Let f:(0, ∞) arrow(0, π) be a function defined by f(x)= cot -1( ln x) and g be the inverse function of f, then

Question

Let f:(0, ∞) arrow(0, π) be a function defined by f(x)= cot -1( ln x) and g be the inverse function of f, then (A) g prime((π/2))+ g prime prime((π/2))=0 (B) g prime((π/2))+ g prime prime((π/2))=-1 (C) g prime((π/2))+ g prime prime((π/2))=1 (D) g prime((π/2))+g prime prime((π/2))=3

Tardigrade Admin · Accepted Answer

f(1)= cot -1(0)=(π/2) ⇒ g ((π/2))=1, f prime( x )=(-1/ x (1+ ln 2 x )) ⇒ f prime(1)=-1 and f prime prime( x )=(1/ x 2(1+ ln 2 x ))+(2 ln x / x 2(1+ ln 2 x )) ⇒ f prime prime(1)=1 g prime((π/2))=(1/ f prime(1))=-1 g prime prime((π/2))=(-( f prime prime(1))/( f prime(1))3)=(-1/-1)=1

Q. Let $f : (0, \infty) \to (0, π)$ be a function defined by $f (x) = cot^{- 1} (ln x)$ and $g$ be the inverse function of $f$ , then

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = 0$

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = - 1$

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = 1$

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = 3$

Q. Let f:(0,∞)→(0,π) be a function defined by f(x)=cot−1(lnx) and g be the inverse function of f, then

g′(2π​)+g′′(2π​)=0

g′(2π​)+g′′(2π​)=−1

g′(2π​)+g′′(2π​)=1

g′(2π​)+g′′(2π​)=3

f(1)=cot−1(0)=2π​⇒g(2π​)=1,f′(x)=x(1+ln2x)−1​⇒f′(1)=−1 and f′′(x)=x2(1+ln2x)1​+x2(1+ln2x)2lnx​⇒f′′(1)=1 g′(2π​)=f′(1)1​=−1 g′′(2π​)=(f′(1))3−(f′′(1))​=−1−1​=1

Q. Let $f : (0, \infty) \to (0, π)$ be a function defined by $f (x) = cot^{- 1} (ln x)$ and $g$ be the inverse function of $f$ , then

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = 0$

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = - 1$

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = 1$

$g^{'} (\frac{π}{2}) + g^{''} (\frac{π}{2}) = 3$