Free Algebra 06 Practice Test - CAT 

Question 1

g(x) is defined as the sum of all the divisors of a natural number n.
If g(x) is less than 2n then it is placed in category A.
If g(x) is greater than 2n it is placed in category B.
Of the following numbers, which pair consists of numbers one of which can be placed in A and the other in B?

A. 32,64
B. 6,64
C. 7,28
D. 14,84

SOLUTION

Solution : D

Option (d)

g(32)= 1+2+4+8+16+32 = 63< 2x32

(Category A)

g(64)=1+2+4+8+16+32+64=127<2x64

(Category A)

g(6)=1+2+3+6=12=2x6 (neither A nor B)

g(7)=1+7=8<2x7 (Category A)

g(28)=1+2+4+7+14+28=56=2x28(Neither A nor B)

g(14)=1+2+7+14=24<2x14 (A)

g(84)= 1+2+3+4+6+7+!2+14+21+28+42+84=224>2x84(B)

Question 2

 Given that a real value function g, is such that  g (a+b) = a+g (g(b))& g(0)=0. Find g(102)?

A. 0
B. 1
C. 99
D. 102
E. 101

SOLUTION

Solution : D

Option (d)

Method 1 –

Assumption is the best way to solve this question. Assume a=0, b=0

g(0)=0+g(g(0))=0

Next assume a=1 and b=0

g(1)=1+g(g(0))=1+0=1

Continuing in the same way, assume a=2, b=0

g(2)=2+g(g(0))=2

Method -2 Shortcut in a shortcut-

Assume a=102 and b=0

Then g(102+0)=102+0= 102!

Question 3

Find the area enclosed by the equation |x| + |y| = 4.

A. 4 squnits
B. 8 squnits
C. 16 squnits
D. 32 squnits

SOLUTION

Solution : D

Given equation, we can write like: 

If x < 0 and y > 0, - x + y = 4

If y < 0 and x > 0 x - y = 4

If x<0 and y<0, - x - y = 4

If x > 0 and y >0, x + y = 4

from the above four equations, we can draw below graph.

Above graph is combination of 4 triangle. Area of one triangle = 1/2 x base x height

 = 1/2 x 4 x 4 = 8

There are four similar triangle, Area = 8 x 4 = 32.

Question 4

f(x) = Nearest prime number to x on the number line and f(x) ≥ x.
g(x) = Nearest even number to x on the number line and g(x) ≥ x.
h(x) = Nearest odd number to x on the number line and h(x) ≤ x.
x is a non-negative number. You do not have to consider zero as an even number.
If x ≤ 25, then for how many x’s f(x) = h(x)?

A. 9
B. 8
C. 10
D. Data insufficient

SOLUTION

Solution : B

Option (b)

If f(x) = h(x), then f(x) = h(x) = x, because f(x) ≥ x and h (x) ≤ x

Now x should be an odd and prime from number below 25 and there are eight such number, i.e.

3, 5, 7, 11, 13, 17, 19, and 23.

Question 5

 Find the domain of

A.
B.
C.
D.

SOLUTION

Solution : B

Option (b)

The given function is defined for those values of x for which |x| -x >0, i.e. |x|>x. this inequality is satisfied only if x<0. Hence the domain of the given function is .

Question 6

Which of the following is an even function?

A. f(x)=
B.
C.
D. f(x)=

SOLUTION

Solution : C

The function f(x) is said to be 'even' if and only if f(x) is a real-valued function of a real variable x, and f(-x) = f(x).

Only at 

F (-x) = f(x), Hence only option c is even.

Question 7

Find the inverse of

A.
B.
C.
D. none of these

SOLUTION

Solution : A

Conventional method

Replacing f(x) with y, the given function is

y= (4-(x-7)3)1/5

To find the inverse function, we need to write an equation for x in terms of y

Raising both sides to the power 5, we get

y5 = 4 - (x-7)3

 (x-7)3 = 4- y5

Taking cube root on both sides

x-7 = (4 - y5)1/3

x = 7 + (4 - y5)1/3

We have an expression for x interms of y. We get f-1(x) if we replace y with x in the above equation

Hence, f-1(x) = 7 + (4 - x5)1/3. Answer option (a)

 

Shortcut

What is the basic definition of an inverse function?

An Inverse function is one which can be expressed as f(y) = x when f(x) = y

Using, this very definition, we can use a common sense approach to arrive at the answer In no time at all

 

1) Substitute a suitable value for x (it can be any value). in this case, for our convenience we will take x=7. Then at x=7, f(x)= 41//5

2) Now use the definition of inverse function. Put x=41/5 in each of the answer options and see where you get f(x)=7.  You are just exchanging "x” and "y” values!

