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: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 $f(x) = ax^2 + bx + c$, where a, b and c are certain constants and $a \neq 0$ ?

It is known that $f(5) = – 3f(2)$. and that 3 is a root of $f(x) = 0$.

What is the other root of f(x) = 0?

[CAT 2008]

a)Â -7

b)Â – 4

c)Â 2

d)Â 6

e)Â cannot be determined

**2)Â AnswerÂ (B)**

**Solution:**

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c

f(2) = 4a + 2b + c

f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c

=> 37a + 11b + 4c = 0 –> (1)

4(9a + 3b + c) = 36a + 12b + 4c = 0 –> (2)

From (1) and (2), a – b = 0 => a = b

=> c = -12a

The equation is, therefore, $ax^2 + ax – 12a = 0 => x^2 + x – 12 = 0$

=> -4 is a root of the equation.

**Question 3: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

**3)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 4:Â **Let $f(x) = ax^2 – b|x|$ , where a and b are constants. Then at x = 0, f(x) is

[CAT 2004]

a)Â maximized whenever a > 0, b > 0

b)Â maximized whenever a > 0, b < 0

c)Â minimized whenever a > 0, b > 0

d)Â minimized whenever a > 0, b< 0

**4)Â AnswerÂ (D)**

**Solution:**

$f(x) = ax^2 – b|x|$. When $x=0, f(x) = 0$

When $a > 0$ and $b < 0$,

For x > 0, $f(x) = ax^2 – bx$, will be greater than 0 as $ax^2 > 0$ and $bx<0$ as $b$ is negative and $x$ is positive.

For x < 0, $f(x) = ax^2 + bx$ will again be greater than 0 as $ax^2 >0$ and $bx>0$ as both $b$ and $x$ are negative.

Therefore, the function $f(x)$ is positive when $x<0$ and when $x>0$ but becomes 0 when $x=0$.

Therefore, for $a > 0$ and $b < 0$, f(x) will attain its minimum value at $x = 0$.

**Question 5:Â **If $f_{1}(x)=x^{2}+11x+n$ and $f_{2}(x)=x$, then the largest positive integer n for which the equation $f_{1}(x)=f_{2}(x)$ has two distinct real roots is

**5)Â Answer:Â 24**

**Solution:**

$f_{1}(x)=x^{2}+11x+n$ andÂ $f_{2}(x) = x$

$f_{1}(x)=f_{2}(x)$

=> $x^{2}+11x+n = x$

=> $ x^2 + 10x + n = 0 $

=> For this equation to have distinct real roots, b$^2$-4ac>0

$ 10^2 > 4n$

=> n < 100/4

=> n < 25

Thus, largest integral value that n can take is 24.

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**Question 6:Â **Let f(x) = min (${2x^{2},52-5x}$) where x is any positive real number. Then the maximum possible value of f(x) is

**6)Â Answer:Â 32**

**Solution:**

f(x) = min (${2x^{2},52-5x}$)

The maximum possible value of this function will be attained at the point in which $2x^2$ is equal to $52-5x$.

$2x^2 = 52-5x$

$2x^2+5x-52=0$

$(2x+13)(x-4)=0$

=> $x=\frac{-13}{2}$ or $x = 4$

It has been given that $x$ is a positive real number. Therefore, we can eliminate the case $x=\frac{-13}{2}$.

$x=4$ is the point at which the function attains the maximum value. $4$ is not the maximum value of the function.

Substituting $x=4$ in the original function, we get, $2x^2 = 2*4^2= 32$.

f(x) = $32$.

Therefore, 32 is the right answer.

**Question 7:Â **If $f(x + 2) = f(x) + f(x + 1)$ for all positive integers x, and $f(11) = 91, f(15) = 617$, then $f(10)$ equals

**7)Â Answer:Â 54**

**Solution:**

$f(x + 2) = f(x) + f(x + 1)$

As we can see, the value of a term is the sum of the 2 terms preceding it.

It has been given thatÂ $f(11) = 91$ and $f(15) = 617$.

We have to find the value ofÂ $f(10)$.

LetÂ $f(10)$ = b

$f(12)$ = b + 91

$f(13)$ = 91 + b +Â 91 = 182 + b

$f(14)$ = 182+b+91+b = 273+2b

$f(15)$ = 273+2b+182+b = 455+3b

It has been given that 455+3b = 617

3b = 162

=> b = 54

Therefore, 54 is the correct answer.

**Question 8: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

**8)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 9: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

**9)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 10:Â **If $f(x)=x^{2}-7x$ and $g(x)=x+3$, then the minimum value of $f(g(x))-3x$ is:

a)Â -20

b)Â -12

c)Â -15

d)Â -16

**10)Â AnswerÂ (D)**

**Solution:**

Now we have :

$f(g(x))-3x$

so we get f(x+3)-3x

=Â $\left(x+3\right)^2-7\left(x+3\right)-3x$

=$x^2-4x-12$

Now minimum value of expression = $-\frac{D}{4a}$Â $ \frac{\left(4ac-b^2\right)}{4a}$

We get – (16+48)/4

= -16