If function f,:,R - ,{ 1, - 1} to A defined by f,(x), = frac{{{x^2}}}{{1 - {

English

Gujarati

Hindi

1.Relation and Function

hard

If the function $f\,:\,R - \,\{ 1, - 1\} \to A$ defined by $f\,(x)\, = \frac{{{x^2}}}{{1 - {x^2}}},$ is surjective, then $A$ is equal to

A

$R\, - \,[ - 1,0)$

B

$R\, - \,( - 1,0)$

C

$R\, - \,\{ - 1\} $

D

$[0,\infty )$

(JEE MAIN-2019)

Solution

$y\, = \frac{{{x^2}}}{{1 – {x^2}}}$

Range of $y : R – [ – 1,0)$ for surjective function, $A$ must be same as above range.

Standard 12

Mathematics

Share

Similar Questions

If $f(x) = \cos (\log x)$, then the value of $f(x).f(4) – \frac{1}{2}\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right]$

medium

Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{\,{x^3} – {x^2} + 10x – 5\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \le 1\,\,\,\,\,\,\,\,\,\,\,\,}\\
{ – 2x + {{\log }_2}({b^2} – 2),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, > 1\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}} \right.$ the set of values of $b$ for which $f(x)$ has greatest value at $x = 1$ is given by

normal

Range of $f(x) = \;[x]\; – x$ is

normal

If $f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _{\varepsilon}(123)}{x \log _{\varepsilon}(1234)-\left(\tan 1^{\circ}\right)}, x > 0$, then the least value of $f(f(x))+f\left(f\left(\frac{4}{x}\right)\right)$ is $………..$.

hard

(JEE MAIN-2023)

If function $f : R \to S, f(x) = (\sin x -\sqrt 3 \cos x+1)$ is onto, then $S$ is equal to

normal