Question

Let $f:\to R\to R,g:R\to Randh:R\to R$ be differentiable functions such that $f\left(x\right)=x^3+3x+2,g\left(f\right(x\left)\right)=xandh\left(g\right(g\left(x\right)\left)\right)=xforallx\epsilon R.$ Then

• A. $g^{\prime }\left(2\right)=\frac{1}{15}$
• B. $h^{\prime }\left(1\right)=666$
• C. $h\left(0\right)=16$
• D. $h\left(g\right(3\left)\right)=36$

Answer 1

Answer: (B), (C)

$h\left(0\right)=16$

$f\left(x\right)=x^3+3x+2\Rightarrow f^{\prime }\left(x\right)=3x^2+3Alsof\left(0\right)=2,f\left(1\right)=6,f\left(2\right)=16,f\left(3\right)=38,f\left(6\right)=236Andg\left(f\right(x\left)\right)=x\Rightarrow g\left(2\right)=0,g\left(6\right)=1,g\left(16\right)=2,g\left(38\right)=3,g\left(236\right)=6$

(a)$g\left(f\right(x\left)\right)=x\Rightarrow g^{\prime }\left(f\right(x\left)\right).f^{\prime }\left(x\right)=1For{g}^{\prime }\left(2\right),f\left(x\right)=2\Rightarrow x=0$
$\therefore$ Putting $x=0$, we get $g^{\prime }\left(f\right(0\left)\right)f^{\prime }\left(0\right)=1$
$\Rightarrow g^{\prime }\left(2\right)=\frac{1}{3}$

(b) $h\left(g\right(g\left(x\right)\left)\right)=x\Rightarrow h^{\prime }\left(g\right(g\left(x\right)\left)\right).g^{\prime }\left(g\right(x\left)\right).g^{\prime }\left(x\right)=1For{h}^{\prime }\left(1\right)$ we need g (g(x)) = 1
$\therefore g\left(x\right)=6\Rightarrow x=236\therefore$ Putting x = 236, we get $h^{\prime }\left[g\right(g\left(236\right)\left)\right]=\frac{1}{g^{\prime }\left(g\right(236\left)\right).g^{\prime }\left(236\right)}\Rightarrow h^{\prime }\left(g\right(6\left)\right)=\frac{1}{g^{\prime }\left(6\right).g^{\prime }\left(236\right)}\Rightarrow h^{\prime }\left(1\right)=\frac{1}{g^{\prime }\left(f\right(1\left)\right).g^{\prime }\left(f\right(6\left)\right)}=f^{\prime }\left(1\right).f^{\prime }\left(6\right)=6\times 111=666$

(c) $h\left[g\right(g\left(x\right)\left)\right]=xForh\left(0\right),g\left(g\right(x\left)\right)=0\Rightarrow g\left(x\right)=2\Rightarrow x=16$
$\therefore$ Putting x = 16, we get
$h\left(g\right(g\left(16\right)\left)\right)=16\Rightarrow h\left(0\right)=16$

(d)$h\left[g\right(g\left(x\right)\left)\right]=xForh\left(g\right(3\left)\right),weneedg\left(x\right)=3\Rightarrow x=38$
$\therefore$ Putting x = 38, we get
$h\left[g\right(g\left(38\right)\left)\right]=38\Rightarrow h\left(g\right(3\left)\right)=38$