If f: R arrow R, g: R arrow R are defined by f(x)=5 x-3, g(x)=x2+3, then g o f-1(3) is equal to

Question

If f: R arrow R, g: R arrow R are defined by f(x)=5 x-3, g(x)=x2+3, then g o f-1(3) is equal to (A) (25/3) (B) (111/25) (C) (9/25) (D) (25/111)

Tardigrade Admin · Accepted Answer

Given, f(x)=5 x-3 Let y=5 x-3 ⇒ y+3=5 x ⇒ x=(y+3/5) ∴ f-1(y)=(y+3/5) ⇒ f-1(x) =(x+3/5) and g(x) =x2+3 Now, g o f-1(3)=g[f-1(3)] =g((3+3/5))=g((6/5)) ⇒ g((6/5))=((6)2/(5)2)+3=(36/25)+3=(111/25)

Q. If $f : R \to R, g : R \to R$ are defined by
$f (x) = 5 x - 3, g (x) = x^{2} + 3$ , then $g o f^{- 1} (3)$ is equal to

$\frac{25}{3}$

$\frac{111}{25}$

$\frac{9}{25}$

$\frac{25}{111}$

Q. If f:R→R,g:R→R are defined by f(x)=5x−3,g(x)=x2+3, then gof−1(3) is equal to

325​

25111​

259​

11125​

Given, f(x)=5x−3 Let y=5x−3 ⇒y+3=5x ⇒x=5y+3​ ∴f−1(y)=5y+3​ ⇒f−1(x)=5x+3​ and g(x)=x2+3 Now, gof−1(3)=g[f−1(3)] =g(53+3​)=g(56​) ⇒g(56​)=(5)2(6)2​+3=2536​+3=25111​

Q. If $f : R \to R, g : R \to R$ are defined by
$f (x) = 5 x - 3, g (x) = x^{2} + 3$ , then $g o f^{- 1} (3)$ is equal to

$\frac{25}{3}$

$\frac{111}{25}$

$\frac{9}{25}$

$\frac{25}{111}$