If f : R to R, f(x) = x^2 + 1 , then f^{-1}(17) and f^{-1}(-3) are

English

Gujarati

Hindi

1.Relation and Function

normal

If $f : R \to R, f(x) = x^2 + 1$, then $f^{-1}(17)$ and $f^{-1}(-3)$ are

A

$\{8, -8\}, \{\sqrt 2 \}$

B

$\{3, -3\}, \phi$

C

$\{4, -4\}, \phi$

D

$\phi ,\{4, -4\}$

Solution

$f^{-1}(17)=x(\text { say })$

$17=\mathrm{f}(\mathrm{x})$

$17=x^{2}+1$

$x=\pm 4$

$f^{-1}(-3)=y(\text { say })$

$-3=f(y)$

$-3=y^{2}+1$

$\mathrm{y}^{2}=-4$ Not possible

Standard 12

Mathematics

Share

Similar Questions

If $a * b=10$ ab on $Q^{+}$ then find the inverse of 0.01

medium

Show that $f:[-1,1] \rightarrow R ,$ given by $f(x)=\frac{x}{(x+2)}$ is one-one. Find the inverse of the function $f:[-1,1] \rightarrow$ Range $f$

$($ Hint: For $y \in $ Range $f$, $y=f(x)=\frac{x}{x+2}$, for some $x$ in $[-1,1]$, i.e., $x=\frac{2 y}{(1-y)})$

hard

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

hard

(IIT-1999)

Which of the following function is invertible

normal

Consider $f: R _{+} \rightarrow[-5, \infty)$ given by $f(x)=9 x^{2}+6 x-5 .$ Show that $f$ is invertible with $f^{-1}(y)=\left(\frac{(\sqrt{y+6})-1}{3}\right)$

hard