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)
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 $f(x) = ax^2 + bx + c$, 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) -7
b) – 4
c) 2
d) 6
e) cannot be determined
2) Answer (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: 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
3) Answer (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: Let $f(x) = ax^2 – b|x|$ , where a and b are constants. Then at x = 0, f(x) is
[CAT 2004]
a) maximized whenever a > 0, b > 0
b) maximized whenever a > 0, b < 0
c) minimized whenever a > 0, b > 0
d) minimized whenever a > 0, b< 0
4) Answer (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: 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) Answer: 24
Solution:
$f_{1}(x)=x^{2}+11x+n$ and $f_{2}(x) = x$
$f_{1}(x)=f_{2}(x)$
=> $x^{2}+11x+n = x$
=> $ 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: 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) Answer: 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: 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) Answer: 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 that $f(11) = 91$ and $f(15) = 617$.
We have to find the value of $f(10)$.
Let $f(10)$ = b
$f(12)$ = b + 91
$f(13)$ = 91 + b + 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: 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) Answer: 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, 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 9: 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) Answer: 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, 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 10: If $f(x)=x^{2}-7x$ and $g(x)=x+3$, then the minimum value of $f(g(x))-3x$ is:
a) -20
b) -12
c) -15
d) -16
10) Answer (D)
Solution:
Now we have :
$f(g(x))-3x$
so we get f(x+3)-3x
= $\left(x+3\right)^2-7\left(x+3\right)-3x$
=$x^2-4x-12$
Now minimum value of expression = $-\frac{D}{4a}$ $ \frac{\left(4ac-b^2\right)}{4a}$
We get – (16+48)/4
= -16
