Let f ( x ) = x 5 + 2 x 3 + 3 x + 1 , x ? R , and g ( x ) be a function such that g ( f ( x ) ) = x for all x ? R . Then g ( 7 ) g ' ( 7 ) is equal to [2024]

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Solution

Q.
$Let f (x) = x^{5} + 2 x^{3} + 3 x + 1, x \in R, and g (x) be a function such that g (f (x)) = x for all x \in R . Then \frac{g (7)}{g' (7)} is equal to$ [2024]

1 1

2 7

3 42

4 14

Ans.
(4)

$We have, g (f (x)) = x$

$On differentiating w.r.t. x, we get$ $g' (f (x)) \cdot f' (x) = 1$

$\Rightarrow g' (f (x)) = \frac{1}{f' (x)}$ ...(i)

$Now, put f (x) = 7, we get$ $x^{5} + 2 x^{3} + 3 x + 1 = 7$

$\Rightarrow x^{5} + 2 x^{3} + 3 x - 6 = 0$

$x = 1$ is the only solution because this function is increasing

$\Rightarrow g (7) = 1$ ...(ii)

$Now, g' (7) = \frac{1}{f' (1)}$ [By (i) and (ii)]

$= \frac{1}{5 + 6 + 3} = \frac{1}{14}$

$∴ \frac{g (7)}{g' (7)} = 14$

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