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Test Level 1: Functions - 2 - CAT MCQ


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20 Questions MCQ Test Level-wise Tests for CAT - Test Level 1: Functions - 2

Test Level 1: Functions - 2 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 1: Functions - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: Functions - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 1: Functions - 2 below.
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Test Level 1: Functions - 2 - Question 1

Directions: Answer the question based on the following information.
A function has been defined for a variable x, where domain of F(x) and F1(x) ∈ [-2, 2].

Choose the correct option.

Detailed Solution for Test Level 1: Functions - 2 - Question 1

F(2) = -2 and F1(-2) = -2
F(-2) = 0 and F1(2) = 0
So, according to this,
F(x) = F1(-x) or F1(x) = F(-x)

Test Level 1: Functions - 2 - Question 2

If f(x) = x3 + y3, g(x) = xy and h(x) = (x + y), then which of the following is the value of f(x)?

Detailed Solution for Test Level 1: Functions - 2 - Question 2

x3 + y3 = (x + y)3 - 3xy (x + y)
= (x + y) [(x + y)2 - 3xy]
= h(x) [ {h(x)}2 - 3g(x) ]

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Test Level 1: Functions - 2 - Question 3

An exponential function is given by f(x) = a3x. If the point  lies on the graph of this function, then the value of 'a' is

Detailed Solution for Test Level 1: Functions - 2 - Question 3

Given function is f(x) = a3x
As point  lies on the given curve, it must satisfy the given equation.

a = 7

Test Level 1: Functions - 2 - Question 4

If f : R - {3} → R - {1} is defined by f(x) = , then f is

Detailed Solution for Test Level 1: Functions - 2 - Question 4

Here, y = f(x)

Which is defined for y ∈ R - {1}
Hence f(x) is invertible.
∴ One - one and onto

Test Level 1: Functions - 2 - Question 5

Which of the following options is necessarily true if f(x) = x3 - 4x + p and f(0) and f(1) have opposite signs?

Detailed Solution for Test Level 1: Functions - 2 - Question 5

f(x) = x3 - 4x + p
f(0) = p
f(1) = 1 - 4 + p = p - 3
∵ f(0) and f(1) have opposite signs
∴ f(0) and f(1) < 0
 p(p - 3) < 0
 0 < p < 3

Test Level 1: Functions - 2 - Question 6

If f(x) = ax2 + bx + c and c = then the graph of y = f(x) will certainly

Detailed Solution for Test Level 1: Functions - 2 - Question 6


This shows that root of equation are real and equal. Thus, the graph touches the x-axis and is hence tangential to the x-axis.
Thus, answer option 3 is correct.

Test Level 1: Functions - 2 - Question 7

If f(x) = 3x - 5 and f(g(x)) = 2x, then g(x) equals

Detailed Solution for Test Level 1: Functions - 2 - Question 7

Given: f(x) = 3x - 5 and f(g(x)) = 2(x)
f(g(x)) = 3g(x) - 5 = 2x
3g(x) - 5 = 2x
g(x) = (2x + 5)/3

Test Level 1: Functions - 2 - Question 8

If f(a, b) = a2 + b3, g(a, b) = a + b, then what is the value of f(3, g(3, 4))?

Detailed Solution for Test Level 1: Functions - 2 - Question 8

f(a, b) = a2 + b3
g(a, b) = a + b
g(3, 4) = 7
f(3, g(3, 4)) = f(3, 7) = 32 + 73 = 352

Test Level 1: Functions - 2 - Question 9

When the curves y = log10 x and y = x-1 are drawn in the x - y plane, how many times do they intersect for the values x ≥ 1?

Detailed Solution for Test Level 1: Functions - 2 - Question 9

It is given that y = log10x & y = 1/x
Draw their graphs approximately and check.

Here, we see the graph has only 1 intersection point.

Test Level 1: Functions - 2 - Question 10

Find the minimum value of x2 - 12x + 27.

Detailed Solution for Test Level 1: Functions - 2 - Question 10

 f (x) = x2 - 12x + 27 = x2 - 12x + 36 - 9
= (x - 6)2 - 9.
Since (x - 6)2 ≥ 0, the minimum value of the expression is - 9.

Test Level 1: Functions - 2 - Question 11

A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with respect to its radius, when the radius is 7 cm.

Detailed Solution for Test Level 1: Functions - 2 - Question 11

Let x be the radius and y be the volume of the balloon.
Then, y = 4/3πx3  dy/dx = 4πx2
 = 4π(7)2 = 196π cm2 
Hence, the volume is increasing with respect to its radius at the rate of 196  π cm2, when the radius is 7 cm.

Test Level 1: Functions - 2 - Question 12

Find the two numbers such that their sum is 12 and the sum of their squares is minimum.

Detailed Solution for Test Level 1: Functions - 2 - Question 12

Let the numbers be x and 12 - x.
∴ Sum of squares = x2 + (12 - x)2 = f(x), say.
∴ f` (x) = 2x + 2 (12 - x) (-1) = 4(x - 6)
∴ f`(x) = 0, only when x = 6.
The numbers are 6 and 6.

Test Level 1: Functions - 2 - Question 13

The sum of two numbers is 2k. The numbers such that the sum of their squares is minimum, will be

Detailed Solution for Test Level 1: Functions - 2 - Question 13

The sum of squares of two numbers is minimum when the numbers are equal.

Test Level 1: Functions - 2 - Question 14

Refer to the following data to answer the question given below:
R(x, y) = and sq(x, y) = 
rem (x, y) = remainder when x is divided by y
q(x, y) = quotient when x is divided by y
sq(5, 10) is equal to

Detailed Solution for Test Level 1: Functions - 2 - Question 14

sq(5, 10) = [R(5, 10)]

Test Level 1: Functions - 2 - Question 15

If f(x) = then f[f[f(x)]] equals

Detailed Solution for Test Level 1: Functions - 2 - Question 15


Test Level 1: Functions - 2 - Question 16

Radius of a circle is increasing uniformly at the rate of 3 cm/sec. The rate of increase in area, when radius is 10 cm will be

Detailed Solution for Test Level 1: Functions - 2 - Question 16

dr/dt = 3, we have A = πr2 ⇒ dA/dt = 2 π r (dr/dt)
⇒ (dA/dt)r = 10 = 2π x 10 x 3 = 60 π cm2/s

Test Level 1: Functions - 2 - Question 17

Differentiate 3y3 + 4y2 - 6y + 7 w.r.t. y.
 

Detailed Solution for Test Level 1: Functions - 2 - Question 17

Let z = 3y3 + 4y2 - 6y + 7
Differentiating both sides w.r.t. y,

= 9y2 + 8y - 6 + 0 = 9y2 + 8y - 6

Test Level 1: Functions - 2 - Question 18

If f(x) is an even function, then the graph y = f(x) will be symmetrical about

Detailed Solution for Test Level 1: Functions - 2 - Question 18

y – axis by definition. 

Test Level 1: Functions - 2 - Question 19

If f (x) is an odd function, then the graph y = f(x) will be symmetrical about

Detailed Solution for Test Level 1: Functions - 2 - Question 19

Origin by definition. 

Test Level 1: Functions - 2 - Question 20

Find the minimum value of the function f (x) = log2 (x2 – 2x + 5).

Detailed Solution for Test Level 1: Functions - 2 - Question 20

The minimum value of the function would occur at the minimum value of (x2 – 2x + 5) as this quadratic function has imaginary roots.
For
y = x2 – 2x + 5
dy/dx = 2x – 2 = 0 ⇒ x = 1
⇒ x2 – 2x + 5 = 4. 

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