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CAT Exam  >  CAT Tests  >  Test Level 1: Inequalities - 2 - CAT MCQ

Test Level 1: Inequalities - 2 - CAT MCQ


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10 Questions MCQ Test - Test Level 1: Inequalities - 2

Test Level 1: Inequalities - 2 for CAT 2024 is part of CAT preparation. The Test Level 1: Inequalities - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: Inequalities - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 1: Inequalities - 2 below.
Solutions of Test Level 1: Inequalities - 2 questions in English are available as part of our course for CAT & Test Level 1: Inequalities - 2 solutions in Hindi for CAT course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt Test Level 1: Inequalities - 2 | 10 questions in 20 minutes | Mock test for CAT preparation | Free important questions MCQ to study for CAT Exam | Download free PDF with solutions
Test Level 1: Inequalities - 2 - Question 1
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3x2 – 7x – 6 < 0

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

At x = 0, inequality is satisfied. Hence, options (b) and (c) are rejected. x = 3 gives LHS = RHS. and x = – 0.66 also does the same. Hence. roots of the equation are 3 and – 0.66.
Thus, option (a) is correct.

Test Level 1: Inequalities - 2 - Question 2
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x2 – 3x + 5 > 0

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

The given quadratic equation has imaginary roots and is hence always positive.
Thus, option (d) is correct 

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Test Level 1: Inequalities - 2 - Question 3
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2 – x – x2 ≥ 0

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

At x = 0, inequality is satisfied.
Thus, options, (c) and (d) are rejected.
At x = 1, inequality is satisfied
Hence, we choose option (a). 

Test Level 1: Inequalities - 2 - Question 4
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|x2 + x| – 5 < 0
Detailed Solution for Test Level 1: Inequalities - 2 - Question 4

At x = 0 inequality is satisfied.
Thus, options (a), (b), are rejected.
Option (c) is obviously not true, as there will be values of x at which the inequality would not be satisfied.
Option (d) is correct.

Test Level 1: Inequalities - 2 - Question 5
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|x2 – 2x| < x
Detailed Solution for Test Level 1: Inequalities - 2 - Question 5

At x = 1 and x = 3 LHS = RHS. At x = 2 inequality is satisfied.
At x = 0.1 inequality is not satisfied.
At x = 2.9 inequality is satisfied.
At x = 3.1 inequality is not satisfied.
Thus, option (a) is correct. 

Test Level 1: Inequalities - 2 - Question 6
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|x2 – 3x| + x – 2 < 0
Detailed Solution for Test Level 1: Inequalities - 2 - Question 6

The options need to be converted to approximate values before you judge the answer. At x = 0, inequality is satisfied.
Thus, option (b) and (d) are rejected. Option (c) is correct. 

Test Level 1: Inequalities - 2 - Question 7
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x2 – |5x – 3| – x < 2
Detailed Solution for Test Level 1: Inequalities - 2 - Question 7

At x = 0, inequality is satisfied, option (a) rejected. At x = 10, inequality is not satisfied, option (c) rejected.
At x = –5, LHS = RHS.
Also at x = 5, inequality is satisfied and at x = 6, inequality is not satisfied.
Thus, option (d) is correct. 

Test Level 1: Inequalities - 2 - Question 8
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3x2 – 7x + 6 < 0
Detailed Solution for Test Level 1: Inequalities - 2 - Question 8

At x = 0, inequality is not satisfied.
Hence, options (b), (c) are rejected. At x = 2, inequality is not satisfied.
Hence, option (a) is rejected.
Thus, option (d) is correct.

Test Level 1: Inequalities - 2 - Question 9
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x2 – 14x – 15 > 0
Detailed Solution for Test Level 1: Inequalities - 2 - Question 9

At x = 0 inequality is not satisfied. Thus option (d) is rejected. 
x = –1 and x = 15 are the roots of the quadratic equation. Thus, option (c) is correct. 

Test Level 1: Inequalities - 2 - Question 10
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|x2 – 4x| < 5
Detailed Solution for Test Level 1: Inequalities - 2 - Question 10

At x  = 0 inequality is satisfied, option (b) is rejected.
At x  = 2, inequality is satisfied, option (c) is rejected.
At x = 5, LHS = RHS.
At x = –1, LHS = RHS.
Thus, option (d) is correct. 

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