cos 2 x}{2} \leq \frac{\sqrt{2}}{2} $
$\Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{2} \leq \frac{1}{2}+\frac{\sin 2 x-\cos 2 x}{2} \leq \frac{1}{2}+\frac{\sqrt{2}}{2} $
$\Rightarrow \frac{1-\sqrt{2}}{2} \leq f(x) \leq \frac{1+\sqrt{2}}{2} $
$ \therefore \frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$