Let f: R-{3} rightarrow R-{1} be defined by f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = \frac{x-2}{x-3}$. Let $y = \frac{x-2}{x-3}$.$y(x-3) = x-2 \implies xy - 3y = x - 2 \implies x(y-1) = 3y - 2 \implies x = \frac{3y-2}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given $f(x) = \frac{x-2}{x-3}$. Let $y = \frac{x-2}{x-3}$.$y(x-3) = x-2 \implies xy - 3y = x - 2 \implies x(y-1) = 3y - 2 \implies x = \frac{3y-2}{y-1}$.Thus, $f^{-1}(x) = \frac{3x-2}{x-1}$.Given $g(x) = 2x-3$. Let $y = 2x-3$.$y+3 = 2x \implies x = \frac{y+3}{2}$.Thus, $g^{-1}(x) = \frac{x+3}{2}$.We are given $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$.$\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$.Multiply by $2(x-1)$: $2(3x-2) + (x+3)(x-1) = 13(x-1)$.$6x - 4 + x^2 + 2x - 3 = 13x - 13$.$x^2 + 8x - 7 = 13x - 13$.$x^2 - 5x + 6 = 0$.$(x-2)(x-3) = 0$.The roots are $x = 2$ and $x = 3$. However, $f^{-1}(x)$ is defined for $x 
eq 1$. Since $x=3$ is in the domain of $f^{-1}$, we check the original function $f(x)$. The domain of $f$ is $R-\{3\}$, so $f^{-1}$ is defined on $R-\{1\}$. Both $x=2$ and $x=3$ are valid. The sum of the roots is $2+3 = 5$.

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

Similar Questions

Which of the following quantities does not depend upon the orbital radius of the satellite?

What is the objective of Systematics?

Which of the following is an acidic salt?

Which oxide of manganese is amphoteric?

Vedclass Test Series

Exam Paper Generator

Online Exam Module