This is the concept of inverse function after all!

Only option (a) gives 7 at x=41/5

 

All other options can be eliminated as for x=41/5, we do not get 7. Answer is option (a)

Question 8

Let hn+1(x) = hn(x) + 1 , if n is odd;
= hn(x) + 2, if n is even;
If the minimum value of h1(x) is 1 , then what will be the minimum value of h100(x)?

A. 100
B. 149
C. 150
D. None of these

SOLUTION

Solution : B

option b

h1(x) = 1                     h2(x) = 2

h3(x) = 4                     h4(x) = 5

h5(x) = 7                      h6(x) = 8

=> hn(x) =, when “n” is even

h100(x) = 149

Question 9

 f(x)= x+1 and g(x)= x2 -2. Find(gof)-1([-2,-1])

A. (0,-2)
B. [0,-1]
C. [-2, 0]
D. none of these

SOLUTION

Solution : C

Option (d)

gof (x) = g(f(x))= g(x+1) = (x+1)2-2= x2 +2x -1

-2 ≤x2 +2x -1≤-1

-1≤x2 +2x≤0

x(x+2) ≤0

- 2 ≤ x ≤ 0

[-2,0]

Question 10

Which of the following functions is odd?

A. f (x)= cosx
B.
C.
D. none of these

SOLUTION

Solution : D

Option (d)

The function f(x) is said to be 'odd' if and only if f(x) is a real-valued function of a real variable x, and f(- x) = - f(x). None of the options satisfies , f(-x)= -f(x), hence answer option d.

Question 11

f (x)= |x|+2; g (x)= |2x|+ 4; and h (x)= |x-4|+2. The reflection of these functions is taken about y=2, y=4, and y=2 respectively. In how many distinct points do these graphs cut the x-axis when plotted together?

A. 6
B. 3
C. 5
D. 4

SOLUTION

Solution : B

Option (b)

First two graphs will be cutting x-axis at +2 and -2.

And third one too will pass through +2 and +6. So there are three ponts on x axis. 

Question 12

 If g(2,3) = 85 and g(5,4)= 281, then find g(7,2)

A. 150
B. 56
C. 90
D. 65

SOLUTION

Solution : D

OPTION (d)

g(2,3)= 85= 22 +34

g(5,4)= 281= 52 +44

therefore g(7,2)= 72+24= 65

Question 13

Find the area enclosed in the first quadrant represented by the function [x] + y= 1 where [t] is the greatest integer less than or equal to x?

A. 1
B. 8
C. infinity
D. 2
E. 4

SOLUTION

Solution : A

Option (a)

For x=0 to x= 1 => y=1

When x=0.9999999, [x] = 0, thus y=1

Thus the area will be very close to, but slightly less than 1

Thus, the area enclosed in the first quadrant can be represented as follows

Required area= 1x1= 1

Question 14

Given p ≤ 4 is a positive real. Let A be the area of the bounded region enclosed by the curves y = 1 - |1-x| and y = |2x-p|. Then which among the following best describes A?

A.
B.
C.
D.

SOLUTION

Solution : C

Option (c)

Case 1: 0 < p ≤ 1The area formed is of a triangle with vertices () and (p, p). Thus, the area is sq. units.

Case 2: 1 ≤ p ≤ 3The figures formed is a quadrilateral and (1, 1).Thus, the area formed is

Case 3: 3 ≤ p ≤ 4The area formed is the image of case 1.

Area =

Thus, min (A) = 0 at p = 4 and max (A) = at p = 2Hence, option (c) is the right answer.

Question 15

How many positive integer values can x take that satisfy the inequality (x - 6)(x - 8)(x - 10) ........ (x - 100) < 0? 

A. 24
B. 20
C. 48
D. 72

SOLUTION

Solution : A

Positive integer values ----> (0, infinity)
Total there are 48 terms.
[101, infinity) ------> product will be positive.
[1, 5] -------> product will be positive.
[6, 8, 10.......100] ------>result zero.
when x = 7,
One positive term and 47 negative terms. So, the product will be negative.
when x = 9,
Two positive terms and 46 negative terms. So, the product will be positive.
.
.
.
and so on.
7, 11, 15, ....... 99 satisfies the given inequality.
Therefore, x can take 24 values.
 
Related Exams