Linear Inequalities MCQ Quiz - Objective Question with Answer for Linear Inequalities - Download Free PDF
Last updated on Apr 16, 2025
Latest Linear Inequalities MCQ Objective Questions
Linear Inequalities Question 1:
(x − 2)(x − 4)(x − 6)(x − 8) ... (x − 20) ≤ 0. How many integer values of x satisfy the given equation?
Answer (Detailed Solution Below) 15
Linear Inequalities Question 1 Detailed Solution
x = 2, 4, 6, 8, ... , 20 satisfy
(x − 2)(x − 4)(x − 6) ... (x − 20) = 0.
Also, for (x − 2)(x − 4)(x − 6) ... (x − 20) < 0, odd number of terms should be negative.
For x > 20 and x < 2 all terms will be positive.
For x = 19, x − 20 < 0, while all other terms are positive.
Thus, x = 19 is one of the solutions.
For x = 17, x − 20 < 0 and x − 18 < 0, while all other terms are positive.
But then (x − 2)(x − 4) ... (x − 20) > 0, hence x = 17 is not the solution.
For x = 15, three terms are negative.
Hence, x = 15 is one of the solutions.
Similarly, x = 11, 7, 3 are the other solutions.
∴ The required number of solutions = 10 + 5 = 15.
Linear Inequalities Question 2:
\( x^2 - 11x + |a| = 0 \) has integer roots. How many possible integer values can 'a' take?
Answer (Detailed Solution Below) 11
Linear Inequalities Question 2 Detailed Solution
To begin, the discriminant should be a perfect square. Let the discriminant be represented by \( D \).
From the quadratic formula:
\( \frac{-11 \pm \sqrt{D}}{2} \)
we can see that the numerator must be an even number for the roots to be integers.
This means that the discriminant should be a perfect square and specifically the square of an odd number. (This is the condition for odd + odd = even in the numerator.)
The discriminant equation is:
\( D = 121 - 4|a| = 121 - 4|a| \)
Since \( 4|a| \) cannot be negative, we conclude that \( D \) can take values such as 121, 81, 49, 25, 9, and 1.
The value of \( |a| \) can be 0, 10, 18, 24, 28, or 30.
Hence, \( a \) can take the values 0, \( \pm 10 \), \( \pm 18 \), \( \pm 24 \), \( \pm 28 \), and \( \pm 30 \).
Linear Inequalities Question 3:
Find the number of integral values of x for which
(5x - 1) < (x + 1)2 < (7x – 3)
Answer (Detailed Solution Below)
Linear Inequalities Question 3 Detailed Solution
Calculation:
Taking
⇒ (x + 1)2 > (5x - 1)
⇒ x2 – 3x + 2 > 0
⇒ (x – 1 )( x – 2 ) > 0
X < 1 or x > 2 ……(i)
Taking
⇒ (x + 1)2 < (7x – 3)
⇒ x2 – 5x + 4 > 0
⇒ (x – 1 )( x – 4 ) > 0
⇒ 1 < x < 4 ……(ii)
Combining (1) and (2) we get 2 < x < 4
Hence, x will only take one integer i.e. 3
Linear Inequalities Question 4:
The value of x satisfies the inequality lx - 1| + lx - 2| ≥ 4 if
Answer (Detailed Solution Below)
Linear Inequalities Question 4 Detailed Solution
Concept:
Calculation :
Case 1: x2
⇒
Simplifying
⇒
case 2:
⇒
Simplifying
⇒
this is contradiction
case 3: x 1
⇒
Simplifying
⇒
From Cases 1 and 3, we get
⇒
⇒
Hence option 1 is correct
Linear Inequalities Question 5:
If a, b and c are the lengths of the sides of a triangle then the range of the values of \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\) is
Answer (Detailed Solution Below)
Linear Inequalities Question 5 Detailed Solution
We know that,
AM (b + c, c + a, a + b) ≥ HM (b + c, c + a, a + b)
\(\Rightarrow \frac{b+c+c+a+a+b}{3} \geq \frac{3}{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}\)
\(\Rightarrow \frac{2}{9}(a+b+c) \geq \frac{1}{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}\)
\(\Rightarrow \frac{9}{2(a+b+c)} \leq \frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\)
\(\Rightarrow \frac{9}{2} \leq \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b} \)
\(\Rightarrow \frac{9}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3 \)
\(\Rightarrow \frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3 \)
\(\Rightarrow \frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\) .........(1)
Now, since a, b and c are the lengths of a triangle b + c > a
2b + 2c > a + b + c
\(\frac{1}{2(b+c)}<\frac{1}{a+b+c} \)
\(\frac{a}{b+c}<\frac{2 a}{a+b+c}\)
Similarly, we have, \(\frac{b}{c+a}<\frac{2 b}{a+b+c}\)
\(\frac{c}{a+b}<\frac{2 c}{a+b+c}\)
Adding these inequalities, we get
\(\frac{\mathrm{a}}{\mathrm{~b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}<2\) .........(2)
Therefore, From (A) and (B), we haved
\(\frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{b+c}+\frac{c}{a+b}<2\)
Top Linear Inequalities MCQ Objective Questions
Calculate the least whole number, which when subtracted from both the terms of the ratio 5 : 6 gives a ratio less than 17 : 22.
