# CAT Functions Questions PDF [Most Important]

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 questions from **CAT Previous year papers**. Practice a good number of questions on CAT **Functions** 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.

Download Function Questutions for CAT

**Question 1:Â **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 $f (9)$ ?

a)Â 80

b)Â 240

c)Â 200

d)Â 100

e)Â 120

**1)Â AnswerÂ (A)**

**Solution:**

According to given conditions we get f(2)=f(1)/3 , thenÂ f(3)=f(1)/6, then Â f(4)=f(1)/10 , thenÂ f(5)=f(1)/15 .

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 .

So f(9)=3600/45 = 80

**Question 2:Â **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 $f(\frac{1}{2})$?

a)Â 0

b)Â 1/4

c)Â 1/2

d)Â 1

e)Â cannot be determined

**2)Â AnswerÂ (B)**

**Solution:**

$f(1)^2$ = f(1) => f(1) = 1

f(2)*(f(1/2) = f(1) => 4x = 1

So, f(1/2) = 1/4

**Question 3:Â **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

**3)Â Answer:Â 12**

**Solution:**

Given, f(mn) = f(m)f(n)

when m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1

when m=1,Â n= 2, f(2) = f(1)*f(2) ==> f(1) = 1

when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $[f(2)]^2$

Similarly f(8) =Â f(4)*f(2) =$[f(2)]^3$

f(24) = 54

$[f(2)]^3$ *Â $[f(3)]$ = $3^3*2$

On comparing LHS and RHS, we get

f(2) = 3 and f(3) = 2

Now we have to find the value of f(18)

f(18) =Â $[f(2)]$ * $[f(3)]^2$

= 3*4=12

**Question 4:Â **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

**4)Â Answer:Â 3**

**Solution:**

f (x + y) = f (x) f (y)

Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4

f(3)=f(2+1)=f(2)*f(1)=4*2=8

f(4)=f(3+1)=f(3)*f(1)=8*2=16

…….=> f(x)=$2^x$

Now,Â f(a + 1) +f (a + 2) + … + f(a + n) = 16 (2$^n$ – 1)

On putting n=1 in the equation we get, f(a+1)=16Â Â => f(a)*f(1)=16Â (It is given thatÂ f (x + y) = f (x) f (y))

=>Â $2^a$*2=16

=> a=3

**Question 5:Â **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

a)Â 12

b)Â 24

c)Â 6

d)Â 36

**5)Â AnswerÂ (B)**

**Solution:**

$f(x) \geq 0$for all real numbers $x$, so D<=0

Since f(2)=0 therefore x=2 is a root of f(x)

Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0

Therefore f(x) =Â $a\left(x-2\right)^2$

f(4)=6

or, 6 =Â $a\left(x-2\right)^2$

a= 3/2

$f\left(-2\right)=\ -\frac{3}{2}\left(-4\right)^2=24$

**Question 6:Â **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

**6)Â Answer:Â 12**

**Solution:**

Given,

$f\left(x\right)+f\left(x-1\right)=1$ …… (1)

$f\left(x^2-x\right)=5$ ……Â (2)

$g\left(x\right)=x^2$

Substituting x = 1 in (1) and (2), we get

f(0) = 5

f(1) + f(0) = 1

f(1) = 1 – 5 = -4

f(2) + f(1) = 1

f(2) = 1 + 4 = 5

f(n) = 5 if n is even and f(n) = -4 if n is odd

f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

**Question 7:Â **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

a)Â $x > r$

b)Â $x \leq r$

c)Â $x \neq r$

d)Â $x \geq r$

**7)Â AnswerÂ (B)**

**Solution:**

When x< r

f(x) = r

f(x) =Â f(f(x))

r = f(r)

r= 2r-r

r=r

When x>=r

f(x) = 2x-r

f(x) = f(f(x))

2x-r = f(2x-r)

2x-r = 2(2x-r) – r

2x-r = 4x-3r

or, x=r

Therefore x<= r

**Question 8:Â **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

a)Â 12

b)Â 24

c)Â 6

d)Â 36

**8)Â AnswerÂ (B)**

**Solution:**

$f(x) \geq 0$for all real numbers $x$, so D<=0

Since f(2)=0 therefore x=2 is a root of f(x)

Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0

Therefore f(x) =Â $a\left(x-2\right)^2$

f(4)=6

or, 6 =Â $a\left(x-2\right)^2$

a= 3/2

$f\left(-2\right)=\ -\frac{3}{2}\left(-4\right)^2=24$

**Question 9:Â **For any real number x, let [x] be the largest integer less than or equal to x. If $\sum_{n=1}^N \left[\frac{1}{5} + \frac{n}{25}\right] = 25$ then N is

**9)Â Answer:Â 44**

**Solution:**

It is given,

$\Sigma_{n=1}^N\ \left[\frac{1}{5}+\frac{n}{25}\right]=25$

$\Sigma_{n=1}^N\ \left[\frac{5+n}{25}\right]=25$

For n = 1 to n = 19, value of function is zero.

For n = 20 to n = 44, value of function will be 1.

44 = 20 + n – 1

n = 25 which is equal to given value.

This implies N = 44