Inequalities in one Variable MCQ Quiz - Objective Question with Answer for Inequalities in one Variable - Download Free PDF

Concept:

Solution Region of an Inequality:

• The solution region of an inequality is the set of all points that satisfy the inequality.
• For linear inequalities, the solution region is typically a half-plane or a region bounded by lines.
• The inequality given is: 2x + 4y ≤ 9.
• We can rewrite the inequality as a line equation: 2x + 4y = 9 and plot it on the coordinate plane.
• The solution region consists of all points that satisfy the inequality, which is typically one side of the line.

Calculation:

Given the inequality: 2x + 4y ≤ 9

First, rewrite the inequality as the equation of a line:

2x + 4y = 9

Now, solve for y:

4y = 9 - 2x

y = (9 - 2x) / 4

y = 9/4 - x/2

The slope of the line is -1/2 and the y-intercept is 9/4.

Plot the line y = (9 - 2x)/4 on the coordinate plane.

The solution region will be the area below this line since the inequality is ≤ (i.e., points that satisfy the inequality lie below or on the line).

Thus, the solution region is the half-plane below the line 2x + 4y = 9, including the line itself.

∴ The solution region is the region below and on the line 2x + 4y = 9.

Calculation

Given;

Inequality: 37 - (3x + 5) ≥ 9x - 8(x - 3)

⇒ 37 - 3x - 5 ≥ 9x - 8x + 24

⇒ 32 - 3x ≥ x + 24

⇒ 32 - 24 ≥ x + 3x

⇒ 8 ≥ 4x

⇒ 4x ≤ 8

⇒ x ≤ 2

∴ The solution set is (-∞, 2].

Hence option 3 is correct.

Calculation

Given:

\(-3 < \frac{1}{2} + \frac{-3x}{2} \leq 6\)

⇒ \(-6 < 1 - 3x \leq 12\)

⇒ \(-7 < -3x \leq 11\)

⇒ \(\frac{-7}{-3} > x \geq \frac{11}{-3}\)

⇒  \(\frac{7}{3} > x \geq -\frac{11}{3}\)

⇒ \(-\frac{11}{3} \leq x < \frac{7}{3}\)

∴ x lies in the interval \(\left[-\frac{11}{3}, \frac{7}{3}\right)\)

Hence option 1 is correct

Concept:

Rules for Operations on Inequalities:

• Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
• Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
• Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

Calculation

Given 6x + 7 ≤ x - 28

⇒ 6x + 7 - x ≤ x - 28 - x 

⇒ 5x + 7 ≤ -28 

⇒ 5x + 7 - 7 ≤ -28 - 7

⇒  5x ≤ -35

⇒  x ≤ -7

Since x is a natural number (which means x must be a positive integer), there are no natural numbers that satisfy (x ≤ -7).

∴ There is no solution 

∴ Option 1 is correct

Concept:

Rules for Operations on Inequalities:

• Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
• Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
• Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

Calculation

Given that, 8x + 2 ≥ 2x + 14

⇒ 8x + 2 - 2x ≥ 2x + 14 - 2x

⇒ 6x + 2 ≥ 14

⇒ 6x + 2 - 2 ≥ 14 - 2

⇒ 6x ≥ 12

⇒ \( x ≥ \frac{12}{6} \)

⇒ x ≥ 2

Hence, x ∈ [2, ∞)

The solution set is [2, ∞)

∴ The correct option is (3)

Concept:

Comparison using inequality:

For any two real numbers \(a\) and \(b\) if \(\rm |a| < |b|\) then \(\rm a^2 < b^2\).

Calculation:

Use the given inequality and proceed as follows:

\(\rm|2x-3| < |x+5|\\ (2x-3)^2 < (x+5)^2\\ 4x^2+9-12x < x^2+25+10x\\ 3x^2-22x-16 < 0\\ 3x^2-24x+2x-16 < 0\\ 3x(x-8)+2(x-8) < 0\\ (3x+2)(x-8) < 0 \)

Therefore, we conclude that \(\left(-\dfrac{2}{3}, 8\right)\).

Calculations:

Given,1.5 ≤ x ≤ 4.5

So critical points are x = 1.5 = \(\frac 3 2\) or x = 4.5 = \(\frac 9 2\)

⇒ 2x - 3 = 0 or 2x - 9 = 0

Changing inequality into equality form

∴ (2x - 3)(2x - 9) = 0

As we can see in the wavy curve value of (2x - 3)(2x - 9) is negative.

(2x - 3)(2x - 9) ≤ 0

Hence option 4 is the correct answer.

Concept:

The Modulus Function '| |' is defined as:

 \(\rm |x|=\left\{\begin{matrix}\rm \ \ \ x, &\rm x \geq 0\\ \rm -x, &\rm x<0\\\end{matrix}\right.\)

Calculation:

We have two cases:

CASE I: If x + 5 ≥ 0, then |x + 5| = x + 5.

⇒ x + 5 ≥ 10

⇒ x ≥ 5

⇒ x ∈ [5, ∞)

CASE II: If x + 5 < 0, then |x + 5| = -(x + 5).

⇒ -(x + 5) ≥ 10

⇒ -x - 5 ≥ 10

⇒ -x ≥ 15

⇒ x ≤ -15

⇒ x ∈ (-∞, -15]

∴ x ∈ (-∞, -15] ∪ [5, ∞).

