Inequalities in one Variable MCQ Quiz in मराठी - Objective Question with Answer for Inequalities in one Variable - मोफत PDF डाउनलोड करा

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 ∈  \(\left[\frac{1}{3},\infty\right)\) .

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 : 

⇒ \( \log_{x+\frac{7}{2}} \left(\frac{x-7}{2x-3}\right)^2 \geq 0 \)

Feasible region: \( x + \frac{7}{2} > 0 \implies x > -\frac{7}{2} \)

Also, \( x + \frac{7}{2} \neq 1 \implies x \neq -\frac{5}{2} \)

And, \( \frac{x-7}{2x-3} \neq 0 \quad \text{and} \quad 2x - 3 \neq 0 \)

∴  \( x \neq 7 \quad \text{and} \quad x \neq \frac{3}{2} \)

Taking intersection: \( x \in \left(-\frac{7}{2}, \infty \right) - \left\{ -\frac{5}{2}, \frac{3}{2}, 7 \right\} \)

Now, \( \log_a b \geq 0 \) if \( a > 1 \) and \( b \geq 1 \)

Or, \( a \in (0,1) \) and \( b \in (0,1) \)

Case I: \( x + \frac{7}{2} > 1 \quad \text{and} \quad \left(\frac{x-7}{2x-3}\right)^2 \geq 1 \)

⇒ \( x > -\frac{5}{2} \)

⇒ \( (2x-3)^2 - (x-7)^2 \leq 0 \)

⇒ \( (2x-3 + x - 7)(2x - 3 - x + 7) \leq 0 \)

⇒ \( (3x - 10)(x + 4) \leq 0 \)

⇒ \( x \in \left[-4, \frac{10}{3}\right] \)

⇒ Intersection with previous feasible region:

 \( x \in \left[-\frac{5}{2}, \frac{10}{3}\right] \)

Case II: \( x + \frac{7}{2} \in (0,1) \quad \text{and} \quad \left(\frac{x-7}{2x-3}\right)^2 \in (0,1) \)

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

⇒ \( (x-7)^2 < (2x-3)^2 \)

⇒ \( x \in (-\infty, -4) \cup \left(\frac{10}{3}, \infty\right) \)

⇒ No common values of \( x \) with feasible region.

Final feasible solution:

⇒ \( x \in \left[-\frac{5}{2}, \frac{10}{3}\right] - \left\{\frac{3}{2}, 7\right\} \)

⇒ Integral values of \( x \) are \( \{-2, -1, 0, 1, 2, 3\} \)

No. of integral values = 6

Hence, the correct answer is Option 1. 

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.