CAT Questions on Functions, Graphs and Statistics [Download PDF] 

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CAT 2023 Functions, Graphs and Statistics Questions PDF
CAT 2023 Functions, Graphs and Statistics Questions PDF

CAT Questions on Functions, Graphs and Statistics [Download PDF]

Functions, Graphs and statistics is the one of the topic in CAT quantitative aptitude section, where the questions are mostly tricky. We have provided one solved set of questions on functions and graphs topic. 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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Let us see CAT Functions, Graphs and Statistics Questions and Answers

Question 1: 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

1) Answer (B)

View Video Solution

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 2: Find the sum $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}} +….+ \sqrt{1+\frac{1}{2007^2}+\frac{1}{2008^2}}$

a) $2008 – \frac{1}{2008}$

b) $2007 – \frac{1}{2007}$

c) $2007 – \frac{1}{2008}$

d) $2008 – \frac{1}{2007}$

e) $2008 – \frac{1}{2009}$

2) Answer (A)

View Video Solution

Solution:

Consider the first term:
$\sqrt{1+1/1^2+1/2^2}$ = $\sqrt{9/4}$ = 3/2
Second term: $\sqrt{1+1/2^2+1/3^2}$ = $\sqrt{49/36}$ = 7/6
First term + Second term = 3/2 + 7/6 = (9+7)/6 = 16/6 = 8/3 = 3 – 1/3
.
.
​Required sum = 2008 – 1/2008

Question 3: Consider the following two curves in the x-y plane:
$y = x^3 + x^2 + 5$
$y = x^2 + x + 5$
Which of following statements is true for $-2 \leq x \leq 2$ ?

a) The two curves intersect once.

b) The two curves intersect twice.

c) The two curves do not intersect

d) The two curves intersect thrice.

3) Answer (D)

View Video Solution

Solution:

Equate the 2 equations we get value of x = 1 and -1 . Also we notice that there is intersection at x=0 . hence D

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)

View Video Solution

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: Let g(x) be a function such that g(x+1) + g(x-1) = g(x) for every real x. Then for what value of p is the relation g(x+p) = g(x) necessarily true for every real x?

a) 5

b) 3

c) 2

d) 6

5) Answer (D)

View Video Solution

Solution:

According to given condition we have , g(x+1) = -g(x-1) + g(x)

Putting x=x+1 we get g(x+2) = g(x+1) – g(x) = -g(x-1)

Putting x=x+2 we get g(x+3)=-g(x)

Similarly g(x+4)=-g(x+1), g(x+5)=-g(x+2)=-g(x+1) + g(x)  and g(x+6) = g(x+1)-g(x+2)=g(x).

So p=6.

Question 6: 
In the above table, for suitably chosen constants a, b and c, which one of the following best describes the relation between y and x?

a) $y=a+bx$

b) $y=a+bx+cx^2$

c) $y=e^{a+bx}$

d) None of the above

6) Answer (B)

View Video Solution

Solution:

y=a+bx is a linear function and clearly the function shown in the table is not linear.
Lets take option B:

Suppose $y=f(x)=a+bx+cx^2$
f(1)=4 => a+b+c=4 — (1)
f(2)=8 => a+2b+4c=8 — (2)
f(3)=14 => a+3b+9c=14 — (3)

Let us solve the equations and see if $f(4), f(5),f(6)$ also satisfy the given equation.

