UPSC NDA I 2016 Mathematics Solved Paper

Concept:The equation is $k \sin x + \cos 2x = 2k - 7$.We convert it into a quadratic in $\sin x$ using the identity $\cos 2x = 1 - 2 \sin^2 x$.Then, for a real $\sin x$ to exist, its value must lie in $[-1, 1]$.This gives a range for $k$, and the minimum $k$ in that range is the answer.Explanation:Start with the given equation: $k \sin x + \cos 2x = 2k - 7$.Substitute $\cos 2x = 1 - 2 \sin^2 x$:$k \sin x + 1 - 2 \sin^2 x = 2k - 7$.Rearrange: $2 \sin^2 x - k \sin x + (2k - 8) = 0$.This is a quadratic in $t = \sin x$: $2t^2 - k t + (2k - 8) = 0$.Solve for $t$ using the quadratic formula:$t = \frac{k \pm \sqrt{k^2 - 4 \cdot 2 \cdot (2k - 8)}}{2 \cdot 2} = \frac{k \pm \sqrt{k^2 - 16k + 64}}{4}$.The radicand simplifies: $\sqrt{(k-8)^2} = |k-8|$.Thus $t = \frac{k \pm |k-8|}{4}$.Two cases: $t_1 = \frac{k - (k-8)}{4} = 2$ (but $\sin x = 2$ is invalid since $\sin x \in [-1,1]$).$t_2 = \frac{k + (k-8)}{4} = \frac{2k-8}{4} = \frac{k-4}{2}$.For a solution to exist, we need $-1 \le \frac{k-4}{2} \le 1$.Multiply by 2: $-2 \le k-4 \le 2$.Add 4: $2 \le k \le 6$.Hence the minimum integer value of $k$ is $2$.Answer:The minimum value of $k$ is $2$, which corresponds to option B.