Answer (Detailed Solution Below)
Linear Inequalities Question 6 Detailed Solution
Download Solution PDFGiven:
Initial ratio = 5 ∶ 6
Final ratio should be less than 17 ∶ 22
Calculation:
Let the least whole number that is needed to be subtracted be a.
According to the question,
(5 - a)/(6 - a) < 17/22
⇒ 5 × 22 - 22a < 17 × 6 - 17a
⇒ 110 - 22a < 102 - 17a
⇒ 110 - 102 < - 17a + 22a
⇒ 8 < 5a
⇒ 8/5 = 1.6 < a
∴ The least whole number must be 2.
If a2 - b2 = 88, a - b = 4 then find the value of ab.
Answer (Detailed Solution Below)
Linear Inequalities Question 7 Detailed Solution
Download Solution PDFGiven:
a2 - b2 = 88
a - b = 4
Formula used:
a2 - b2 = (a - b)(a + b)
Calculation:
a - b = 4 ----(1)
(a - b)(a + b) = 88
⇒ 4 × (a + b) = 88
⇒ a + b = 88/4
⇒ a + b = 22 ----(2)
Adding equation (1) and equation (2), we get
⇒ a - b + a + b = 4 + 22
2a = 26
⇒ a = 13
Put the value of a in equation (2), we get
13 + b = 22
⇒ b = 9
value of ab = 13 × 9
⇒ ab = 117
∴ The value of ab is 117.
If \(x = 8 - 2\sqrt {15}\), then find the value of \({\left( {\frac{{{\rm{x}} + {\rm{}}1}}{{\sqrt {\rm{x}} }}} \right)^2}\)
Answer (Detailed Solution Below)
Linear Inequalities Question 8 Detailed Solution
Download Solution PDFGiven:
x = 8 – 2√(15)
Calculation:
x = (√5)2 + (√3)2 – 2√(15)
⇒ x = (√5 - √3)2
⇒ √x = √5 - √3
And 1/√x = (√5 + √3)/2
According to question,
\({\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2} = {\left[ {\sqrt 5 - \sqrt 3 + \left( {\frac{{\sqrt 5 + \sqrt 3 \;}}{2}} \right)} \right]^2}\)
\(\Rightarrow {\left[ {\frac{{3\sqrt 5 - \sqrt 3 }}{2}} \right]^2}\)
\(\Rightarrow \left[ {\frac{{45 + 3 - 6\sqrt {15} }}{4}} \right]\)
\(\Rightarrow \frac{{48 - 6\sqrt {15} }}{4}\)
⇒ \(12 - \frac{{3\sqrt {15} }}{2}\)Consider the following inequalities:
1. \(\frac{a^2 - b^2 }{a^2 + b^2 } > \frac{a - b}{a + b} \) where a > b > 0
2. \(\frac{a^3 + b^3 }{a^2 + b^2 } > \frac{a^2 + b^2}{a + b} \) only when a > b > 0
Which of the above is/are correct?
Answer (Detailed Solution Below)
Linear Inequalities Question 9 Detailed Solution
Download Solution PDFCalculation:
Statement:1 \(\frac{a^2 - b^2 }{a^2 + b^2 } > \frac{a - b}{a + b} \) where a > b > 0
Let a = 2 and b = 1
⇒ \(\frac{2^2 - 1^2 }{2^2 + 1^2 } > \frac{2 - 1}{2 + 1} \)
⇒ \(\frac{3 }{5 } > \frac{1}{3} \) which is correct.
Hence, statement 1 is correct.
Statemnt:2 \(\frac{a^3 + b^3 }{a^2 + b^2 } > \frac{a^2 + b^2}{a + b} \) only when a > b > 0
Let a = 2 and b = 1
⇒ \(\frac{2^3 + 1^3 }{2^2 + 1^2 } > \frac{2^2 + 1^2}{2 + 1} \)
⇒ \(\frac{9 }{5 } > \frac{5}{3} \) Which is also correct.
This inequality also holds for a = 1 and b = 2 i.e.
Statement 2 is correct for b > a > 0 as well.
But according to statement 2, inequality will be correct only if a > b > 0
So, statement 2 is incorrect.
Hence, statement (1) is correct but (2) statement (2) is incorrect.
Mistake PointsPlease note the language of the second statement
"only when a > b > 0"
We have proved thet, statement is correct for
a = 2 & b = 1 (i.e. a > b > 0) and b = 2 & a = 1 (i.e. b > a > 0)
Hence, due to only word, statement 2 will become incorrect.
