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

x = 2, 4, 6, 8, ... , 20 satisfy

(x − 2)(x − 4)(x − 6) ... (x − 20) = 0.

Also, for (x − 2)(x − 4)(x − 6) ... (x − 20) < 0, odd number of terms should be negative.

For x > 20 and x < 2 all terms will be positive.

For x = 19, x − 20 < 0, while all other terms are positive.

Thus, x = 19 is one of the solutions.

For x = 17, x − 20 < 0 and x − 18 < 0, while all other terms are positive.

But then (x − 2)(x − 4) ... (x − 20) > 0, hence x = 17 is not the solution.

For x = 15, three terms are negative.

Hence, x = 15 is one of the solutions.

Similarly, x = 11, 7, 3 are the other solutions.

∴ The required number of solutions = 10 + 5 = 15.

To begin, the discriminant should be a perfect square. Let the discriminant be represented by \( D \).

From the quadratic formula:

\( \frac{-11 \pm \sqrt{D}}{2} \)

we can see that the numerator must be an even number for the roots to be integers.

This means that the discriminant should be a perfect square and specifically the square of an odd number. (This is the condition for odd + odd = even in the numerator.)

The discriminant equation is:

\( D = 121 - 4|a| = 121 - 4|a| \)

Since \( 4|a| \) cannot be negative, we conclude that \( D \) can take values such as 121, 81, 49, 25, 9, and 1.

The value of \( |a| \) can be 0, 10, 18, 24, 28, or 30.

Hence, \( a \) can take the values 0, \( \pm 10 \), \( \pm 18 \), \( \pm 24 \), \( \pm 28 \), and \( \pm 30 \).

Calculation:

Taking

⇒ (x + 1)²  > (5x - 1)

⇒ x² – 3x + 2 > 0

⇒ (x – 1 )( x – 2 ) > 0

X < 1 or x > 2        ……(i)

Taking

⇒ (x + 1)² < (7x – 3)

⇒ x² – 5x + 4 > 0

⇒ (x – 1 )( x – 4 ) > 0

⇒ 1 < x < 4         ……(ii)

Combining (1) and (2) we get 2 < x < 4

Hence, x will only take one integer i.e. 3

Concept:

Calculation :

Case 1: x2

⇒ 

Simplifying

⇒ 

case 2: 

⇒ 

Simplifying

⇒ 

this is contradiction

case 3:  x   1 

⇒ 

Simplifying

⇒ 

From Cases 1 and 3, we get

⇒ 

⇒

Hence option 1 is correct

We know that,

AM (b + c, c + a, a + b) ≥ HM (b + c, c + a, a + b)

\(\Rightarrow \frac{b+c+c+a+a+b}{3} \geq \frac{3}{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}\)

\(\Rightarrow \frac{2}{9}(a+b+c) \geq \frac{1}{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}\)

\(\Rightarrow \frac{9}{2(a+b+c)} \leq \frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\)

\(\Rightarrow \frac{9}{2} \leq \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b} \)

\(\Rightarrow \frac{9}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3 \)

\(\Rightarrow \frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3 \)

\(\Rightarrow \frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\) .........(1)

Now, since a, b and c are the lengths of a triangle b + c > a

2b + 2c > a + b + c

\(\frac{1}{2(b+c)}<\frac{1}{a+b+c} \)

\(\frac{a}{b+c}<\frac{2 a}{a+b+c}\)

Similarly, we have, \(\frac{b}{c+a}<\frac{2 b}{a+b+c}\)

\(\frac{c}{a+b}<\frac{2 c}{a+b+c}\)

Adding these inequalities, we get 

\(\frac{\mathrm{a}}{\mathrm{~b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}<2\) .........(2)

Therefore, From (A) and (B), we haved

\(\frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{b+c}+\frac{c}{a+b}<2\)

Given:

Initial ratio = 5 ∶ 6

Final ratio should be less than 17 ∶ 22

Calculation:

Let the least whole number that is needed to be subtracted be a.

According to the question,

(5 - a)/(6 - a) < 17/22

⇒ 5 × 22 - 22a < 17 × 6 - 17a 

⇒ 110 - 22a < 102 - 17a 

⇒ 110 - 102 < - 17a + 22a 

⇒ 8 < 5a 

⇒ 8/5 = 1.6 < a 

∴ The least whole number must be 2.

Given:

a² - b² = 88

a - b = 4

Formula used:

a² - b² = (a - b)(a + b)

Calculation:

a - b = 4      ----(1)

(a - b)(a + b) = 88

⇒ 4 × (a + b) = 88

⇒ a + b = 88/4

⇒ a + b = 22      ----(2)

Adding equation (1) and equation (2), we get

⇒ a - b + a + b = 4 + 22

2a = 26

⇒ a = 13

Put the value of a in equation (2), we get

13 + b = 22

⇒ b = 9

value of ab = 13 × 9

⇒ ab = 117

∴ The value of ab is 117.

