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Test Level 1: Functions - 1 - CAT MCQ


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10 Questions MCQ Test - Test Level 1: Functions - 1

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

If f(x) = y and g(x) = y2 + 1, find g(f(x)).

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

 f(x) = y
g(f(x)) = g(y) = y2 + 1
The dependent variable always remains constant, irrespective of the independent variable.

Test Level 1: Functions - 1 - Question 2

If , then

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

Given f(x) = 
Let us suppose, y = 
Or, (5x – 3)y = 3x + 2
 5xy – 3y = 3x + 2
(5xy – 3x) = 2 + 3y
x(5y – 3) = 2 + 3y
Or, x = 
For finding inverse, we need to interchange x and y.
So, 
So, the function is equal to its inverse.

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Test Level 1: Functions - 1 - Question 3

The graph of the function y = 5x6 + 3x4 - x2 + 8

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

If y = f (x), f (x) = f (- x).
So the y values are the same for +ve and -ve values of x, and the graph is symmetric about the Y-axis.

Test Level 1: Functions - 1 - Question 4

If f(x, y) = 3x2 - 2xy - y2 + 4, find f(1, -1).

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

Substituting x = 1 and y = -1 in f(x, y), we get 
f(1, -1) = 3 (1)2 - 2 × (1) (-1) - (-1)2 + 4 = 8

Test Level 1: Functions - 1 - Question 5

The given graph is

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

Graph is symmetric about Y - axis;
y = f(x) = I x I
f(-x) = I- x I = I x I
f(-x) = f(x)
therefore it is an even function.

Test Level 1: Functions - 1 - Question 6

A function is defined as f(x) =  + 1 + x + 2x2 + 3x3 + 4x4 + 5x5. If f(2) = 260.78, what is the value of f?

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

f(x) = + 1 + x + 2x2 + 3x3 + 4x4 + 5x5
f(1/x) = 5x5 + 4x4 + 3x3 + 2x2 + x + 1 + 
f(x) = f(1/x)
Therefore, f(1/2) = f(2) = 260.78.

Test Level 1: Functions - 1 - Question 7

Find the range of the function f(x) = 

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

f(x) = 
Let, y = f(x)

⇒ y2 = x - 3
⇒ x = y2 + 3
f-1(x) = x2 + 3
Domain of f-1(x) = R ....(i)
f(x) > 0 ......(ii)
Hence from (i) and (ii)
Range of f(x) = (0, ∞)

Test Level 1: Functions - 1 - Question 8

f : → R and g : R → R are defined as f (x) = x2 and g (x) = x + 3, x ∈ R, then (fog) (x) = ?

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

f(x) = x2, g(x) = x + 3
(fog) (x) = f(x + 3) = (x + 3)2 = x2 + 6x + 9.

Test Level 1: Functions - 1 - Question 9

Let a function be fn+1(x) = fn (x) + 3 . If f2(2) = 4, find the value of f6(2).

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

We have, fn+1(x) = fn (x) + 3
So, f3(2) = f2(2) + 3 = 7
f4(2) = f3(2) + 3 = 10
f5(2) = f4(2) + 3 = 13
f6(2) = f5(2) + 3 = 16

Test Level 1: Functions - 1 - Question 10

Let A = {a, b, c}. Then, the range of the relation R = {(a, b), (a, c), (b, c)} defined on A is

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

Range of R = {y ∈  A : (x, y) ∈ R for some x ∈ A} = {b, c}

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