Let R denote the set of all real numbers and | vedclass

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

Question

(D) The function is defined as $f(x) = x^2$.$A$. $f$ is one-one and onto,if $A = B = R^{+}$.For $A = R^{+}$,$f(x) = x^2$ is strictly increasing,so it

Vedclass · Accepted Answer

Vedclass · Answer

(D) The function is defined as $f(x) = x^2$.$A$. $f$ is one-one and onto,if $A = B = R^{+}$.For $A = R^{+}$,$f(x) = x^2$ is strictly increasing,so it is one-one. Since $B = R^{+}$,for every $y \in R^{+}$,there exists $x = \sqrt{y} \in R^{+}$ such that $f(x) = y$,so it is onto. Thus,$A \rightarrow 4$.$B$. $f$ is one-one but not onto,if $A = R^{+}, B = R$.For $A = R^{+}$,$f(x) = x^2$ is one-one. Since $B = R$,the range is $R^{+}$,which is a proper subset of $B$,so it is not onto. Thus,$B \rightarrow 1$.$C$. $f$ is onto but not one-one,if $A = R, B = R^{+}$.For $A = R$,$f(1) = 1$ and $f(-1) = 1$,so it is not one-one. Since $B = R^{+}$,for every $y \in R^{+}$,there exists $x = \pm \sqrt{y} \in R$ such that $f(x) = y$,so it is onto. Thus,$C \rightarrow 3$.$D$. $f$ is neither one-one nor onto,if $A = B = R$.For $A = R$,$f(1) = f(-1) = 1$,so it is not one-one. Since $B = R$,the range is $R^{+}$,which is a proper subset of $B$,so it is not onto. Thus,$D \rightarrow 2$.Therefore,the correct matching is $A \rightarrow 4, B \rightarrow 1, C \rightarrow 3, D \rightarrow 2$.

Similar Questions

If $f(x) = 2x$ and $g$ is the identity function,then:

If the function $f:(-\infty,-1] \rightarrow(a, b]$ defined by $f(x)=e^{x^3-3 x+1}$ is one-one and onto,then the distance of the point $P(2 b+4, a+2)$ from the line $x+e^{-3} y=4$ is:

$f: R \rightarrow R$ is a function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$. Assertion $(A): f(c) = \frac{1}{3}$ for some values of $c \in R$. Reason $(R): 0 < f(x) \leq \frac{1}{2\sqrt{2}}$ for all $x \in R$. Then which of the following options is correct?

Let $f: R \to R$ be a function. Define $g: R \to R$ by $g(x) = |f(x)|$ for all $x$. Then $g$ is

If $f(x) = \cos (\log x)$,then $f(x^2)f(y^2) - \frac{1}{2}\left[ f\left( \frac{x^2}{y^2} \right) + f(x^2y^2) \right]$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module