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One-one, into and onto functions Mock Test for JEE

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Syllabus context

Part of Sets, Relations and Functions in JEE Main Mathematics.

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Q1MathsUnit 1: Sets, Relations and Functions
If f:[0,∞)→[0,∞),\mathbf{f}:[\mathbf{0}, \infty) \rightarrow[\mathbf{0}, \infty), and f(x)=x1+x\mathbf{f}(\mathbf{x})=\frac{\mathbf{x}}{\mathbf{1}+\mathbf{x}} then f\boldsymbol{f} is
Q2MathsUnit 1: Sets, Relations and Functions
Let f:R→R\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R} where f(x)=2∣x∣−2−x\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{2}^{|\boldsymbol{x}|}-\boldsymbol{2}^{-\boldsymbol{x}} then f(x)f(x) is
Q3MathsUnit 1: Sets, Relations and Functions
The function f:(−∞,−1)→(0,e5]\boldsymbol{f}:(-\infty,-1) \rightarrow\left(\mathbf{0}, e^{5}\right] defined by f(x)=ex3−3x+2f(x)=e^{x^{3}-3 x+2} is
Q4MathsUnit 1: Sets, Relations and Functions
If a function f:(2,∞)→B\boldsymbol{f}:(2, \infty) \rightarrow \boldsymbol{B} defined by f(x)=x2−4x+5f(x)=x^{2}-4 x+5 is a bijection, then B=\boldsymbol{B}=
Q5MathsUnit 1: Sets, Relations and Functions
Let A={x,y,z},B={a,b,c}\boldsymbol{A}=\{\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}\}, \boldsymbol{B}=\{\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\} and f\boldsymbol{f} A→BA \rightarrow B be defined by f(x)=a,f(y)=f(x)=a, f(y)= b,f(z)=c.\boldsymbol{b}, \boldsymbol{f}(\boldsymbol{z})=\boldsymbol{c} . This function is
Q6MathsUnit 1: Sets, Relations and Functions
If f(A)f(A) is a proper subset of BB, then ff A→BA \rightarrow B is called a/an ……..\ldots \ldots . . function
Q7MathsUnit 1: Sets, Relations and Functions
An identity function is a?
Q8MathsUnit 1: Sets, Relations and Functions
f:Q→Q\boldsymbol{f}: \boldsymbol{Q} \rightarrow \boldsymbol{Q} is defined by f(x)=15x+7\boldsymbol{f}(\boldsymbol{x})=\mathbf{1 5} \boldsymbol{x}+\mathbf{7} is
Q9MathsUnit 1: Sets, Relations and Functions
The function f:R→R\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R} defined by f(x)=x21+x2∀x∈Rf(x)=\frac{x^{2}}{1+x^{2}} \forall x \in R is
Q10MathsUnit 1: Sets, Relations and Functions
Let f:N→Nf: N \rightarrow N defined by f(n)=f(n)= {n+12 if n is odd n2 if n is even \left\{\begin{array}{cc}\frac{n+1}{2} & \text { if } n \text { is odd } \\ \frac{n}{2} & \text { if } n \text { is even }\end{array}\right. then f\boldsymbol{f} is.
Q11MathsUnit 1: Sets, Relations and Functions
The function f:N→N\boldsymbol{f}: \boldsymbol{N} \rightarrow \boldsymbol{N} defined by f(x)=x−5[x5],\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}-\mathbf{5}\left[\frac{\boldsymbol{x}}{\mathbf{5}}\right], where N\mathrm{N} is the set of natural numbers and [x][x] denotes the greatest integer less than or equal to xx is
Q12MathsUnit 1: Sets, Relations and Functions
Assertion f:R→[0,π2)f: R \rightarrow\left[0, \frac{\pi}{2}\right) defined by f(x)=f(x)= tan⁡−1(x2+x+a)\tan ^{-1}\left(x^{2}+x+a\right) is onto for all a∈a \in (−∞,14)\left(-\infty, \frac{1}{4}\right) Reason For onto function codomain of f=\boldsymbol{f}= Range of f\boldsymbol{f}
Q13MathsUnit 1: Sets, Relations and Functions
Let Z\mathbb{Z} be the set of integers and f:Z→\boldsymbol{f}: \mathbb{Z} \rightarrow Z\mathbb{Z} be defined as f(x)=x2,x∈Z\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}^{2}, \boldsymbol{x} \in \mathbb{Z} then function is

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