MHTCET 23 Apr 2025 Shift 1 Paper

Concept:Simplify the trigonometric expression inside $\tan^{-1}$ using standard identities, then differentiate.Explanation:Step 1: Use $\sin 2x = 2\sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$.The denominator becomes $\cos^2 x - \sin^2 x - 6\sin^2 x = \cos^2 x - 7\sin^2 x$.So $y = \tan^{-1}\left(\frac{8\sin x \cos x}{\cos^2 x - 7\sin^2 x}\right)$.Step 2: Divide the numerator and denominator by $\cos^2 x$.$y = \tan^{-1}\left(\frac{8\tan x}{1 - 7\tan^2 x}\right)$.Step 3: Write $8\tan x$ as $7\tan x + \tan x$ and use $\tan^{-1}\left(\frac{a+b}{1-ab}\right) = \tan^{-1}a + \tan^{-1}b$.Here, $a = 7\tan x$ and $b = \tan x$, with $ab = 7\tan^2 x$.Thus $y = \tan^{-1}(7\tan x) + \tan^{-1}(\tan x)$.For the principal branch near $x=0$, $\tan^{-1}(\tan x) = x$, so $y = \tan^{-1}(7\tan x) + x$.Step 4: Differentiate with respect to $x$.$\frac{dy}{dx} = \frac{1}{1 + (7\tan x)^2} \cdot 7\sec^2 x + 1$.At $x = 0$: $\tan 0 = 0$ and $\sec 0 = 1$.$\left(\frac{dy}{dx}\right)_{x=0} = \frac{7(1)}{1 + 0} + 1 = 8$.Answer:$\frac{dy}{dx}$ at $x=0$ is $8$.Hence, the correct option is C. $8$.