Chapter 2
71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers · 14 exercises
Problem 1
Let \(R_{1}\) and \(R_{2}\) be two relations defined as follows: \(R_{1}=\left\\{(a, b) \in \mathbf{R}^{2}: a^{2}+b^{2} \in Q\right\\}\) and \(R_{2}=\left\\{(a, b) \in \mathbf{R}^{2}: a^{2}+b^{2} \notin Q\right\\}\), where \(Q\) is the set of all rational numbers. Then : (a) Neither \(R_{1}\) nor \(R_{2}\) is transitive. (b) \(R_{2}\) is transitive but \(R_{1}\) is not transitive. (c) \(R_{1}\) is transitive but \(R_{2}\) is not transitive. (d) \(R_{1}\) and \(R_{2}\) are both transitive.
4 step solution
Problem 2
The domain of the function \(f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^{2}+1}\right)\) is \((-\infty,-a] \cup[a, \infty] .\) Then \(a\) is equal to : [Sep. 02,2020 (I)] (a) \(\frac{\sqrt{17}}{2}\) (b) \(\frac{\sqrt{17}-1}{2}\) (c) \(\frac{1+\sqrt{17}}{2}\) (d) \(\frac{\sqrt{17}}{2}+1\)
8 step solution
Problem 3
If \(R=\left\\{(x, y): x, y \in \mathbf{Z}, x^{2}+3 y^{2} \leq 8\right\\}\) is a relation on the set of integers \(\mathbf{Z}\), then the domain of \(R^{-1}\) is : [Sep. \(02,2020(\mathrm{I})]\) (a) \(\\{-2,-1,1,2\\}\) (b) \(\\{0,1\\}\) (c) \(\\{-2,-1,0,1,2\\}\) (d) \(\\{-1,0,1\\}\)
5 step solution
Problem 4
Let \(f: R \rightarrow R\) be defined by \(f(x)=\frac{x}{1+x^{2}}, x \in R\). Then the range of \(f\) is : (a) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) (b) \(R-[-1,1]\) (c) \(R-\left[-\frac{1}{2}, \frac{1}{2}\right]\) (d) \((-1,1)-\\{0\\}\)
5 step solution
Problem 5
The domain of the definition of the function \(f(x)=\frac{1}{4-x^{2}}+\log _{10}\left(x^{3}-x\right)\) is: \(\quad\) [April. 09,2019 (II)] (a) \((-1,0) \cup(1,2) \cup(3, \infty)\) (b) \((-2,-1) \cup(-1,0) \cup(2, \infty)\) (c) \((-1,0) \cup(1,2) \cup(2, \infty)\) (d) \((1,2) \cup(2, \infty)\)
6 step solution
Problem 6
The range of the function \(f(x)=\frac{x}{1+|x|}, x \in R\), is [Online May 7, 2012] (a) \(R\) (b) \((-1,1)\) (c) \(R-\\{0\\}\) (d) \([-1,1]\)
4 step solution
Problem 7
The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is (a) \((0, \infty)\) (b) \((-\infty, 0)\) (c) \((-\infty, \infty)-\\{0\\}\) (d) \((-\infty, \infty)\)
5 step solution
Problem 8
Domain of definition of the function \(f(x)=\frac{3}{4-x^{2}}+\log _{10}\left(x^{3}-x\right)\), is (a) \((-1,0) \cup(1,2) \cup(2, \infty)\) (b) \((a, 2)\) (c) \((-1,0) \cup(a, 2)\) (d) \((1,2) \cup(2, \infty)\)
4 step solution
Problem 9
Let \([t]\) denote the greatest integer \(\leq t\). Then the equation in \(x,[x]^{2}+2[x+2]-7=0\) has: \(\quad\) [Sep. 04, 2020 (I)] (a) exactly two solutions (b) exactly four integral solutions (c) nointegral solution (d) infinitely many solutions
7 step solution
Problem 10
Let \(f(x)\) be a quadratic polynomial such that \(f(-1)+f(2)=\) 0 . If one of the roots of \(f(x)=0\) is 3, then its other root lies in : (a) \((-1,0)\) (b) \((1,3)\) (c) \((-3,-1)\) (d) \((0,1)\)
7 step solution
Problem 12
If \(f(x)=\log _{e}\left(\frac{1-x}{1+x}\right),|x|<1\), then \(f\left(\frac{2 x}{1+x^{2}}\right)\) is equal to: [April 8, 2019 (I)] (a) \(2 f(x)\) (b) \(2 f\left(x^{2}\right)\) (c) \((f(x))^{2}\) (d) \(-2 f(x)\)
6 step solution
Problem 14
Let \(f(n)=\left[\frac{1}{3}+\frac{3 n}{100}\right] n\), where \([\mathrm{n}]\) denotes the greatest integer less than or equal to \(n\). Then \(\sum_{n=1}^{56} f(n)\) is equal to: [Online April 19, 2014] (a) 56 (b) 689 (c) 1287 (d) 1399
6 step solution
Problem 17
The graph of the function \(y=f(x)\) is symmetrical about the line \(x=2\), then \(\quad\) [2004] (a) \(f(x)=-f(-x)\) (b) \(f(2+x)=f(2-x)\) (c) \(f(x)=f(-x)\) (d) \(f(x+2)=f(x-2)\)
4 step solution
Problem 18
If \(f: R \rightarrow R\) satisfies \(f(x+y)=f(x)+f(y)\), for all \(x\), \(y \in R\) and \(f(1)=7\), then \(\sum_{r=1}^{n} f(r)\) is (a) \(\frac{7 n(n+1)}{2}\) (b) \(\frac{7 n}{2}\) (c) \(\frac{7(n+1)}{2}\) (d) \(7 n+(n+1)\)
5 step solution