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

Last updated on Apr 17, 2025

पाईये Inequalities in one Variable उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Inequalities in one Variable एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Inequalities in one Variable MCQ Objective Questions

Top Inequalities in one Variable MCQ Objective Questions

Inequalities in one Variable Question 1:

Find the solution of the inequality:

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

  1. \((4,\infty )\)
  2. \([4,\infty )\)
  3. (2, 4)
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : \((4,\infty )\)

Inequalities in one Variable Question 1 Detailed Solution

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 \((4,\infty )\).

Inequalities in one Variable Question 2:

The solution set of the inequality 17 - (2x + 4) ≤ 9x - 4(2x - 3) is

  1. \(\left[\frac{-1}{3},\infty\right)\)
  2. \(\left(-\infty,\frac{1}{3}\right]\)
  3. \(\left[\frac{1}{3},\infty\right)\)
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : \(\left[\frac{1}{3},\infty\right)\)

Inequalities in one Variable Question 2 Detailed Solution

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

Inequalities in one Variable Question 3:

The number of positive integral solutions of the inequation \(\frac{x + 2}{x + 3} > 1 \)

  1. Infinite 
  2. 4
  3. 3
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Inequalities in one Variable Question 3 Detailed Solution

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).

Inequalities in one Variable Question 4:

The number of integral solutions x of \(\log_{x + \frac{7}{2}} \left( \frac{x - 7}{2x - 3} \right)^2 \geq 0 \quad \text{is} \)

  1. 6
  2. 8
  3. 5
  4. 7

Answer (Detailed Solution Below)

Option 1 : 6

Inequalities in one Variable Question 4 Detailed Solution

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. 

Inequalities in one Variable Question 5:

The solution set of the inequality 37 − (3x + 5) ≥ 9x − 8(x − 3) is

  1. (−∞, 2)
  2. (−∞, −2) 
  3. (−∞, 2]
  4. (−∞, −2] 

Answer (Detailed Solution Below)

Option 3 : (−∞, 2]

Inequalities in one Variable Question 5 Detailed Solution

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.

Inequalities in one Variable Question 6:

If x satisfies the inequality \(-3<\frac{1}{2}+\frac{-3 x}{2} \leq 6\), then x lies in the interval

  1. \(\left[\frac{-11}{3}, \frac{7}{3}\right)\)
  2. \(\left(\frac{-11}{3}, \frac{7}{3}\right]\)
  3. \(\left(\frac{7}{3}, \frac{11}{3}\right]\)
  4. \(\left[\frac{-10}{3}, \frac{7}{3}\right)\)
  5. \(\left[\frac{7}{3}, \frac{10}{3}\right)\)

Answer (Detailed Solution Below)

Option 1 : \(\left[\frac{-11}{3}, \frac{7}{3}\right)\)

Inequalities in one Variable Question 6 Detailed Solution

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

Inequalities in one Variable Question 7:

The solution set of the inequality 17 - (2x + 4) ≤ 9x - 4(2x - 3) is

  1. \(\left[\frac{-1}{3},\infty\right)\)
  2. \(\left(-\infty,\frac{1}{3}\right]\)
  3. \(\left[\frac{1}{3},\infty\right)\)
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : \(\left[\frac{1}{3},\infty\right)\)

Inequalities in one Variable Question 7 Detailed Solution

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

Inequalities in one Variable Question 8:

If |3x - 5| ≤ 2 then

  1. \(\rm -1 \le x \le \dfrac{7}{3}\)
  2. \(\rm 1 \le x \le \dfrac{7}{3}\)
  3. \(\rm 1 \le x \le \dfrac{9}{3}\)
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : \(\rm 1 \le x \le \dfrac{7}{3}\)

Inequalities in one Variable Question 8 Detailed Solution

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 \(\rm 1 \le x \le \dfrac{7}{3}\)

Inequalities in one Variable Question 9:

The solution set of the inequality 17 - (2x + 4) ≤ 9x - 4(2x - 3) is

  1. \(\left[\frac{-1}{3},\infty\right)\)
  2. \(\left(-\infty,\frac{1}{3}\right]\)
  3. \(\left[\frac{1}{3},\infty\right)\)
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : \(\left[\frac{1}{3},\infty\right)\)

Inequalities in one Variable Question 9 Detailed Solution

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

Inequalities in one Variable Question 10:

On the number line, the solution of system of inequalities \(\left\{ \begin{matrix} 5+x > 3x - 7 \\\ 11 - 5x \le 1 \end{matrix} \right.\) is represented

  1. F1 Vinanti Teaching 09.01.23 D2
  2. F1 Vinanti Teaching 09.01.23 D3
  3. F1 Vinanti Teaching 09.01.23 D4
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : F1 Vinanti Teaching 09.01.23 D3

Inequalities in one Variable Question 10 Detailed Solution

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.

F1 Vinanti Teaching 09.01.23 D3

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