UPSC NDA I 2016 Mathematics Solved Paper

Concept:The equation is a trigonometric identity that can be rewritten as a quadratic in $\sin x$.We use the condition $-1 \le \sin x \le 1$ to find the maximum value of $k$ for which a real solution exists.Explanation:Start with $k \sin x + \cos 2x = 2k - 7$.Substitute $\cos 2x = 1 - 2\sin^2 x$: $k \sin x + 1 - 2\sin^2 x = 2k - 7$.Rearrange: $-2\sin^2 x + k \sin x + 1 - 2k + 7 = 0$ → $-2\sin^2 x + k \sin x - 2k + 8 = 0$.Multiply by $-1$: $2\sin^2 x - k \sin x + 2k - 8 = 0$.This is quadratic in $\sin x$ with $a = 2$, $b = -k$, $c = 2k - 8$.Solve: $\sin x = \frac{k \pm \sqrt{k^2 - 8(2k - 8)}}{4}$.Simplify discriminant: $k^2 - 16k + 64 = (k-8)^2$, so $\sin x = \frac{k \pm (k-8)}{4}$.Thus two possible values: $\sin x = \frac{2k-8}{4} = \frac{k-4}{2}$ and $\sin x = \frac{8}{4} = 2$.Since $\sin x$ cannot exceed $1$, $\sin x = 2$ is invalid.So we require $\frac{k-4}{2} \le 1$ → $k-4 \le 2$ → $k \le 6$.Also $\frac{k-4}{2} \ge -1$ gives $k \ge 2$, but the question asks for maximum $k$.Hence the maximum possible value of $k$ is $6$.Answer:$6$ (Option D).