(2)-(1) => b+3c=4
(3)-(2) => b+5c=6
This gives c=1 and b=1 ==> a=2

So, $f(x)=2+x+x^2$
$f(4)=2+4+4^2 = 22$
$f(5)=2+5+5^2 = 32$
$f(6)=2+6+6^2 = 44$

As all the three also satisfy the numbers given in the table, it can be inferred that the relationship between x and y is quadratic and the correct option is option (b)

Question 7: A function can sometimes reflect on itself, i.e. if y = f(x), then x = f(y). Both of them retain the same structure and form. Which of the following functions has this property?

a) $y=\frac{2x+3}{3x+4}$

b) $y=\frac{2x+3}{3x-2}$

c) $y=\frac{3x+4}{4x-5}$

d) None of the above

7) Answer (B)

View Video Solution

Solution:

Putting y as x and x as y in given options, only options B satisfies the function property as follows.

x = $\frac{2y+3}{3y-2}$

or 3xy – 2x = 2y+3

or y(3x-2) = 2x+3

or $y = \frac{2x+3}{3x-2}$

Checkout: CAT Free Practice Questions and Videos

Question 8: 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:

a) 1/2 n

b) (n – 1)

c) n

d) None of these

8) Answer (B)

View Video Solution

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

Question 9: Let $f(x) = x^{2}$ and $g(x) = 2^{x}$, for all real x. Then the value of f[f(g(x)) + g(f(x))] at x = 1 is

a) 16

b) 18

c) 36

d) 40

9) Answer (C)

View Video Solution

Solution:

$f[f(g(1)) + g(f(1))]$
= $f[f(2^1) + g(1^2)]$
= $f[f(2) + g(1)]$
= $f[2^2 + 2^1]$
= $f(6)$
= $6^2 = 36$

Question 10: Let f(x)= $\max(5x, 52-2x^2)$, where x is any positive real number. Then the minimum possible value of f(x)

10) Answer: 20

View Video Solution

Solution:

The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$x$ = $52 – 2x^2$
or, $2x^2 + 5x – 52$ = 0
On solving, we get $x$ = 4 or $-\dfrac{13}{2}$
But, it is given that $x$ is a positive number.
So, $x$ = 4
And the minimum value = 5*4 = 20
Hence, 20 is the correct answer.

Question 11: Let S be the set of all points (x, y) in the x-y plane such that $\mid x \mid + \mid y \mid \leq 2$ and $\mid x \mid \geq 1.$ Then, the area, in square units, of the region represented by S equals

11) Answer: 2

View Video Solution

Solution:

Sum of the area of region I and II is the required area.

Now, required area =  $\ 4\times\frac{\ 1}{2}\times\ 1\times\ 1$ = 2

Question 12: If $f(5+x)=f(5-x)$ for every real x, and $f(x)=0$ has four distinct real roots, then the sum of these roots is

a) 0

b) 40

c) 10

d) 20

12) Answer (D)

View Video Solution

Solution:

Let ‘r’ be the root of the function. It follows that f(r) = 0. We can represent this as $f\left(r\right)=f\left\{5-\left(5-r\right)\right\}$

Based on the relation: $f\left(5-x\right)=f\left(5+x\right)$; $f\left(r\right)=f\left\{5-\left(5-r\right)\right\}=f\left\{5+\left(5-r\right)\right\}$

$\therefore\ f\left(r\right)=f\left(10-r\right)$

Thus, every root ‘r’ is associated with another root ‘(10-r)’ [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $r_1,\ \left(10-r_1\right),\ r_2,\ \left(10-r_2\right)$

The sum of these roots = $r_1\ +\left(10-r_1\right)+\ r_2+\ \left(10-r_2\right)\ =\ 20$

Hence, Option D is the correct answer.

Question 13: In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

a) 5

b) 4

c) 3

d) 6

13) Answer (B)

View Video Solution

Solution:

Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say that

x(2)+x(3)+x(4)+……..x(10)= 47*9=423……………….(1)

Similarly x(1)+x(2)+x(3)+x(4)…………….+x(9)= 42*9=378……………(2)

Subtracting both the equations we get x(10)-x(1)=45

Now, the sum of the 10 observations from equation (1) is 423+x(1)

Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5

Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5

Hence difference in average will be 46.5-42.5=4 which is the correct answer

Question 14: 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

14) 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

 

Question 15: 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) $x \geq r$

15) 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

 

 

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