Inequalities in one Variable MCQ Quiz in मल्याळम - Objective Question with Answer for Inequalities in one Variable - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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 $(4,\mathrm{\infty })$.

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 ∈  $[\frac{1}{3},\mathrm{\infty })$ .

Given:

The inequation: (x + 2)/(x + 3) > 1

Formula used:

For a rational inequality of the form (a/b) > 1, we analyze critical points and test values between intervals.

Calculation:

(x + 2)/(x + 3) > 1

⇒ (x + 2) - (x + 3) > 0

⇒ x + 2 - x - 3 > 0

⇒ -1 > 0

This is not possible.

Since the inequality is never satisfied, there are no positive integral solutions.

∴ The correct answer is option (4).

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:

$-3<\frac{1}{2}+\frac{-3x}{2}\le 6$

⇒ $-6<1-3x\le 12$

⇒ $-7<-3x\le 11$

⇒ $\frac{-7}{-3}>x\ge \frac{11}{-3}$

⇒  $\frac{7}{3}>x\ge -\frac{11}{3}$

⇒ $-\frac{11}{3}\le x<\frac{7}{3}$

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

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 ∈  $[\frac{1}{3},\mathrm{\infty })$ .

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

⇒$1\le \mathrm{x}\le {\displaystyle \frac{7}{3}}$ 

Hence, if |3x - 5| ≤ 2 then then $1\le \mathrm{x}\le {\displaystyle \frac{7}{3}}$

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 ∈  $[\frac{1}{3},\mathrm{\infty })$ .

Explanation:

Given system of linear inequality is 

$\{\begin{array}{c}5+x>3x-7\\ 11-5x\le 1\end{array}$

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

⇒$1\le \mathrm{x}\le {\displaystyle \frac{7}{3}}$ 

Hence, if |3x - 5| ≤ 2 then then $1\le \mathrm{x}\le {\displaystyle \frac{7}{3}}$