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:
- 7
- 42
- 1
- 14
The Correct Option is D
Solution and Explanation
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, \[ \frac{g(7)}{g'(7)} = \frac{1}{\frac{1}{14}} = 14 \]
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