Q53 / 155 · WBJEE 2017 · Mathematics
The equation $\sin x(\sin x+\cos x)=k$ has real solutions, where k is a real number. Then,
- A$0 \leq k \leq \frac{1+\sqrt{2}}{2}$
- B$2-\sqrt{3} \leq k \leq 2+\sqrt{3}$
- C$0 \leq k \leq 2-\sqrt{3}$
- Correct answerD$\frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$
Solution
Let, $f(x)=\sin x(\sin x+\cos x)$
$=\sin ^{2} x+\sin x \cos x$
$=\frac{1-\cos 2 x}{2}+\frac{2 \sin x \cos x}{2}$
$=\frac{1}{2}-\frac{1}{2} \cos 2 x+\frac{1}{2} \sin 2 x$
$=\frac{1}{2}+\frac{1}{2}(\sin 2 x-\cos 2 x)$
$\begin{aligned} \text { Now, }-\sqrt{(1)^{2}+(1)^{2}} & \leq \sin 2 x-\cos 2 x \\ & \leq \sqrt{(1)^{2}+(1)^{2}} \end{aligned}$
$\left[\because \sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\right]$
$\Rightarrow \quad-\sqrt{2} \leq \sin 2 x-\cos 2 x \leq \sqrt{2}$
$\Rightarrow \quad-\frac{\sqrt{2}}{2} \leq \frac{\sin 2 x-\cos 2 x}{2} \leq \frac{\sqrt{2}}{2}$
$\Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{2} \leq \frac{1}{2}+\frac{\sin 2 x-\cos 2 x}{2} \leq \frac{1}{2}+\frac{\sqrt{2}}{2}$
$\Rightarrow \frac{1-\sqrt{2}}{2} \leq f(x) \leq \frac{1+\sqrt{2}}{2}$
$\therefore \quad \frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$
Hence, the correct answer is option: 'D'.