If f(x) and g(x) are functions satisfying f(g(x)) = x^3 + 3x^2 + 3x + 4 f(x) = log^3x + 3 , t

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1.Relation and Function

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If $f(x)$ and $g(x)$ are functions satisfying $f(g(x))$ = $x^3 + 3x^2 + 3x + 4$ $f(x)$ = $log^3x + 3$, then slope of the tangent to the curve $y = g(x)$ at $x = \ -1$ is

A

$0$

B

$-1$

C

$1$

D

$e$

Solution

$f(g(x))=(x+1)^{3}+3$

$f(x)=(\log x)^{3}+3$

$\therefore \quad g(x)=e^{x+1}$

$\therefore \quad \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}+1}$

$g^{\prime}(-1)=1$

Standard 12

Mathematics

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