Let \(f: R-\{3\} \rightarrow R-\{1\}\) be defined by \(f(x)=\frac{x-2}{x-3}\). Let \(g: R \rightarrow R\) be given as \(g(x)=2x-3\). Then, the sum of all the values of \(x\) for which \(f^{-1}(x)+g^{-1}(x)=\frac{13}{2}\) is equal to :

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Sets Relations and Functions

Let

f : R - {3} \to R - {1}

be defined by

f (x) = \frac{x - 2}{x - 3}

.
Let

g : R \to R

be given as

g (x) = 2 x - 3

. Then, the sum of all the values of

x

for which

f^{- 1} (x) + g^{- 1} (x) = \frac{13}{2}

is equal to :

Answer: (C)

Inverting

y = \frac{x - 2}{x - 3}

gives

y (x - 3) = x - 2

, so

x (y - 1) = 3 y - 2

and

f^{- 1} (x) = \frac{3 x - 2}{x - 1}

with domain

R - {1}

. Inverting

y = 2 x - 3

gives

g^{- 1} (x) = \frac{x + 3}{2}

. The equation becomes

\frac{3 x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2}

Multiplying through by

2 (x - 1)

2 (3 x - 2) + (x + 3) (x - 1) = 13 (x - 1) ⟹ x^{2} - 5 x + 6 = 0

x = 2

and

x = 3

. Both lie in

R - {1}

and both satisfy the original equation, so the sum of all such values is

5

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option