MHTCET 15 May 2024 Shift 2 Paper

Concept:This question tests the chain rule for differentiating composite functions and the derivative of a squared function.Explanation:Let $y = f(f(f(x))) + (f(x))^2$.Differentiate both terms with respect to $x$.For the first term, apply the chain rule repeatedly:$\frac{d}{dx} f(f(f(x))) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)$.For the second term: $\frac{d}{dx}(f(x))^2 = 2 f(x) f'(x)$.So, $\frac{dy}{dx} = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2 f(x) f'(x)$.At $x=1$, we know $f(1)=1$ and $f'(1)=3$.Hence, $f(f(1)) = f(1) = 1$ and $f(f(f(1))) = f(1) = 1$.Substitute into the derivative:$\left.\frac{dy}{dx}\right|_{x=1} = f'(1) \cdot f'(1) \cdot f'(1) + 2 \cdot 1 \cdot 3$.$= 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.Answer:The required derivative is $33$, so the correct option is D.