Inequalities in one Variable MCQ Quiz in தமிழ் - Objective Question with Answer for Inequalities in one Variable - இலவச PDF ஐப் பதிவிறக்கவும்

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:

Let us rewrite the given inequality, we get

\(\frac{x}{4}< \frac{(5x-2)}{3}-\frac{(7x-3)}{5}\)

\(\Rightarrow \frac{x}{4}< \frac{5(5x-2)-3(7x-3)}{15}\)

\(\Rightarrow \frac{x}{4}< \frac{25x-10-21x+9}{15}\)

\(\Rightarrow \frac{x}{4}< \frac{4x-1}{15}\)

⇒ 15x < 4(4x - 1)

⇒ 15x < 16x - 4

⇒ 15x - 16x < 16x - 16x - 4

⇒ -x < -4

⇒ x > 4

Therefore, we can say that the solution set for the given inequality is \((4,\infty )\).

Calculation:

We have, 17 - (2x + 4) ≤ 9x - 4(2x - 3)

⇒ 17 - 2x - 4 ≤ 9x - 8x + 12

⇒ 17 - 4 - 12 ≤ 9x - 8x + 2x

⇒ 1 ≤ 3x

⇒ x ≥ \(\frac{1}{3}\)

∴ The solution set is x ∈  \(\left[\frac{1}{3},\infty\right)\) .

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

Calculation:

We have, 17 - (2x + 4) ≤ 9x - 4(2x - 3)

⇒ 17 - 2x - 4 ≤ 9x - 8x + 12

⇒ 17 - 4 - 12 ≤ 9x - 8x + 2x

⇒ 1 ≤ 3x

⇒ x ≥ \(\frac{1}{3}\)

∴ The solution set is x ∈  \(\left[\frac{1}{3},\infty\right)\) .

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 \(\rm 1 \le x \le \dfrac{7}{3}\)

Calculation:

We have, 17 - (2x + 4) ≤ 9x - 4(2x - 3)

⇒ 17 - 2x - 4 ≤ 9x - 8x + 12

⇒ 17 - 4 - 12 ≤ 9x - 8x + 2x

⇒ 1 ≤ 3x

⇒ x ≥ \(\frac{1}{3}\)

∴ The solution set is x ∈  \(\left[\frac{1}{3},\infty\right)\) .

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.

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 \(\rm 1 \le x \le \dfrac{7}{3}\)

Concept:

Modulus function: It gives the absolute value of a variable (x). It is denoted by |x|

• \(|x|=\left\{\begin{matrix} x, x≥0\\ -x, x<0 \end{matrix}\right. \)
• |x| ≥ 0 for x ∈ R

For any real number a:

• |x| ≤ a  - a ≤ x ≤ a
• |x| ≥ a  - x ≤ -a or x ≥ a

Calculation:

We have, ||x - 2| - \(\frac{k}{2}\)| = 3

⇒ |x - 2| - \(\frac{k}{2}\) = ±3

⇒ |x - 2| = 3 + \(\frac{k}{2}\) or -3 + \(\frac{k}{2}\)

Now, for real values of x,

3 + \(\frac{k}{2}\) ≥ 0 or -3 + \(\frac{k}{2}\) ≥ 0

⇒ k ≥ - 6 or k ≥ 6

∴ k ∈ \([-6, \infty)\) or k ∈ \([6, \infty)\)

Since \([6, \infty)\) is included in the interval \([-6, \infty)\).

∴ k ∈ \([-6, \infty)\).

The correct answer is Option 2.