If g be the inverse of function f, then  g(x)=f−1(x)⇒f(g(x))=x⇒f|(g(x))g|(x)=1⇒g|(x)=1f|(g(x)) also , if f(α)=β, then g(β)=αFrom (1), g|(β)=1f|[g(β)]∴g|(β)=1f|(α), where  f(α)=βIf  f:R→R defined by f(x)=x5+e2x and g is the inverse of f , then g|(1)=

# If $g$ be the inverse of function $f$, then  $g\left(x\right)={f}^{-1}\left(x\right)⇒f\left(g\left(x\right)\right)=x⇒{f}^{|}\left(g\left(x\right)\right){g}^{|}\left(x\right)=1⇒{g}^{|}\left(x\right)=\frac{1}{{f}^{|}\left(g\left(x\right)\right)}$ also , if $f\left(\alpha \right)=\beta$, then $g\left(\beta \right)=\alpha$From (1), ${g}^{|}\left(\beta \right)=\frac{1}{{f}^{|}\left[g\left(\beta \right)\right]}\therefore {g}^{|}\left(\beta \right)=\frac{1}{{f}^{|}\left(\alpha \right)}$, where  $f\left(\alpha \right)=\beta$If  $f:R\to R$ defined by $f\left(x\right)={x}^{5}+{e}^{2x}$ and $g$ is the inverse of $f$ , then ${g}^{|}\left(1\right)=$

1. A

2

2. B

1

3. C

$\frac{1}{2}$

4. D

none of these

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### Solution:

$f\left(x\right)={x}^{5}+{e}^{2x}$

$f\left(0\right)=0+{e}^{0}=1$

${f}^{|}\left(x\right)=5.{x}^{4}+2.{e}^{2x}⇒{f}^{|}\left(0\right)=5\left(0\right)+2\left(1\right)=2$

$\left(gof\right)\left(x\right)=x⇒g\left[f\left(x\right)\right]=x$

$⇒{g}^{|}f\left(x\right).{f}^{|}\left(x\right)=1$

$⇒{g}^{|}f\left(x\right)=\frac{1}{{f}^{|}\left(x\right)}$

$\text{put\hspace{0.17em}\hspace{0.17em}}x=0$

$⇒{g}^{|}\left(f\left(0\right)\right)=\frac{1}{{f}^{|}\left(0\right)}$

${g}^{|}\left(1\right)=\frac{1}{2}$

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