Question 11 Mark
If x < 5, then.
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$-\text{x} < – 5 $
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$-\text{x}\leq-5$
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$-\text{x} > – 5 $
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$-\text{x}\leq-5$
Answer
View full question & answer→- $-\text{x} > – 5 $
Solution:
If x > 5 then - x > - 5.
Questions
M.C.Q (1 Marks)
Take a timed test
11 questions · self-marked practice — reveal the answer and mark yourself.
Question 21 Mark
If $|\text{x}+2|\leq9,$ then:
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$\text{x}\in(-7,11)$
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$\text{x}\in[-11, 7]$
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$\text{x}\in[-\infty,-7)\cup(11,\infty) $
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$\text{x}\in(-\infty,-7)\cup[11,\infty) $
Answer
View full question & answer→- $\text{x}\in[-11, 7]$
Solution:
Given that $|\text{x}+2|\leq9$
$\Rightarrow-9\leq\text{x}+2\leq9$
$\Rightarrow-9-2\leq\text{x}\leq9-2[|\text{x}\leq\text{a}|]$
$\Rightarrow-11\leq\text{x}\leq7$
$\Rightarrow\text{x}\in[-11, 7]$
Question 31 Mark
If – 3x + 17 < – 13, then:
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$\text{x}\in(10, \infty)$
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$\text{x}\in[10, \infty)$
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$\text{x}\in(-\infty\text{j},10]$
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$\text{x}\in[-10, 10)$
Answer
View full question & answer→- $\text{x}\in(10, \infty)$
Solution:
Given that - 3x + 17 < - 13
⇒ - 3x < - 17 - 13
⇒ -3x < - 30
⇒ 3x > 30
⇒ x > 10
$\Rightarrow\text{x}\in(10, \infty)$
Question 41 Mark
Given that x, y and b are real numbers and x < y, b < 0, then:
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$\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
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$\frac{\text{x}}{\text{b}}\leq\frac{\text{y}}{\text{b}}$
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$\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
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$\frac{\text{x}}{\text{b}}\geq\frac{\text{y}}{\text{b}}$
Answer
View full question & answer→- $\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
Solution:
Given that x < y, b < 0
$\Rightarrow\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}},\text{b}<0$
Question 51 Mark
If x is a real number and |x| < 3, then:
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$\text{x}\geq3$
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$-3<\text{x}<3$
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$\text{x}\leq-3$
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$-3\leq\text{x}\leq3$
Answer
View full question & answer→- $-3\leq\text{x}\leq3$
Solution:
Given that |x| < 3
⇒ -3 < x < 3 | x | < a
⇒ -a < x < a.
Question 61 Mark
Solution of a linear inequality in variable x is represented on number line.
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$\text{x}\in[-\infty,5) $
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$\text{x}\in(-\infty,5) $
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$\text{x}\in(5,\infty) $
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$\text{x}\in[5,\infty) $
Answer
View full question & answer→- $\text{x}\in[5,\infty) $
Solution:
The given graph represents all value of x greater than 5 including 5 on the real number line.
So, $\text{x}\in[5,\infty). $
Question 71 Mark
Solution of a linear inequality in variable x is represented on number line.
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$\text{x}\in\big(-\infty,-2\big)$
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$\text{x}\in\big[\infty,-2\big]$
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$\text{x}\in\big(-2,-\infty\big)$
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$\text{x}\in\big(-2,-\infty\big)$
Answer
View full question & answer→- $\text{x}\in\big[\infty,-2\big]$
Solution:
The given graph has all real values of x greater than and equal to -2.
So, $\text{x}\in\big[\infty,-2\big]$
Question 81 Mark
Solution of a linear inequality in variable x is represented on number line.
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$\text{x}\in\big(\frac{9}{2},\infty\big)$
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$\text{x}\in\big[\frac{9}{2},\infty\big]$
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$\text{x}\in\big(-\infty,\frac{9}{2}\big)$
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$\text{x}\in\big[\frac{9}{2},\infty\big)$
Answer
View full question & answer→- $\text{x}\in\big[\frac{9}{2},\infty\big]$
Solution:
The given graph has all real values of x greater than and equal to $\frac{9}{2}.$
So, $\text{x}\in\big[\frac{9}{2}\infty\big]$
Question 91 Mark
Solution of a linear inequality in variable x is represented on number line.
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$\text{x}\in\big(-\infty,\frac{7}{2}\big)$
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$\text{x}\in\big(-\infty,\frac{7}{2}\big]$
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$\text{x}\in\big(\frac{7}{2},-\infty\big)$
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$\text{x}\in\big(\frac{7}{2},\infty\big)$
Answer
View full question & answer→- $\text{x}\in\big(-\infty,\frac{7}{2}\big)$
Solution:
The given graph all real values of x greater than and equal $\frac{7}{2}$ on real number line.
So, $\text{x}\in\big(-\infty,\frac{7}{2}\big)$
Question 101 Mark
If |x - 1| > 5, then:
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$\text{x}\in(-4, 6)$
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$\text{x}\in[-4,6]$
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$\text{x}\in[-\infty,-4)\cup(6,\infty) $
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$\text{x}\in[-\infty,-4)\cup[6,\infty) $
Answer
View full question & answer→- $\text{x}\in[-\infty,-4)\cup(6,\infty) $
Solution:
Given that |x - 1| > 5
⇒ (x - 1) < -5 or (x - 1) > 5
⇒ x < -5 + 1 or x > 5 + 1
⇒ x < -4 or x > 6
$\Rightarrow\text{x}\in[-\infty,-4)\cup(6,\infty) $
Question 111 Mark
x and b are real numbers. If b > 0 and |x| > b, then:
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$\text{x}\in(-\text{b},\infty)$
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$\text{x}\in(\infty,-\text{b})$
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$\text{x}\in(-\text{b},\text{b})$
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$\text{x}\in(-\infty,-\text{b})\cup(\text{b},\infty)$
Answer
View full question & answer→- $\text{x}\in(-\infty,-\text{b})\cup(\text{b,}\infty)$
Solution:
Given that |x| > b, b > 0
⇒ x < -b or x > b
$\Rightarrow\text{x}\in(-\infty,-\text{b})\cup(\text{b,}\infty)$