If x - 1/x = 10, then x3 - 1/x3 = ?
Answer (Detailed Solution Below)
Linear Inequalities Question 10 Detailed Solution
Download Solution PDFGiven:
x - 1/x = 10
Formula:
(a - b)3 = a3 - b3 - 3ab(a - b)
Calculation:
(x - 1/x)3 = x3 - 1/x3 - 3(x)(1/x)(x - 1/x)
⇒ 103 = x3 - 1/x3 - 3 × 10
⇒ x3 - 1/x3 = 1000 + 30
⇒ x3 - 1/x3 = 1030
∴ The value of x3 - 1/x3 is 1030.
If x - 1/x = n,
x3 - 1/x3 = n3 + 3n
Here, n = 10
x3 - 1/x3 = 103 + 3 × 10
⇒ x3 - 1/x3 = 1000 + 30
⇒ x3 - 1/x3 = 1030
∴ The value of x3 - 1/x3 is 1030.
If a + b = 5 and 2a – b = 4. Find the relation between a and b.
Answer (Detailed Solution Below)
Linear Inequalities Question 11 Detailed Solution
Download Solution PDFa + b = 5 ----(1)
2a – b = 4 ----(2)
Adding Eq. (1) and (2), we get
⇒ a + b + 2a – b = 5 + 4
⇒ 3a = 9
⇒ a = 3
Putting a = 3 in eq. (1) we get b = 2
∴ a > bThe equations ax + 9y = 1 and 9y - x - 1 = 0 represent the same line if a =
Answer (Detailed Solution Below)
Linear Inequalities Question 12 Detailed Solution
Download Solution PDFGiven:
Equation1 = ax + 9y = 1
Equation2 = 9y - x - 1 = 0
Concept used:
If linear equations are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Here, the equations have an infinite number of solutions, if
a1/a2 = b1/b2 = c1/c2
Calculation:
We have equations,
ax + 9y = 1
⇒ ax + 9y - 1 = 0
and, 9y - x - 1 = 0
⇒ x - 9y + 1 = 0
Here, a1 = a, b1 = 9, c1 = -1
and, a2 = 1, b2 = -9, c2 = 1
As we know that
a1/a2 = b1/b2 = c1/c2
⇒ a/1 = 9/-9 = -1/1
⇒ a = -1 = -1
⇒ a = -1
∴ The value of a is -1.
Alternate Method
ax + 9y = 1 ----(i)
9y - x - 1 = 0 ----(ii)
From equation (ii) we get,
-x + 9y = 1 ----(iii)
On comparing eq(iii) and eq (i)
We get, a = -1
For what value of 'a', does the inequality 9a - a2 \(\leq\) 17a + 15 holds
Answer (Detailed Solution Below)
Linear Inequalities Question 13 Detailed Solution
Download Solution PDFCalculation:
Given: 9a - a2 \(\leq\) 17a + 15
On rearranging
-a2 + 9a \(\leq\) 17a + 15
Shifting the sign
a2 - 9a \(≥\) - 17a - 15
a2 - 9a + 17a + 15 \(≥\) 0
a2 + 8a + 15 \(≥\) 0
a2 + 5a + 3a + 15 \(≥\) 0
a(a+ 5) + 3(a + 5) \(≥\) 0
(a + 3)(a + 5) \(≥\) 0
So, -3 and -5 are the roots of the equation.
Now look at the diagram given below,
We see that all the numbers less than -5 and all the numbers greater than -3 will give us positive result. While numbers between -5 and -3 will give us negative results
So, all the above value holds for the above equation.
Solve the inequality: \(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1, x ≠ 4
Answer (Detailed Solution Below)
Linear Inequalities Question 14 Detailed Solution
Download Solution PDFCalculation-
\(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 ⇒\(\rm \begin {vmatrix}\frac{x-4}{2} \end{vmatrix}\) < 1 ⇒|x - 4| < 2
⇒ -2 < x - 4 < 2 ⇒ 2 < x < 6
As x ≠ 4 hence x ϵ (2, 4) ∪ (4, 6)
Since, (4, 6) is lying in the interval for all the acceptable values of x ϵ (2, 4) ∪ (4, 6) so this will be the best option.
∴ The solution of the equality \(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 is x ϵ (4, 6).
If 2 (3x + 5) > 4x - 5 < 3x + 2; then x can take which of the following values?
Answer (Detailed Solution Below)
Linear Inequalities Question 15 Detailed Solution
Download Solution PDF2 (3x + 5) > 4x - 5 < 3x + 2
2 (3x + 5) > 4x - 5
⇒ 6x + 10 > 4x – 5
⇒ 2x > -15
⇒ x > -15/2
4x - 5 < 3x + 2
⇒ x < 7
So, -15/2 < x < 7
∴ The required value of x = 6.