MHTCET 17 Apr 2026 Shift 2 Paper

Concept:Differentiate the given integral expression and compare the result with the integrand to determine the unknown coefficients.Explanation:Rewrite $\sin 4x = 4\sin x \cos x \cos 2x$.So, $\frac{\sin x}{\sin 4x} = \frac{1}{4\cos x \cos 2x}$.Differentiate the first log term: $\frac{d}{dx}\log\left|\frac{1+\sin x}{1-\sin x}\right| = \frac{2}{\cos x}$.Differentiate the second log term: $\frac{d}{dx}\log\left|\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right| = \frac{2\sqrt{2}\cos x}{\cos 2x}$.Thus, the derivative of the RHS is $\frac{2\alpha}{\cos x} + \frac{2\sqrt{2}\beta\cos x}{\cos 2x}$.Equate this to the integrand: $\frac{1}{4\cos x \cos 2x} = \frac{2\alpha}{\cos x} + \frac{2\sqrt{2}\beta\cos x}{\cos 2x}$.Multiply both sides by $\cos x \cos 2x$: $\frac{1}{4} = 2\alpha\cos 2x + 2\sqrt{2}\beta\cos^2 x$.Using $\cos 2x = 2\cos^2 x - 1$ gives $\frac{1}{4} = (4\alpha + 2\sqrt{2}\beta)\cos^2 x - 2\alpha$.Compare coefficients: $-2\alpha = \frac{1}{4}$, hence $\alpha = -\frac{1}{8}$.Also, $4\alpha + 2\sqrt{2}\beta = 0$, giving $\beta = \frac{1}{4\sqrt{2}}$.Therefore, $32(\alpha + \beta^2) = 32\left(-\frac{1}{8} + \frac{1}{32}\right) = -3$.Answer:$-3$ (Option D).