CAT Functions, Graphs and Statistics Questions & Solutions

A sample of real CAT Functions, Graphs and Statistics past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 93 Functions, Graphs and Statistics questions in all — sign up free to practise them timed.

Q1.CAT 2025
If $f(x)= (x^{2} + 3x)(x^{2}+ 3x+2)$ then the sum of all real roots of the equation $\sqrt{f(x)+1}= 9701$ , is
- -6
- 6
- 3
- -3
Show solution
Let $(x^2+3x)$ be equal to $k$ . We have,

$f(x)= k(k+2) = k^2+2k$

Therefore, $\sqrt{f(x)+1} = \sqrt{k^2+2k+1} = \sqrt{(k+1)^2} = k+1 = 9701$

We get $k=9700$ .

Thus, $x^2+3x=9700$ or $x^2+3x-9700 = 0$

Since $x$ is real, the discriminant of the above quadratic has to be greater than or equal to zero.

We find that $3^2 + 4*9700 \geq 0$ and therefore the quadratic has real roots.

The sum of the roots will be $-\dfrac{b}{a} = -\dfrac{3}{1} = -3$

Option D is the correct answer.
Q2.CAT 2023
Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x - 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is
Answer: 3
Show solution
Given that f(3x + 2y, 2x - 5y) = 19x.

Let us assume the function f(a,b) is a linear combination of a and b.

=> f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x

=> 3m + 2n = 19 and 2m - 5n = 0

Solving we get m = 5 and n = 2

=> f(a,b) = 5a+2b

=> f(x,2x) = 5x + 2(2x) = 9x = 27 => x = 3.
Q3.CAT 2020
The number of real-valued solutions of the equation $2^{x}+2^{-x}=2-(x-2)^{2}$ is:
- 1
- 2
- infinite
- 0
Show solution

The graphs of $2^{x}+2^{-x} and 2-(x-2)^{2}$ never intersect. So, number of solutions=0.
Alternate method:
We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x=0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.
Q4.CAT 2019
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
Answer: 12
Show 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
Q5.CAT 2017
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
Answer: 24
Show 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.
Q6.CAT 2008
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 $f(\frac{1}{2})$ ?
- 0
- 1/4
- 1/2
- 1
- cannot be determined
Show solution
$f(1)^2$ = f(1) => f(1) = 1

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

So, f(1/2) = 1/4
Q7.CAT 2006
The graph of y - x (on the y axis) against y + x (on the x axis) is as shown below. (All graphs in this question are drawn to scale and the same scale and the same scale has been used on each axis.)

Which of the following shows the graph of y against x?
Show solution
For a normal graph with y and x-axis, the equation of the line passing through the origin is y =mx where m is the slope of the line.

m is +ve if the angle made by the line with the x-axis is < $90^{\circ\ }$

$\therefore\$ The equation of the line in the given graph would be y-x = k( y+ x) since the axes are y-x and y+x and the line is passing through the origin.

k > 1 because the angle is greater than 45 $^{\circ\ }$

$y=\dfrac{x\left(k+1\right)}{1-k}$

Since k>1

Therefore y<0 for x>-1 and y>0 for x<-1

Option d correctly satisfy this condition
Q8.CAT 2004
If $\frac{a}{b+c}=\frac{b}{a+c} =\frac{c}{b+a} =r$ , then r cannot take any value except
- 1/2
- -1
- 1/2 or -1
- -1/2 or -1
Show solution
a = r(b+c)

b = r(a+c)

c = r(a+b)

On adding all the equations,

a+b+c = 2r(a+b+c)

If r = 1/2, a+b+c = a+b+c (valid)

If r = -1, a+b+c = -2(a+b+c) => a+b+c = 0 => b+c = -a and a/(b+c) = a/(-a) = -1 (valid)

So, r can take the values 1/2 or -1
Q9.CAT 2002
Suppose for any real number x, [x] denotes the greatest integer less than or equal to x. Let L(x, y) = [x] + [y] + [x + y] and R(x, y) = [2x] + [2y]. Then it is impossible to find any two positive real numbers x and y for which
- L(x, y) = R(x, y)
- L(x, y) $\neq$ R(x, y)
- L(x, y) < R(x, y)
- L(x, y) > R(x, y)
Show solution
Consider different values of x and y:

x = -1.5 and y = -1.5; x = 1.5 and y = -1.5; x = -1.5 and y = 1.5; x = 1.5 and y = 1.5.

