Relations & Functions 1 Welcome to your IIT JEE MOCK TEST: Relations and Functions 1 Chapter Wise Test Topic - Relations & Functions Maximum Marks : 120 Marking Scheme: (+4) for Correct & (-1) for incorrect answer Time: 60Mins Please Don't Cheat! INSTRUCTIONS : This Daily Practice Problem Sheet contains 30 MCQs. For each question only one option is correct. Select the correct option in the Response Grid provided on each page. Name Email Phone For the following relation $$ R=\{(0,0),(0,1),(1,1),(2,1),(2,2),(2,0),(1,0),(0,2),(0,1)\} $$ domain ={0,1} both correct range ={0,1,2} None of these The domain of the function $$\sqrt{x^{2}-5 x+6}+\sqrt{2 x+8-x^{2}}$$ is [2,3] [-2,4] $$[-2,2] \cup[3,4]$$ $$[-2,1] \cup[2,4]$$ If $$3 f(x)-f\left(\frac{1}{x}\right)=\log x^{4}$$, then $$f\left(e^{-x}\right)$$ is $$1+x$$ $$1 / x$$ $$x$$ $$-x$$ The domain of the function $$f(x)=\frac{1}{\sqrt{|x|-x}}$$ is $$(0, \infty)$$ $$(-\infty, \infty)-\{0\}$$ $$(-\infty, 0)$$ $$(-\infty, \infty)$$ $$f(x)=\sqrt{|x|^{2}-5|x|+6}+\sqrt{8+2|x|-|x|^{2}}$$ is real for all $$x$$ in $$[-4,-3]$$ $$[-3,-2]$$ $$[-2,2]$$ $$[3,4]$$ $$f(x)=\frac{x(x-p)}{q-p}+\frac{x(x-q)}{p-q}, p \neq q$$. What is the value of $$f(p)+f(q) ?$$ $$f(p-q)$$ $$\mathrm{f}(\mathrm{p}(\mathrm{p}+\mathrm{q}))$$ $$f(p+q)$$ $$f(q(p-q))$$ A real valued function $$f(x)$$ satisfies the functional equation $$ f(x-y)=f(x) f(y)-f(a-x) f(a+y) $$ where $$a$$ is a given constant and $$f(0)=1, f(2 a-x)$$ is equal to $$-f(x)$$ $$f(a)+f(a-x)$$ $$f(x)$$ $$f(-x)$$ Domain of definition of the function $$f(x)=\frac{3}{4-x^{2}}+\log _{10}\left(x^{3}-x\right)$$, is $$(-1,0) \cup(1,2) \cup(2, \infty)$$ $$(a, 2)$$ $$\quad(-1,0) \cup(a, 2)$$ $$(1,2) \cup(2, \infty)$$. Let $$\mathrm{A}=\{1,2,3,4,5\} ; \mathrm{B}=\{2,3,6,7\}$$. Then the number of elements in $$(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})$$ is 18 4 6 0 A relation $$R$$ is defined in the set $$Z$$ of integers as follows $$(x, y) \in R$$ iff z$$x^{2}+y^{2}=9$$. Which of the following is false? $$\mathrm{R}=\{(0,3),(0,-3),(3,0),(-3,0)\}$$ Domain of $$\mathrm{R}=\{-3,0,3\}$$ Range of $$\mathrm{R}=\{-3,0,3\}$$ None of these Let $$f(x)=\sqrt{1+x^{2}}$$, then $$f(x y)=f(x) \cdot f(y)$$ $$f(x y) \geq f(x) \cdot f(y)$$ $$f(x y) \leq f(x) . f(y)$$ None of these The domain of the function $$f(x)=\sqrt{x-\sqrt{1-x^{2}}}$$ is $$\left[-1,-\frac{1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, 1\right]$$ $$[-1,1]$$ $$\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{\sqrt{2}},+\infty\right)$$ $$\left[\frac{1}{\sqrt{2}}, 1\right]$$ Period of the function $$\left|\sin ^{3} \frac{x}{2}\right|+\left|\cos ^{5} \frac{x}{5}\right|$$ is : $$2 \pi$$ $$10 \pi$$ $$8 \pi$$ $$5 \pi$$ If $$n(\mathrm{~A})=4, n(\mathrm{~B})=3, n(\mathrm{~A} \times \mathrm{B} \times \mathrm{C})=24$$, then $$n(\mathrm{C})=$$ 288 1 12 2 If S={1,2,3,4,5} and R={(x, y): x+y<6} then n(R)= 8 10 6 5 The function $$f(x)=\log \left(x+\sqrt{x^{2}+1}\right)$$, is neither an even nor an odd function an even function an odd function a periodic function Let $$f(x)=\frac{x}{1-x}$$ and ' $$a$$ ' be a real number. If $$x_{0}=a$$, $$x_{1}=f\left(x_{0}\right), x_{2}=f\left(x_{1}\right), x_{3}=f\left(x_{2}\right) \ldots \ldots$$. If $$x_{2009}=1$$, then the value of $$a$$ is 0 $$\frac{2009}{2010}$$ $$\frac{1}{2009}$$ $$\frac{1}{2010}$$ The domain of the function $$f(x)=\log _{2}\left(-\log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right)-1\right)$$ is $$(0,1)$$ $$[1, \infty)$$ $$(0,1]$$ $$(1, \infty)$$ The domain of the function $$f(x)=\frac{1}{\sqrt{x^{2}-3 x+2}}$$ is $$(-\infty, 1)$$ $$(-\infty, 1) \cup(2, \infty)$$ $$(-\infty, 1] \cup[2, \infty)$$ $$(2, \infty)$$ If $$(1,3),(2,5)$$ and $$(3,3)$$ are three elements of $$A \times B$$ and the total number of elements in $$\mathrm{A} \times \mathrm{B}$$ is 6 , then the remaining elements of $$\mathrm{A} \times \mathrm{B}$$ are $$\quad(1,5) ;(2,3) ;(3,5)$$ $$(1,5) ;(2,3) ;(5,3)$$ $$(5,1) ;(3,2) ;(5,3)$$ None of these If $$f(x)=\ln \left(\frac{x^{2}+e}{x^{2}+1}\right)$$, then range of $$f(x)$$is $$(0,1)$$ $$[0,1)$$ $$(0,1]$$ $$\{0,1\}$$ The function $$f(x)=\log \left(\frac{1+x}{1-x}\right)$$ satisfies the equation $$f(x+2)-2 f(x+1)+f(x)=0$$ $$f(x+1)+f(x)=f(x(x+1))$$ $$f\left(x_{1}\right) \cdot f\left(x_{2}\right)=f\left(x_{1}+x_{2}\right)$$ $$f\left(x_{1}\right)+f\left(x_{2}\right)=f\left(\frac{x_{1}+x_{2}}{1+x_{1} x_{2}}\right)$$ 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 $$\frac{7 n(n+1)}{2}$$ $$\frac{7 n}{2}$$ $$\frac{7(n+1)}{2}$$ $$7 n+(n+1)$$ If $$\{\}$$ denotes the fractional part of $$x$$, the range of the function $$f(x)=\sqrt{\{x\}^{2}-2\{x\}}$$ is $$\phi$$ $$\{0,1 / 2\}$$ $$[0,1 / 2]$$ $$\{0\}$$ If $$\mathrm{f}(x)=\frac{x-1}{x+1}$$, then $$f(2 x)$$ is equal to $$\frac{f(x)+1}{f(x)+3}$$ $$\frac{3 f(x)+1}{f(x)+3}$$ $$\frac{f(x)+3}{f(x)+1}$$ $$\frac{f(x)+3}{3 f(x)+1}$$ The range of the function $$f(x)=\frac{x^{2}-x+1}{x^{2}+x+1}$$ where $$x \in R$$, is $$(-\infty, 3]$$ $$(-\infty, \infty)$$ $$[3, \infty)$$ $$\left[\frac{1}{3}, 3\right]$$ The domain of the function $$f(x)=\exp \left(\sqrt{5 x-3-2 x^{2}}\right)$$ is $$[3 / 2, \infty)$$ $$\quad(-\infty, 1]$$ $$[1,3 / 2]$$ $$(1,3 / 2)$$ If $$f(x+y)=f(x)+2 y^{2}+k x y$$ and $$f(a)=2, f(b)=8$$, then $$f(x)$$ is of the form $$2 x^{2}$$ $$2 x^{2}-1$$ $$2 x^{2}+1$$ $$x^{2}$$ The relation $$\mathrm{R}$$ defined on the set $$\mathrm{A}=\{1,2,3,4,5\}$$ by $$\mathrm{R}=\left\{(x, y):\left|x^{2}-y^{2}\right|<16\right\}$$ is given by $$\{(1,1),(2,1),(3,1),(4,1),(2,3)\}$$ $$\{(2,2),(3,2),(4,2),(2,4)\}$$ $$\{(3,3),(3,4),(5,4),(4,3),(3,1)\}$$ None of these Which of the following relation is NOT a function $$f=\{(x, x) \mid x \in \mathrm{R}\}$$ $$g=\{(x, 3) \mid x \in \mathrm{R}\}$$ $$h=\left\{\left(n, \frac{1}{n}\right) \mid n \in \mathrm{I}\right\}$$ $$t=\left\{\left(n, n^{2}\right) \mid n \in \mathrm{N}\right\}$$
