Let f: R rightarrow R be given by f(x) = tan | vedclass

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

Question

(B) Given $f(x) = 	an x$.The inverse function $f^{-1}(y)$ is defined such that $f(f^{-1}(y)) = y$.Here,we need to find $f^{-1}(1)$,so we set $f(x) =

Vedclass · Accepted Answer

$\{n \pi + \frac{\pi}{4} : n \in Z\}$

Vedclass · Answer

(B) Given $f(x) = 	an x$.The inverse function $f^{-1}(y)$ is defined such that $f(f^{-1}(y)) = y$.Here,we need to find $f^{-1}(1)$,so we set $f(x) = 1$.$	an x = 1$.The general solution for $	an x = 	an \alpha$ is $x = n \pi + \alpha$,where $n \in Z$.Since $	an(\frac{\pi}{4}) = 1$,the set of values for $x$ is $\{n \pi + \frac{\pi}{4} : n \in Z\}$.Therefore,$f^{-1}(1) = \{n \pi + \frac{\pi}{4} : n \in Z\}$.

Let $f: R \rightarrow R$ be given by $f(x) = \tan x$. Then,$f^{-1}(1)$ is

Similar Questions

If $g$ is the inverse of a function $f$ and $f'(x) = \frac{1}{1 + x^5}$,then $g'(x)$ is equal to:

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

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 $f: R - \{\frac{\alpha}{6}\} \rightarrow R$ 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:

If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = 3x - 4$,then ${f^{ - 1}}: \mathbb{R} \to \mathbb{R}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module