For these possibilities, options A,B and C gets satisfied , but it is impossible to find any two positive real numbers x and y for which L(x, y) > R(x, y).
Passage
Directions for the next 3 questions: For three distinct real positive numbers x, y and z, let

f(x, y, z) = min (max(x, y), max (y, z), max (z, x))

g(x, y, z) = max (min(x, y), min (y, z), min (z, x))

h(x, y, z) = max (max(x, y), max(y, z), max (z, x))

j(x, y, z) = min (min (x, y), min(y, z), min (z, x))

m(x, y, z) = max (x, y, z)

n(x, y, z) = min (x, y, z)
Q10.CAT 2000
Which of the following expressions is indeterminate?
- (f(x, y, z)-h(x, y, z))/(g(x, y, z) —j(x, y, z))
- [f(x,y,z)+h(x,y,z)+g(x,y,z)+j(x,y, z)]/[j(x, y, z) + h(x, y, z)-m(x, y, z) - n(x, y, z)]
- [g(x, y, z)-j(x, y, z)]/[f(x, y, z)-h(x, y, z)]
- [h(x, y, z) fix, y, z)]/[n(x, y, z) - g(x, y, z)]
Show solution

From the given functions we can make out that function h and m give max value , function n and j give min value , function f and g give middle value.So in option B , j cancels out n and h cancels out m . So the denominator becomes 0 and value is indeterminable.

Suppose x>y>z

f(x,y,z) = y

g(x,y,z) = y

h(x,y,z) = x

j(x,y,z) = z

m(x,y,z) = x

n(x,y,z) = z

The denominator of the second option becomes 0, hence making it indeterminate.
Q11.CAT 1999
For two positive integers a and b define the function h(a,b):as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the GCF of the elements of set A is computed by repeatedly using the function h.
The minimum number of times h is required to be used to compute G is:
- 1/2 n
- (n - 1)
- n
- None of these
Show solution
Let p and q be any two elements of the set A.
For the computation of the GCF of elements of the set A, we can replace both p and q by just the GCF(p,q) and the result is unchanged.

So, for every application of the function h, we are reducing the number of elements of the set A by 1. (In this case two numbers p and q are replaced by one number GCF(p,q)).
Expanding this concept further, the minimum number of times the function h should be called is n-1
Passage
DIRECTIONS for the following questions: These questions are based on the situation given below: Let x and y be real numbers

f(x, y) = | x + y |

F(f(x, y)) = -f(x, y)

G(f(x, y)) = -F(f(x, y))
Q12.CAT 1999
What is the value of f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2))))?
- 3
- 2
- 1
- 0
Show solution
F(f(x,y)) = -f(x,y)

G(f(x,y)) = -F(f(x,y)) = f(x,y)

G(f(1,0)) = 1

F(f(1,2)) = -3

G(f(1,2)) = 3

f(F(f(1,2)),G(f(1,2))) = 0

f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2)))) = 1 + 0 = 1

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CAT Functions, Graphs and Statistics Questions & Solutions

Q1.CAT 2025

If $f(x)= (x^{2} + 3x)(x^{2}+ 3x+2)$ then the sum of all real roots of the equation $\sqrt{f(x)+1}= 9701$ , is

-6
6
3
-3

Show solution

Let $(x^2+3x)$ be equal to $k$ . We have,

$f(x)= k(k+2) = k^2+2k$

Therefore, $\sqrt{f(x)+1} = \sqrt{k^2+2k+1} = \sqrt{(k+1)^2} = k+1 = 9701$

We get $k=9700$ .

Thus, $x^2+3x=9700$ or $x^2+3x-9700 = 0$

Since $x$ is real, the discriminant of the above quadratic has to be greater than or equal to zero.

We find that $3^2 + 4*9700 \geq 0$ and therefore the quadratic has real roots.

The sum of the roots will be $-\dfrac{b}{a} = -\dfrac{3}{1} = -3$

Option D is the correct answer.

Q2.CAT 2023

Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x - 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is

Answer: 3

Show solution

Given that f(3x + 2y, 2x - 5y) = 19x.

Let us assume the function f(a,b) is a linear combination of a and b.

=> f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x

=> 3m + 2n = 19 and 2m - 5n = 0

Solving we get m = 5 and n = 2

=> f(a,b) = 5a+2b

=> f(x,2x) = 5x + 2(2x) = 9x = 27 => x = 3.

Q3.CAT 2020

The number of real-valued solutions of the equation $2^{x}+2^{-x}=2-(x-2)^{2}$ is:

1
2
infinite
0

Show solution

The graphs of $2^{x}+2^{-x} and 2-(x-2)^{2}$ never intersect. So, number of solutions=0.

Alternate method:

We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x=0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

Q4.CAT 2019

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

Answer: 12

Show 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

Q5.CAT 2017

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

Answer: 24

Show 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.

Q6.CAT 2008

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 $f(\frac{1}{2})$ ?

0
1/4
1/2
1
cannot be determined

Show solution

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

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

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

Q7.CAT 2006

The graph of y - x (on the y axis) against y + x (on the x axis) is as shown below. (All graphs in this question are drawn to scale and the same scale and the same scale has been used on each axis.)

Which of the following shows the graph of y against x?

