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

If the function $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$,then the value of $\frac{1}{g'(1 + e^{1/5})}$ is

If $f(x) = x^2 + 1$,then $f^{-1}(17)$ and $f^{-1}(-3)$ will be

Let $f : A \to B$ be a function defined as $f(x) = \frac{x - 1}{x - 2}$,where $A = R - \{2\}$ and $B = R - \{1\}$. Then $f$ is

Let $f(x) > 0$ for all $x$ and $f^{\prime}(x)$ exists for all $x$. If $f$ is the inverse function of $h$ and $h^{\prime}(x) = \frac{1}{1 + \log x}$,then $f^{\prime}(x)$ will be

If $g(x)$ is the inverse function of $f(x)$ and $f^{\prime}(x) = \frac{1}{1+x^4}$,then $g^{\prime}(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module