Q.12. Let . Then is equal to :

Question

Q.12. Let . Then  is equal to :

FirstInTest · Accepted Answer

Correct answer: D. Sol. The equation is $\sin^2 x + \sin x \cos x = a$.
We can rewrite this using double angle formulas:
$\frac{1 - \cos 2x}{2} + \frac{\sin 2x}{2} = a$
$1 - \cos 2x + \sin 2x = 2a$
$\sin 2x - \cos 2x = 2a - 1$
Let $2x = y$. The interval for $x$ is $[-\pi, \pi]$, so for $y = 2x$, the interval is $[-2\pi, 2\pi]$.
We have $\sin y - \cos y = 2a - 1$.
Multiply by $\frac{1}{\sqrt{2}}$: $\frac{1}{\sqrt{2}} \sin y - \frac{1}{\sqrt{2}} \cos y = \frac{2a-1}{\sqrt{2}}$
$\sin y \cos \frac{\pi}{4} - \cos y \sin \frac{\pi}{4} = \frac{2a-1}{\sqrt{2}}$
$\sin (y - \frac{\pi}{4}) = \frac{2a-1}{\sqrt{2}}$.
Since $a \in \mathrm{Z}$, the possible values for $2a-1$ are integers. The range of $\sin (y - \frac{\pi}{4})$ is $[-1, 1]$.
So, $-1 \le \frac{2a-1}{\sqrt{2}} \le 1 \Rightarrow -\sqrt{2} \le 2a-1 \le \sqrt{2}$.
Approximately, $-1.414 \le 2a-1 \le 1.414$.
Since $2a-1$ must be an integer, $2a-1$ can be $-1$, $0$, or $1$.
Case 1: $2a-1 = -1 \Rightarrow 2a = 0 \Rightarrow a = 0$.
Then $\sin (y - \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.
For $y \in [-2\pi, 2\pi]$, $y - \frac{\pi}{4} \in [-2\pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4}] = [-\frac{9\pi}{4}, \frac{7\pi}{4}]$.
In this interval, $y - \frac{\pi}{4}$ can be $-\frac{5\pi}{4}$, $-\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{7\pi}{4}$.
$y = -\pi$, $0$, $\pi$, $2\pi$.
$x = -\frac{\pi}{2}$, $0$, $\frac{\pi}{2}$, $\pi$. (4 solutions)
Case 2: $2a-1 = 0 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2}$. Not an integer, so this case is not possible.
Case 3: $2a-1 = 1 \Rightarrow 2a = 2 \Rightarrow a = 1$.
Then $\sin (y - \frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
In the interval $[-\frac{9\pi}{4}, \frac{7\pi}{4}]$, $y - \frac{\pi}{4}$ can be $-\frac{7\pi}{4}$, $-\frac{3\pi}{4}$, $\frac{\pi}{4}$, $\frac{5\pi}{4}$.
$y = -\frac{3\pi}{2}$, $-\frac{\pi}{2}$, $\frac{\pi}{2}$, $\frac{3\pi}{2}$.
$x = -\frac{3\pi}{4}$, $-\frac{\pi}{4}$, $\frac{\pi}{4}$, $\frac{3\pi}{4}$. (4 solutions)
Total number of solutions $n(S) = 4 + 4 = 8$.

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