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MHTCET 16 May 2024 Shift 1 Paper

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Question : 114 of 150

Marks: +1, -0

If

\mathrm{f}(1)=1, \mathrm{f}^{\prime}(1)=3

, then the derivative of

\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^{2}

x=1

is

12
9
15
33

Solution:

Concept:

Use the chain rule for composite functions and evaluate at

x = 1

using the given values.

Explanation:

Let

y = \mathrm{f}(\mathrm{f}(\mathrm{f}(x))) + (\mathrm{f}(x))^2

.

Differentiate with respect to

x

:

\frac{dy}{dx} = \mathrm{f}'(\mathrm{f}(\mathrm{f}(x))) \cdot \mathrm{f}'(\mathrm{f}(x)) \cdot \mathrm{f}'(x) + 2\mathrm{f}(x)\mathrm{f}'(x)

Given

\mathrm{f}(1)=1

and

\mathrm{f}'(1)=3

.

So,

\mathrm{f}(\mathrm{f}(1)) = \mathrm{f}(1) = 1

, and

\mathrm{f}(\mathrm{f}(\mathrm{f}(1))) = \mathrm{f}(1) = 1

.

Now evaluate at

x=1

:

\left.\frac{dy}{dx}\right|_{x=1} = \mathrm{f}'(1) \cdot \mathrm{f}'(1) \cdot \mathrm{f}'(1) + 2\mathrm{f}(1)\mathrm{f}'(1)

= 3 \cdot 3 \cdot 3 + 2 \cdot 1 \cdot 3 = 27 + 6 = 33

Answer:

The derivative at

x=1

is

33

, so the correct option is D.

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