Let f: R - {frac{alpha}{6}} rightarrow R be d | vedclass

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

Question

(B) Given $f(x) = \frac{5x + 3}{6x - \alpha}$.For $(f \circ f)(x) = x$,the function $f$ must be its own inverse,i.e.,$f(x) = f^{-1}(x)$.Let $y = f(x)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given $f(x) = \frac{5x + 3}{6x - \alpha}$.For $(f \circ f)(x) = x$,the function $f$ must be its own inverse,i.e.,$f(x) = f^{-1}(x)$.Let $y = f(x) = \frac{5x + 3}{6x - \alpha}$.Then $y(6x - \alpha) = 5x + 3$.$6xy - \alpha y = 5x + 3$.$6xy - 5x = \alpha y + 3$.$x(6y - 5) = \alpha y + 3$.$x = \frac{\alpha y + 3}{6y - 5}$.Thus,$f^{-1}(x) = \frac{\alpha x + 3}{6x - 5}$.Since $f(x) = f^{-1}(x)$,we have $\frac{5x + 3}{6x - \alpha} = \frac{\alpha x + 3}{6x - 5}$.Comparing the coefficients,we get $\alpha = 5$.

Let $f: R - \{\frac{\alpha}{6}\} \rightarrow R$ be defined by $f(x) = \frac{5x + 3}{6x - \alpha}$. Then the value of $\alpha$ for which $(f \circ f)(x) = x$,for all $x \in R - \{\frac{\alpha}{6}\}$,is:

Similar Questions

If $f:[1, \infty) \rightarrow [1, \infty)$ is defined by $f(x) = \frac{1+\sqrt{1+4 \log_2 x}}{2}$,then $f^{-1}(3) =$

Let $f(x) = (x + 2)^2 - 2, x \geq - 2$. Then $f^{-1}(x) =$

Inverse of the function $y = 2x - 3$ is

Which of the following functions is an invertible function?

If the function $f:[1, \infty) \to [1, \infty)$ is defined by $f(x) = 2^{x(x - 1)}$,then $f^{-1}(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module