Given:

 x = 8 – 2√(15)

Calculation:

 x = (√5)² + (√3)² – 2√(15)

⇒ x = (√5 - √3)²

⇒ √x = √5 - √3

And 1/√x = (√5 + √3)/2

According to question,

\({\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2} = {\left[ {\sqrt 5 - \sqrt 3 + \left( {\frac{{\sqrt 5 + \sqrt 3 \;}}{2}} \right)} \right]^2}\)

\(\Rightarrow {\left[ {\frac{{3\sqrt 5 - \sqrt 3 }}{2}} \right]^2}\)

\(\Rightarrow \left[ {\frac{{45 + 3 - 6\sqrt {15} }}{4}} \right]\)

\(\Rightarrow \frac{{48 - 6\sqrt {15} }}{4}\)

Calculation:

Statement:1    \(\frac{a^2 - b^2 }{a^2 + b^2 } > \frac{a - b}{a + b} \) where a > b > 0

Let a = 2 and b = 1

⇒ \(\frac{2^2 - 1^2 }{2^2 + 1^2 } > \frac{2 - 1}{2 + 1} \)

⇒ \(\frac{3 }{5 } > \frac{1}{3} \) which is correct.

Hence, statement 1 is correct.

Statemnt:2   \(\frac{a^3 + b^3 }{a^2 + b^2 } > \frac{a^2 + b^2}{a + b} \) only when a > b > 0

Let a = 2 and b = 1

⇒ \(\frac{2^3 + 1^3 }{2^2 + 1^2 } > \frac{2^2 + 1^2}{2 + 1} \)

⇒ \(\frac{9 }{5 } > \frac{5}{3} \) Which is also correct.

This inequality also holds for a = 1 and b = 2 i.e.

Statement 2 is correct for b > a > 0 as well.

But according to statement 2, inequality will be correct only if a > b > 0

So, statement 2 is incorrect.

Hence, statement (1) is correct but (2) statement (2) is incorrect.

Mistake PointsPlease note the language of  the second statement

"only when a > b > 0"

We have proved thet, statement is correct for 

a = 2 & b = 1 (i.e. a > b > 0) and b = 2 & a = 1 (i.e. b > a > 0)

Hence, due to only word, statement 2 will become incorrect.

Given:

x - 1/x = 10

Formula:

(a - b)³ = a³ - b³ - 3ab(a - b)

Calculation:

(x - 1/x)³ = x³ - 1/x³ - 3(x)(1/x)(x - 1/x)

⇒ 10³ = x3 - 1/x3 - 3 × 10

⇒  x3 - 1/x3 = 1000 + 30

⇒  x3 - 1/x3 = 1030

∴ The value of x3 - 1/x³ is 1030.

If x - 1/x = n,

x3 - 1/x3 = n³ + 3n

Here, n = 10

x³ - 1/x³ = 10³ + 3 × 10

⇒ x3 - 1/x3 = 1000 + 30

⇒  x3 - 1/x3 = 1030

∴ The value of x3 - 1/x3 is 1030.

a + b = 5      ----(1)

2a – b = 4      ----(2)

Adding Eq. (1) and (2), we get

⇒ a + b + 2a – b = 5 + 4

⇒ 3a = 9

⇒ a = 3

Putting a = 3 in eq. (1) we get b = 2

Given:

Equation₁ = ax + 9y = 1

Equation₂ = 9y - x - 1 = 0 

Concept used:

If linear equations are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. Here, the equations have an infinite number of solutions, if

a₁/a₂ = b₁/b₂ = c₁/c₂

Calculation:

We have equations,

ax + 9y = 1 

⇒ ax + 9y - 1 = 0

and, 9y - x - 1 = 0

⇒ x - 9y + 1 = 0

Here, a₁ = a, b₁ = 9, c₁ = -1

and, a₂ = 1, b₂ = -9, c₂ = 1

As we know that 

a1/a2 = b1/b2 = c1/c₂

⇒ a/1 = 9/-9 = -1/1

⇒ a = -1 = -1

⇒ a = -1

∴ The value of a is -1.

Alternate Method

ax + 9y = 1    ----(i)

9y - x - 1 = 0    ----(ii)

From equation (ii) we get,

-x + 9y = 1    ----(iii)

On comparing eq(iii) and eq (i)

We get, a = -1

Calculation: 

Given: 9a - a2 \(\leq\) 17a + 15

On rearranging

-a² + 9a \(\leq\) 17a + 15

Shifting the sign

a² - 9a \(≥\) - 17a - 15

a2 - 9a + 17a + 15 \(≥\) 0

a² + 8a + 15 \(≥\) 0

a² + 5a + 3a + 15 \(≥\) 0

a(a+ 5) + 3(a + 5)  \(≥\) 0

(a + 3)(a + 5)  \(≥\) 0

So, -3 and -5 are the roots of the equation.

Now look at the diagram given below,

We see that all the numbers less than -5 and all the numbers greater than -3 will give us positive result. While numbers between -5 and -3 will give us negative results

So, all the above value holds for the above equation.

Calculation-

\(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 ⇒\(\rm \begin {vmatrix}\frac{x-4}{2} \end{vmatrix}\) < 1 ⇒|x - 4| < 2

⇒ -2 < x - 4 < 2 ⇒ 2 < x < 6 

As x ≠ 4 hence x ϵ (2, 4) ∪ (4, 6) 

Since, (4, 6) is lying in the interval for all the acceptable values of x ϵ (2, 4) ∪ (4, 6) so this will be the best option.  

∴ The solution of the equality \(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 is x ϵ (4, 6).

2 (3x + 5) > 4x - 5 < 3x + 2

2 (3x + 5) > 4x - 5

⇒ 6x + 10 > 4x – 5

⇒ 2x > -15

⇒ x > -15/2

4x - 5 < 3x + 2

⇒ x < 7

So, -15/2 < x < 7