Let f: R + arrow(-7, ∞) be a function defined by f(x)=( x 3-7/ x 2+1) and g be the inverse of f such that (1/α) g ((1/α2+1))=1 for some α. If k =( g (α+1)/ g ( g (α))) then find the value of k.

Question

Let f: R + arrow(-7, ∞) be a function defined by f(x)=( x 3-7/ x 2+1) and g be the inverse of f such that (1/α) g ((1/α2+1))=1 for some α. If k =( g (α+1)/ g ( g (α))) then find the value of k. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

g ((1/α2+1))=α ⇒ f (α)=(1/α2+1) ⇒ α=2 k =( g (α+1)/ g ( g (α)))=( g (3)/ g ( g (2)))=( g (3)/ g (3))=1

Q. Let $f : R^{+} \to (- 7, \infty)$ be a function defined by $f (x) = \frac{x ^{3} - 7}{x ^{2} + 1}$ and $g$ be the inverse of $f$ such that $\frac{1}{α} g (\frac{1}{α ^{2} + 1}) = 1$ for some $α$ . If $k = \frac{g ( α + 1 )}{g ( g ( α ))}$ then find the value of $k$ .

$g (\frac{1}{α ^{2} + 1}) = α \Rightarrow f (α) = \frac{1}{α ^{2} + 1} \Rightarrow α = 2$
$k = \frac{g ( α + 1 )}{g ( g ( α ))} = \frac{g ( 3 )}{g ( g ( 2 ))} = \frac{g ( 3 )}{g ( 3 )} = 1$

Q. Let f:R+→(−7,∞) be a function defined by f(x)=x2+1x3−7​ and g be the inverse of f such that α1​g(α2+11​)=1 for some α. If k=g(g(α))g(α+1)​ then find the value of k.

g(α2+11​)=α⇒f(α)=α2+11​⇒α=2 k=g(g(α))g(α+1)​=g(g(2))g(3)​=g(3)g(3)​=1

Q. Let $f : R^{+} \to (- 7, \infty)$ be a function defined by $f (x) = \frac{x ^{3} - 7}{x ^{2} + 1}$ and $g$ be the inverse of $f$ such that $\frac{1}{α} g (\frac{1}{α ^{2} + 1}) = 1$ for some $α$ . If $k = \frac{g ( α + 1 )}{g ( g ( α ))}$ then find the value of $k$ .

$g (\frac{1}{α ^{2} + 1}) = α \Rightarrow f (α) = \frac{1}{α ^{2} + 1} \Rightarrow α = 2$
$k = \frac{g ( α + 1 )}{g ( g ( α ))} = \frac{g ( 3 )}{g ( g ( 2 ))} = \frac{g ( 3 )}{g ( 3 )} = 1$