Question

Let \( f(x) = x^5 + 2x^3 + 3x + 1 \), \( x \in \mathbb{R} \), and \( g(x) \) be a function such that \( g(f(x)) = x \) for all \( x \in \mathbb{R} \). Then \( \frac{g(7)}{g'(7)} \) is equal to:

• A. 7
• B. 42
• C. 1
• D. 14

Answer 1

The Correct Option is D

Given:

\( f(x) = x^5 + 2x^3 + 3x + 1 \)

Then,

\( f'(x) = 5x^4 + 6x^2 + 3 \)

Calculate \( f'(1) \):

\( f'(1) = 5 \cdot 1^4 + 6 \cdot 1^2 + 3 = 14 \)

Since \( g(f(x)) = x \), by differentiation, we get:

\( g'(f(x))f'(x) = 1 \)

For \( f(x) = 7 \):

\( x^5 + 2x^3 + 3x + 1 = 7 \)

This implies \( x = 1 \), so \( f(1) = 7 \).

Then \( g(7) = 1 \).

Now,

\( g'(7)f'(1) = 1 \Rightarrow g'(7) = \frac{1}{f'(1)} = \frac{1}{14} \)

Thus, \[ \frac{g(7)}{g'(7)} = \frac{1}{\frac{1}{14}} = 14 \]