Functions are one of the important topics in the Quantitative Ability section of the CAT. It is an easy topic and so one must not avoid this topic. Every year 1-2 questions are asked on Functions. You can check out these Functions<\/span> questions from CAT Previous year papers<\/a><\/strong>. Practice a good number of questions on CAT Functions<\/strong> questions so that you don’t miss out on the easy questions from this topic. In this article, we will look into some important Functions Questions for CAT Quants. These are a good source for practice; If you want to practice these questions, you can download this CAT Functions Questions PDF below, which is completely Free.<\/p>\n Download Function Questutions for CAT <\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>A function $f (x)$ satisfies $f(1) = 3600$, and $f (1) + f(2) + … + f(n) =n^2f(n)$, for all positive integers $n > 1$. What is the value of $f (9)$ ?<\/p>\n a)\u00a080<\/p>\n b)\u00a0240<\/p>\n c)\u00a0200<\/p>\n d)\u00a0100<\/p>\n e)\u00a0120<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n According to given conditions we get f(2)=f(1)\/3 , then\u00a0f(3)=f(1)\/6, then \u00a0f(4)=f(1)\/10 , then\u00a0f(5)=f(1)\/15 .<\/p>\n We can see the pattern here that the denominator goes on increasing from 3,3+3,6+4,10+5,15+6,.. so for the f(9) the denominator will be same as 15+6+7+8+9=45 .<\/p>\n So f(9)=3600\/45 = 80<\/p>\n Question 2:\u00a0<\/b>Let $f(x)\\neq0$ for any ‘x’ be a function satisfying $f(x)f(y) = f(xy)$ for all real x, y. If $f(2) = 4$, then what is the value of $f(\\frac{1}{2})$?<\/p>\n a)\u00a00<\/p>\n b)\u00a01\/4<\/p>\n c)\u00a01\/2<\/p>\n d)\u00a01<\/p>\n e)\u00a0cannot be determined<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $f(1)^2$ = f(1) => f(1) = 1<\/p>\n f(2)*(f(1\/2) = f(1) => 4x = 1<\/p>\n So, f(1\/2) = 1\/4<\/p>\n Question 3:\u00a0<\/b>Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f (1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals<\/p>\n 3)\u00a0Answer:\u00a012<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given, f(mn) = f(m)f(n) when m=1,\u00a0 n= 2, f(2) = f(1)*f(2) ==> f(1) = 1<\/p>\n when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $[f(2)]^2$<\/p>\n Similarly f(8) =\u00a0f(4)*f(2) =$[f(2)]^3$<\/p>\n f(24) = 54<\/p>\n $[f(2)]^3$ *\u00a0$[f(3)]$ = $3^3*2$<\/p>\n On comparing LHS and RHS, we get<\/p>\n f(2) = 3 and f(3) = 2<\/p>\n Now we have to find the value of f(18)<\/p>\n f(18) =\u00a0$[f(2)]$ * $[f(3)]^2$<\/p>\n = 3*4=12<\/p>\n Question 4:\u00a0<\/b>Consider a function f satisfying f (x + y) = f (x) f (y) where x,y are positive integers, and f(1) = 2. If f(a + 1) +f (a + 2) + … + f(a + n) = 16 (2$^n$ – 1) then a is equal to<\/p>\n 4)\u00a0Answer:\u00a03<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n f (x + y) = f (x) f (y)<\/p>\n Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4<\/p>\n f(3)=f(2+1)=f(2)*f(1)=4*2=8<\/p>\n f(4)=f(3+1)=f(3)*f(1)=8*2=16<\/p>\n …….=> f(x)=$2^x$<\/p>\n Now,\u00a0f(a + 1) +f (a + 2) + … + f(a + n) = 16 (2$^n$ – 1)<\/p>\n On putting n=1 in the equation we get, f(a+1)=16\u00a0 \u00a0=> f(a)*f(1)=16\u00a0 (It is given that\u00a0f (x + y) = f (x) f (y))<\/p>\n =>\u00a0$2^a$*2=16<\/p>\n => a=3<\/p>\n Question 5:\u00a0<\/b>Let $f(x)$ be a quadratic polynomial in $x$ such that $f(x) \\geq 0$ for all real numbers $x$. If f(2) = 0 and f( 4) = 6, then f(-2) is equal to<\/p>\n a)\u00a012<\/p>\n b)\u00a024<\/p>\n c)\u00a06<\/p>\n d)\u00a036<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $f(x) \\geq 0$for all real numbers $x$, so D<=0<\/p>\n Since f(2)=0 therefore x=2 is a root of f(x)<\/p>\n Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0<\/p>\n Therefore f(x) =\u00a0$a\\left(x-2\\right)^2$<\/p>\n f(4)=6<\/p>\n or, 6 =\u00a0$a\\left(x-2\\right)^2$<\/p>\n a= 3\/2<\/p>\n $f\\left(-2\\right)=\\ -\\frac{3}{2}\\left(-4\\right)^2=24$<\/p>\n Question 6:\u00a0<\/b>Suppose for all integers x, there are two functions f and g such that $f(x) + f (x – 1) – 1 = 0$ and $g(x ) = x^{2}$. If $f\\left(x^{2} – x \\right) = 5$, then the value of the sum f(g(5)) + g(f(5)) is<\/p>\n 6)\u00a0Answer:\u00a012<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n $f\\left(x\\right)+f\\left(x-1\\right)=1$ …… (1)<\/p>\n $f\\left(x^2-x\\right)=5$ ……\u00a0 (2)<\/p>\n $g\\left(x\\right)=x^2$<\/p>\n Substituting x = 1 in (1) and (2), we get<\/p>\n f(0) = 5<\/p>\n f(1) + f(0) = 1<\/p>\n f(1) = 1 – 5 = -4<\/p>\n f(2) + f(1) = 1<\/p>\n f(2) = 1 + 4 = 5<\/p>\n f(n) = 5 if n is even and f(n) = -4 if n is odd<\/p>\n f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12<\/p>\n Question 7:\u00a0<\/b>Let r be a real number and $f(x) = \\begin{cases}2x -r & ifx \\geq r\\\\ r &ifx < r\\end{cases}$. Then, the equation $f(x) = f(f(x))$ holds for all real values of $x$ where<\/p>\n a)\u00a0$x > r$<\/p>\n b)\u00a0$x \\leq r$<\/p>\n c)\u00a0$x \\neq r$<\/p>\n d)\u00a0$x \\geq r$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n When x< r<\/p>\n f(x) = r<\/p>\n f(x) =\u00a0f(f(x))<\/p>\n r = f(r)<\/p>\n r= 2r-r<\/p>\n r=r<\/p>\n When x>=r<\/p>\n f(x) = 2x-r<\/p>\n f(x) = f(f(x))<\/p>\n 2x-r = f(2x-r)<\/p>\n 2x-r = 2(2x-r) – r<\/p>\n 2x-r = 4x-3r<\/p>\n or, x=r<\/p>\n Therefore x<= r<\/p>\n Question 8:\u00a0<\/b>Let $f(x)$ be a quadratic polynomial in $x$ such that $f(x) \\geq 0$ for all real numbers $x$. If f(2) = 0 and f( 4) = 6, then f(-2) is equal to<\/p>\n a)\u00a012<\/p>\n b)\u00a024<\/p>\n c)\u00a06<\/p>\n d)\u00a036<\/p>\n
\nwhen m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1<\/p>\n