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)=

$$f(x)=x5+e2xf(0)=0+e0=1f$$|(x)=5$$.$$x4+2$$.$$e2x$$⇒f|$$(0)=5(0)+2(1)=2(gof)(x)=x$$⇒g[$$f(x)]$$=x$$⇒g|$$f(x)$$.f|$$(x)=1$$⇒g|$$f(x)=1f$$|$$(x)put$$ $$x=0$$⇒g|$$(f(0))=1f$$|$$(0)g$$|$$(1)=12$$

Q.

If $g$ be the inverse of function $f$ , then $g (x) = f^{- 1} (x) \Rightarrow f (g (x)) = x \Rightarrow f^{|} (g (x)) g^{|} (x) = 1 \Rightarrow g^{|} (x) = \frac{1}{f^{|} (g (x))}$ also , if $f (α) = β$ , then $g (β) = α$

From (1), $g^{|} (β) = \frac{1}{f^{|} [g (β)]} ∴ g^{|} (β) = \frac{1}{f^{|} (α)}$ , where $f (α) = β$

If $f : R \to R$ defined by $f (x) = x^{5} + e^{2 x}$ and $g$ is the inverse of $f$ , then $g^{|} (1) =$

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$\frac{1}{2}$

none of these

answer is C.

(Detailed Solution Below)

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Detailed Solution

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

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

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

$(g o f) (x) = x \Rightarrow g [f (x)] = x$

$\Rightarrow g^{|} f (x) . f^{|} (x) = 1$

$\Rightarrow g^{|} f (x) = \frac{1}{f^{|} (x)}$

$put x = 0$

$\Rightarrow g^{|} (f (0)) = \frac{1}{f^{|} (0)}$

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

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