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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