Test Index

MHTCET 22 Apr 2025 Shift 2 Paper

Question : 105 of 150

Marks: +1, -0

\mathrm{f}(1)=3, \mathrm{f}^{\prime}(1)=2

\frac{d}{\mathrm{dx}}\left\{\log \left[\mathrm{f}\left(e^{x}+2 x\right)\right]\right\}

x=0

Solution:

Concept:

Apply the chain rule to differentiate a composite logarithmic function, then substitute the given values.

Explanation:

Let

y = \log\left[f\left(e^x + 2x\right)\right]

Differentiate with respect to

x

using the chain rule:

\frac{dy}{dx} = \frac{1}{f\left(e^x + 2x\right)} \cdot f'\left(e^x + 2x\right) \cdot \left(e^x + 2\right)

x = 0

, we have

e^0 + 2(0) = 1

Given

f(1) = 3

and

f'(1) = 2

Substitute

x = 0

into the derivative:

\left.\frac{dy}{dx}\right|_{x=0} = \frac{1}{f(1)} \cdot f'(1) \cdot \left(e^0 + 2\right)

= \frac{1}{3} \cdot 2 \cdot (1 + 2) = \frac{1}{3} \cdot 2 \cdot 3 = 2

Answer:

2

, which matches option C.

Go to Question:

Prev Question

Next Question