Let f : R - 3 [ ] R - 1 be defined by f(x) = [x - 2 x - 3].

Question

Let f : R - {3} $$ \to $$ R - {1} be defined by f(x) = $${{x - 2} \over {x - 3}}$$.

Let g : R $$ \to $$ 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) = $${{13} \over 2}$$ is equal to :

Accepted Answer

Finding inverse of f(x)

$$y = {{x - 2} \over {x - 3}} \Rightarrow xy - 3y = x - 2 \Rightarrow x(y - 1) = 3y - 2$$

$$ \therefore $$ $${f^{ - 1}}(x) = {{3x - 2} \over {x - 1}}$$

Similarly for ${g^{ - 1}}(x)$

$$y = 2x - 3 \Rightarrow x = {{y + 3} \over 2} \Rightarrow {g^{ - 1}}(x) = {{x + 3} \over 2}$$

$$ \therefore $$ $${{3x - 2} \over {x - 1}} + {{x + 3} \over 2} = {{13} \over 2}$$

$$ \Rightarrow 6x - 4 + {x^2} + 2x - 3 = 13x - 13$$

$$ \Rightarrow {x^2} - 5x + 6 = 0$$

$$ \Rightarrow (x - 2)(x - 3) = 0$$

$$ \Rightarrow $$ x = 2 or 3

Solutions · 1