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MHTCET 16 May 2024 Shift 2 Paper

Question : 101 of 150

Marks: +1, -0

\mathrm{g}(x)=[\mathrm{f}(2 \mathrm{f}(x)+2)]^{2}

\mathrm{f}(0)=-1, \mathrm{f}^{\prime}(0)=1

g^{\prime}(0)

Solution:

Concept:

Use the chain rule twice to differentiate a composite function, then substitute the given values.

Explanation:

Given:

\mathrm{g}(x) = \left[\mathrm{f}(2\mathrm{f}(x)+2)\right]^2

Differentiate using the chain rule:

\mathrm{g}'(x) = 2\,\mathrm{f}(2\mathrm{f}(x)+2) \cdot \mathrm{f}'(2\mathrm{f}(x)+2) \cdot 2\mathrm{f}'(x)

Now substitute

x = 0

\mathrm{g}'(0) = 2\,\mathrm{f}(2\mathrm{f}(0)+2) \cdot \mathrm{f}'(2\mathrm{f}(0)+2) \cdot 2\mathrm{f}'(0)

Using

\mathrm{f}(0) = -1

and

\mathrm{f}'(0) = 1

2\mathrm{f}(0)+2 = 2(-1)+2 = 0

So,

\mathrm{g}'(0) = 2\,\mathrm{f}(0) \cdot \mathrm{f}'(0) \cdot 2(1)

= 4(-1)(1) = -4

Answer:

\mathrm{g}'(0) = -4

, which matches option A.

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