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(D) Given $f(x)=x^3+3x+2$. Since $g(f(x))=x$,$g$ is the inverse of $f$. Differentiating $g(f(x))=x$ with respect to $x$,we get $g'(f(x)) \cdot f'(x)=1

Vedclass · Accepted Answer

$BC$

Vedclass · Answer

(D) Given $f(x)=x^3+3x+2$. Since $g(f(x))=x$,$g$ is the inverse of $f$. Differentiating $g(f(x))=x$ with respect to $x$,we get $g'(f(x)) \cdot f'(x)=1$. For $g'(2)$,we set $f(x)=2 \implies x^3+3x+2=2 \implies x^3+3x=0 \implies x(x^2+3)=0$,so $x=0$. Thus $g'(2) \cdot f'(0)=1$. Since $f'(x)=3x^2+3$,$f'(0)=3$. So $g'(2) = \frac{1}{3}$. Option $A$ is incorrect. Given $h(g(g(x)))=x$. Since $g(f(x))=x$,$g(g(f(f(x))))=x$,which implies $g(g(x))=f(f(x))$. Thus $h(f(f(x)))=x$. This means $h$ is the inverse of $f(f(x))$. However,the problem states $h(g(g(x)))=x$. Since $g(g(x))=f^{-1}(f^{-1}(x))$,we have $h(f^{-1}(f^{-1}(x)))=x$. Let $f^{-1}(x)=y$,then $f(y)=x$. The equation becomes $h(f^{-1}(y))=f(y)$. Let $f^{-1}(y)=z$,then $y=f(z)$. So $h(z)=f(f(z))$. Thus $h(x)=f(f(x))$. Now,$h'(x)=f'(f(x)) \cdot f'(x)$. $h'(1)=f'(f(1)) \cdot f'(1)$. Since $f(1)=1^3+3(1)+2=6$ and $f'(x)=3x^2+3$,$f'(1)=6$ and $f'(6)=3(6^2)+3=111$. $h'(1)=111 	imes 6=666$. Option $B$ is correct. $h(0)=f(f(0))=f(2)=2^3+3(2)+2=16$. Option $C$ is correct. $h(g(3))=f(f(g(3)))=f(3)=3^3+3(3)+2=38$. Option $D$ is incorrect. Therefore,the correct options are $B$ and $C$.

Let $f: R \rightarrow R, g: R \rightarrow R$ and $h: R \rightarrow R$ be differentiable functions such that $f(x)=x^3+3x+2, g(f(x))=x$ and $h(g(g(x)))=x$ for all $x \in R$. Then

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