Linear Inequations MCQ Quiz - Objective Question with Answer for Linear Inequations - Download Free PDF

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,

The inequality is: \( k < (√{2} + 1)^3 < k + 2 \), where k is a natural number.

We need to compute \((√{2} + 1)^3\)

\( (√{2} + 1)^3 = 2√{2} + 6 + 3√{2} + 1 = 5√{2} + 7 \)

Approximating √ 2≈ 1.414 , we find:

\( 5√{2} + 7 ≈ 5 \times 1.414 + 7 = 7.07 + 7 = 14.07 \)

The inequality becomes:

\( k < 14.07 < k + 2 \)

This implies that k > 12.07 \), so the smallest integer value of k  is:

\( k = 13 \)

∴ The value of k  is 13.

Hence, the correct answer is Option 2. 

Calculation: 

x + y = 5λ

Cases :

\( \begin{array}{|c|c|c|} \hline \mathbf{x} & \mathbf{y} & \text{Number of ways} \\ \hline 5\lambda & 5\lambda & 20 \\ 5\lambda+1 & 5\lambda+4 & 25 \\ 5\lambda+2 & 5\lambda+3 & 25 \\ 5\lambda+3 & 5\lambda+2 & 25 \\ 5\lambda+4 & 5\lambda+1 & 25 \\ \hline \end{array} \)

Total = 120

Hence, the correct answer is 120. 

Calculation: 

(p ∧ ~ q) → (p → ~ q) 

≡ (~ (p ∧ ~q)) ∨ (~ p ∨ ~q)

≡ (~ (p ∨ ~q)) ∨ (~ p ∨ ~q)

≡ ~ p ∨ t ≡ t

∴ The statement (p ∧ (~q)) ⇒ (p ⇒ (~q)) is tautology 

Hence, the correct answer is Option 2. 

Calculation

x² - 4x - 21 > 0:

(x-7)(x+3) > 0.

Solutions: x < -3 or x > 7.

x² - 9x + 8 < 0:

(x-1)(x-8) < 0.

Solutions: 1 < x < 8.

Combine solutions: x < -3 or x > 7, and 1 < x < 8.

Intersection is 7 < x < 8.

Integral values between 7 and 8: None.

Therefore, no integral values satisfy both inequalities.

Hence option 4 is correct

CALCULATIONS:

Given \(\frac{1}{5}\left( {\frac{{3x}}{5} + 4} \right) \ge \frac{1}{3}(x - 6)\)

\(\begin{array}{l} \Rightarrow 3\left( {\frac{{3x}}{5} + 4} \right) \ge 5(x - 6)\\ \Rightarrow \left( {\frac{{9x}}{5} + 12} \right) \ge 5x - 30\\ \Rightarrow \left( {30 + 12} \right) \ge - \frac{{9x}}{5} + 5x\\ \Rightarrow 42 \ge \frac{{ - 9x + 25x}}{5}\\ \Rightarrow 42 \ge \frac{{16x}}{5}\\ \Rightarrow \frac{{42 \times 5}}{{16}} \ge x\\ \Rightarrow x \le \frac{{105}}{8} \end{array}\)

Therefore, option (1) is the correct answer.

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:

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: 0 < − x/2 < 3

Multiply by 2 in above inequality (Here 2 is a positive number so the direction of the inequality does not change)

⇒ 0 < −x < 6

⇒ −6 < x < 0

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:

\(\frac{2x+~4}{x-1}\ge 5\)

\(\Rightarrow \frac{2x+~4}{x-1}-5\ge 0\)

\(\Rightarrow ~\frac{2x+4-5\left( x-1 \right)}{x-1}\ge 0\)

\(\Rightarrow ~\frac{2x+4-5x+5}{x-1}\ge 0\)

\(\Rightarrow \frac{9-3x}{x-1}\ge 0\)

\(\Rightarrow ~\frac{-3~\left( x-3 \right)}{x-1}\ge 0\)

Multiplying each side of an inequality by a negative number (-1/3) reverses the direction of the inequality symbol.

\(\Rightarrow ~\frac{~\left( x-3 \right)}{x-1}\le 0\), Here x – 1 ≠ 0 ⇒ x ≠ 1

Concept:

Solution Set: The set of all possible values of x.

When (a > 0 and b < 0) or (a < 0 and b > 0) then \(\rm \frac a b <0\)

When (a > 0 and b > 0) or (a < 0 and b < 0) then \(\rm \frac a b >0\)

Calculation:

Here, \( {1 \over x - 2} < 0\)

So, x - 2 < 0

⇒ x < 2

∴ x ∈ (-∞ , 2)

Calculation:

Here, constraints: x ≤ 40, y ≤  20 and x, y ≥ 0 

We get minimum value of P, only when x = 0 and y = 0

So, P = 6(0) + 16(0) = 0

Hence, option (3) is correct.

CALCULATION:

Given \(\frac{x}{4} > \frac{x}{2} + 1\)

\( \Rightarrow \frac{x}{4} - \frac{x}{2} > 1\)

\(\Rightarrow \frac{{x - 2x}}{4} > 1\)

\( \Rightarrow \frac{{ - x}}{4} > 1\)

\( \Rightarrow \frac{{ - x}}{4} > 1\)

\( \Rightarrow - x > 4\)

\( \Rightarrow x < - 4\)

Therefore option (3) is the correct answer.

Concept:

If f (x) = |x|, then \(f\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} {x\;,\;x \ge 0}\\ { - x,\;x < 0} \end{array}} \right.\)

Calculation:

Given: |x² - 3x + 2| > x² - 3x + 2

Case-1: If x ≥ 0

⇒ x² - 3x + 2 > x² - 3x + 2

∴ No real value of x satisfies the above equation.

Case-2:- If x < 0

⇒ - (x² - 3x + 2) > x² - 3x + 2

⇒ x² - 3x + 2 < 0

⇒ (x - 1) (x - 2) < 0

Concept:

First, graph the "equals" line, then shade in the correct area.

There are three steps:

• Rearrange the equation so "y" is on the left and everything else on the right.
• Plot the "y = " line (make it a solid line for y ≤ or y ≥, and a dashed line for y < or y >)
• Shade above the line for a "greater than" (y > or y ≥) or below the line for a "less than" (y < or y ≤).

Calculation:

Given: The constraints –x + y ≤ 1, −x + 3y ≤ 9 and x, y ≥ 0

–x + y ≤ 1

⇒ y ≤ 1 + x

Hence shade below the line.

−x + 3y ≤ 9

3y ≤ 9 + x

⇒ y ≤ 3 + x/3

Hence shade below the line.

We can see through above graph region is unbounded feasible space

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:

\(\frac{4}{x}<\frac{1}{3}\)

\(\Rightarrow \frac{4}{x}-~\frac{1}{3}<0\)

\(\Rightarrow \frac{12-x}{3x}<0\)

\(\Rightarrow \frac{-~\left( x-12 \right)}{3x}<0\)

Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.

\(\Rightarrow \frac{\left( x-12 \right)}{3x}>0\), Here x ≠ 0

∴ x < 0 OR x > 12