Concept:

If |x| ≤ a then - a ≤ x ≤ a

Calculations:

Given , |3x - 5| ≤ 2 

⇒ - 2 ≤ 3x - 5 ≤ 2

⇒ - 2 + 5 ≤ 3x  ≤ 2 + 5 

⇒ 3 ≤ 3x  ≤ 7

⇒\(\rm 1 \le x \le \dfrac{7}{3}\) 

Hence, if |3x - 5| ≤ 2 then then \(\rm 1 \le x \le \dfrac{7}{3}\)

Explanation:

Given system of linear inequality is 

\(\left\{ \begin{matrix} 5 + x > 3x - 7 \\\ 11 - 5x \le 1 \end{matrix} \right.\)

When 5 + x > 3x - 7 ⇒ 2x < 12 ⇒ x < 6

When 11 - 5x ≤ 1 ⇒ 5x ≥ 10 ⇒ x ≥ 2

On the number line, the solution is represented as below.

\(\frac{{11 - 2x}}{5} \ge \frac{{9-3x}}{8} + \frac{3}{4}\) x ∈ n

\(\frac{{11}}{5} - \frac{{2x}}{5} \ge \frac{9}{8} - \frac{{3x}}{8} + \frac{3}{4}\)

\(\frac{{11}}{5} - \frac{9}{8} - \frac{3}{4} \ge \frac{{2x}}{5} - \frac{{3x}}{8}\)

\(\frac{{16x - 15x}}{{40}}\)\( \le \frac{{88 - 45 - 30}}{{40}}\)

\(\frac{x}{{40}} \le \frac{{13}}{{40}}\)

x ≤ 13

Since x ∈ N, Natural number

x > 0

so x ∈ (1, 2, 3, ...13)

Concept:

Modulus Function:

The function f(x) = |x|, defined as |x| = \(\left\{\begin{matrix} x,x≥0\\ -x, x<0 \end{matrix}\right.\), is called the Modulus function.

Calculation:

Let f(x) = \(\frac{|x-2|}{x-2}\)

Given f(x) ≥ 0

For f(x) to be defined, x - 2 ≠ 0

⇒ x ≠ 2

⇒ f(x) = \(\left\{\begin{matrix} \frac{x-2}{x-2},x>2\\ \frac{-(x-2)}{x-2}, x<2 \end{matrix}\right.\) [ x ≠ 2 ]

⇒ f(x) = \(\left\{\begin{matrix} 1,x≥2\\ -1, x<2 \end{matrix}\right.\)

∴ For f(x) ≥ 0

⇒ x > 2

∴ x ∈ (2, ∞)

Concept:

Comparison using inequality:

For any two real numbers \(a\) and \(b\) if \(\rm |a| < |b|\) then \(\rm a^2 < b^2\).

Calculation:

Use the given inequality and proceed as follows:

\(\rm|2x-3| < |x+5|\\ (2x-3)^2 < (x+5)^2\\ 4x^2+9-12x < x^2+25+10x\\ 3x^2-22x-16 < 0\\ 3x^2-24x+2x-16 < 0\\ 3x(x-8)+2(x-8) < 0\\ (3x+2)(x-8) < 0 \)

Therefore, we conclude that \(\left(-\dfrac{2}{3}, 8\right)\).

Concept:

Comparison using inequality:

For any two real numbers \(a\) and \(b\) if \(\rm |a| < |b|\) then \(\rm a^2 < b^2\).

Calculation:

Use the given inequality and proceed as follows:

\(\rm|2x-3| < |x+5|\\ (2x-3)^2 < (x+5)^2\\ 4x^2+9-12x < x^2+25+10x\\ 3x^2-22x-16 < 0\\ 3x^2-24x+2x-16 < 0\\ 3x(x-8)+2(x-8) < 0\\ (3x+2)(x-8) < 0 \)

Therefore, we conclude that \(\left(-\dfrac{2}{3}, 8\right)\).

Given:

\(\left| {\frac{{{x^2}}}{{x - 1}}} \right| \le 1\)

Concept:

(i) \(|\frac{a}{b}|= \frac{|a|}{|b|}\)

(ii) |x - a| = (x - a) if x ≥ a and -(x - a) if x < a

Calculation:

\(\left| {\frac{{{x^2}}}{{x - 1}}} \right| \le 1\)

⇒ \( \frac{|x^2|}{|x - 1|} \le 1\)

⇒ |x²| ≤ |x - 1|

⇒ x2 ≤ |x - 1| as x2 is always positive for real x

(i) Now for x > 1, |x - 1| = (x - 1)

∴ above inequality will be x2 ≤ x - 1

⇒ x2 - x + 1 ≤ 0

Which gives no solution as for x2 - x + 1, (a > 0 and D < 0) which makes it always positive.

Hence no solution for x > 1

(ii) Now for x ≤ 1, |x - 1| = - (x - 1)

∴ above inequality will be x2 ≤ -(x - 1)

⇒ x2 + x - 1 ≤ 0

which gives \(\frac{-1-\sqrt5}{2} ≤ x≤\frac{-1+\sqrt5}{2}\)

Taking the intersection of  x ≤ 1 and \(\frac{-1-\sqrt5}{2} ≤ x≤\frac{-1+\sqrt5}{2}\)

We get \(\frac{-1-\sqrt5}{2} ≤ x≤\frac{-1+\sqrt5}{2}\)