Limit Continuity And Differentiability Ques 34

Answer:

Correct Answer: 34.(d)

Solution: (d) Let $y=f(f(f(x)))+(f(x))^2$

On differentiating both sides w.r.t. $x$, we get

$ \frac{d y}{d x}=f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)+2 f(x) f^{\prime}(x) $

[by chain rule]

So, $\left.\frac{d y}{d x}\right|_{\text {at } x=1}=f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)+2 f(1) f^{\prime}(1)$

$\left.\therefore \quad \frac{d y}{d x}\right|_{x=1}=f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot(3)+2(1)(3)$

$ \left [\because\quad f(1)=1 \text { and } f^{\prime}(1)=3\right] $

$ =f^{\prime}(1) \cdot(3) \cdot(3)+6 $

$ =(3 \times 9)+6=27+6=33$