Show solution

For a normal graph with y and x-axis, the equation of the line passing through the origin is y =mx where m is the slope of the line.

m is +ve if the angle made by the line with the x-axis is < $90^{\circ\ }$

$\therefore\$ The equation of the line in the given graph would be y-x = k( y+ x) since the axes are y-x and y+x and the line is passing through the origin.

k > 1 because the angle is greater than 45 $^{\circ\ }$

$y=\dfrac{x\left(k+1\right)}{1-k}$

Since k>1

Therefore y<0 for x>-1 and y>0 for x<-1

Option d correctly satisfy this condition

Q8.CAT 2004

If $\frac{a}{b+c}=\frac{b}{a+c} =\frac{c}{b+a} =r$ , then r cannot take any value except

1/2
-1
1/2 or -1
-1/2 or -1

Show solution

a = r(b+c)

b = r(a+c)

c = r(a+b)

On adding all the equations,

a+b+c = 2r(a+b+c)

If r = 1/2, a+b+c = a+b+c (valid)

If r = -1, a+b+c = -2(a+b+c) => a+b+c = 0 => b+c = -a and a/(b+c) = a/(-a) = -1 (valid)

So, r can take the values 1/2 or -1

Q9.CAT 2002

Suppose for any real number x, [x] denotes the greatest integer less than or equal to x. Let L(x, y) = [x] + [y] + [x + y] and R(x, y) = [2x] + [2y]. Then it is impossible to find any two positive real numbers x and y for which

L(x, y) = R(x, y)
L(x, y) $\neq$ R(x, y)
L(x, y) < R(x, y)
L(x, y) > R(x, y)

Show solution

Consider different values of x and y:

x = -1.5 and y = -1.5; x = 1.5 and y = -1.5; x = -1.5 and y = 1.5; x = 1.5 and y = 1.5.

For these possibilities, options A,B and C gets satisfied , but it is impossible to find any two positive real numbers x and y for which L(x, y) > R(x, y).

Passage

Directions for the next 3 questions: For three distinct real positive numbers x, y and z, let

f(x, y, z) = min (max(x, y), max (y, z), max (z, x))

g(x, y, z) = max (min(x, y), min (y, z), min (z, x))

h(x, y, z) = max (max(x, y), max(y, z), max (z, x))

j(x, y, z) = min (min (x, y), min(y, z), min (z, x))

m(x, y, z) = max (x, y, z)

n(x, y, z) = min (x, y, z)

Q10.CAT 2000

Which of the following expressions is indeterminate?

(f(x, y, z)-h(x, y, z))/(g(x, y, z) —j(x, y, z))
[f(x,y,z)+h(x,y,z)+g(x,y,z)+j(x,y, z)]/[j(x, y, z) + h(x, y, z)-m(x, y, z) - n(x, y, z)]
[g(x, y, z)-j(x, y, z)]/[f(x, y, z)-h(x, y, z)]
[h(x, y, z) fix, y, z)]/[n(x, y, z) - g(x, y, z)]

Show solution

From the given functions we can make out that function h and m give max value , function n and j give min value , function f and g give middle value.So in option B , j cancels out n and h cancels out m . So the denominator becomes 0 and value is indeterminable.

Suppose x>y>z

f(x,y,z) = y

g(x,y,z) = y

h(x,y,z) = x

j(x,y,z) = z

m(x,y,z) = x

n(x,y,z) = z

The denominator of the second option becomes 0, hence making it indeterminate.

Q11.CAT 1999

For two positive integers a and b define the function h(a,b):as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the GCF of the elements of set A is computed by repeatedly using the function h.
The minimum number of times h is required to be used to compute G is:

1/2 n
(n - 1)
n
None of these

Show solution

Let p and q be any two elements of the set A.
For the computation of the GCF of elements of the set A, we can replace both p and q by just the GCF(p,q) and the result is unchanged.

So, for every application of the function h, we are reducing the number of elements of the set A by 1. (In this case two numbers p and q are replaced by one number GCF(p,q)).
Expanding this concept further, the minimum number of times the function h should be called is n-1

Passage

DIRECTIONS for the following questions: These questions are based on the situation given below: Let x and y be real numbers

f(x, y) = | x + y |

F(f(x, y)) = -f(x, y)

G(f(x, y)) = -F(f(x, y))

Q12.CAT 1999

What is the value of f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2))))?

Show solution

F(f(x,y)) = -f(x,y)

G(f(x,y)) = -F(f(x,y)) = f(x,y)

G(f(1,0)) = 1

F(f(1,2)) = -3

G(f(1,2)) = 3

f(F(f(1,2)),G(f(1,2))) = 0

f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2)))) = 1 + 0 = 1

81+ more Functions, Graphs and Statistics questions inside

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CAT Functions, Graphs and Statistics previous year questions with solutions — AthenaPrep