CAT Functions Questions PDF [Most Important]

0
704
CAT Functions Questions PDF
CAT Functions Questions PDF

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

Join CAT 2023 Online Coaching

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)

View Video Solution

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)

View Video Solution

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

View Video Solution

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

View Video Solution

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)

View Video Solution

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

View Video Solution

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)

View Video Solution

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)

View Video Solution

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

View Video Solution

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

Download CAT 2023 Study Plan PDF

LEAVE A REPLY

Please enter your comment!
Please enter your name here