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.

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Question 1: 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) 80

b) 240

c) 200

d) 100

e) 120

1) Answer (A)

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Solution:

According to given conditions we get f(2)=f(1)/3 , then f(3)=f(1)/6, then  f(4)=f(1)/10 , then 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: 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) 0

b) 1/4

c) 1/2

d) 1

e) cannot be determined

2) Answer (B)

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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: 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) Answer: 12

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Solution:

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

when m=1,  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) = f(4)*f(2) =$[f(2)]^3$

f(24) = 54

$[f(2)]^3$ * $[f(3)]$ = $3^3*2$

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) = $[f(2)]$ * $[f(3)]^2$

= 3*4=12

Question 4: 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) Answer: 3

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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, 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)=16   => f(a)*f(1)=16  (It is given that f (x + y) = f (x) f (y))

=> $2^a$*2=16

=> a=3

Question 5: 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) 12

b) 24

c) 6

d) 36

5) Answer (B)

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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) = $a\left(x-2\right)^2$

f(4)=6

or, 6 = $a\left(x-2\right)^2$

a= 3/2

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

Question 6: 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) Answer: 12

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Solution:

Given,

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

$f\left(x^2-x\right)=5$ ……  (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: 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) $x > r$

b) $x \leq r$

c) $x \neq r$

d) $x \geq r$

7) Answer (B)

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Solution:

When x< r

f(x) = r

f(x) = 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: 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) 12

b) 24

c) 6

d) 36

8) Answer (B)

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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) = $a\left(x-2\right)^2$

f(4)=6

or, 6 = $a\left(x-2\right)^2$

a= 3/2

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

Question 9: 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) Answer: 44

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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

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