If f: R - {frac{3}{5}} rightarrow R - {frac{3 | vedclass

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

Question

(B) To find the inverse of the function $f(x) = \frac{3x+1}{5x-3}$,let $y = f(x)$.So,$y = \frac{3x+1}{5x-3}$.Cross-multiplying gives $y(5x-3) = 3x+1$,

Vedclass · Accepted Answer

$f^{-1}(x) = f(x)$

Vedclass · Answer

(B) To find the inverse of the function $f(x) = \frac{3x+1}{5x-3}$,let $y = f(x)$.So,$y = \frac{3x+1}{5x-3}$.Cross-multiplying gives $y(5x-3) = 3x+1$,which simplifies to $5xy - 3y = 3x + 1$.Rearranging the terms to solve for $x$: $5xy - 3x = 3y + 1$.Factoring out $x$: $x(5y - 3) = 3y + 1$.Thus,$x = \frac{3y+1}{5y-3}$.By definition,$f^{-1}(y) = x = \frac{3y+1}{5y-3}$.Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{3x+1}{5x-3}$.Since $f^{-1}(x) = f(x)$,the correct option is $B$.

If $f: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+1}{5x-3}$,then which of the following is true?

Similar Questions

Let $g$ be the inverse function of $f$ and $f'(x) = \frac{x^{10}}{1 + x^2}$. If $g(2) = a$,then $g'(2)$ is equal to:

Let $g(x)$ be the inverse of the function $f(x)$ and $f'(x) = \frac{1}{1 + x^3}$. Then $g'(x)$ is equal to

$f: R \rightarrow R$,$f(x) = 4x + 3$ is defined,then $f^{-1}(x) =$ . . . . . . .

If $f(x) = x + \tan x$ and $g(x)$ is the inverse of $f(x)$,then $g'(x)$ is equal to

Let $x \neq 0$ and $|x| < \frac{1}{2}$. If $f(x) = 1 + 2x + 4x^2 + 8x^3 + \ldots$, then $f^{-1}(x) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module