(Each question 1 marks)

Question 11 Mark

Write the solution set of the inequation $\Big|\frac{1}{\text{x}}-1\Big|<4$

Answer

$\Big|\frac{1}{\text{x}}-2\Big|>4$ $\Leftrightarrow\frac{1}{\text{x}}<2-4$or $\frac{1}{\text{x}}>2+4$ $\Leftrightarrow\frac{-1}{\text{x}}>2$ or $\frac{1}{\text{x}}>6$ $\Leftrightarrow\frac{-1}{\text{x}}>2$ or $\frac{1}{\text{x}}>6$ $\Leftrightarrow\frac{-1}{2}>\text{x}$ or $\frac{1}{6}>\text{x}$ $\Leftrightarrow\text{x}\in\Big(-\infty,\frac{-1}{2}\Big)$or $\text{x}\in\Big(\frac{1}{6},\infty\Big)$ $\Leftrightarrow\text{x}\in\Big(-\infty,\frac{-1}{2}\Big)\cup\Big(\frac{1}{6},\infty\Big)$ Hence, the solution set of the given system of inequation is $\Big(-\infty,\frac{-1}{2}\Big)\cup\Big(\frac{1}{6},\infty\Big)$

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Question 21 Mark

Write the solution set of the inequation $|\text{x}-1|\geq|\text{x}-3|$

Answer

$|\text{x}-1|\geq|\text{x}-3|$ $\Rightarrow|\text{x}-1|-|\text{x}-3|\geq0$ By equating the expression within the modulus to zero, we get x = 1, 3. These point divide real line in three parts viz. $(-\infty,1),[1,3)$ and $[3\infty).$ Case 1: When $-\infty<\text{x}<1$ |x - 1| = -(x - 1) and |x - 3| = -(x - 3) $\therefore|\text{x}-1|-|\text{x}-3|\geq0$ $\Rightarrow-2\geq0$ which is not true. So, the given inequation has no solution for $\text{x}\in(-\infty,1)$ Case 2: When $1\geq\text{x}<3$ |x - 1| = -(x - 1) and |x - 3| = -(x - 3) $\therefore|\text{x}-1|-|\text{x}-3|\geq0$ ⇒ (x - 1) + (x - 3) $\geq$ 0 $\Rightarrow2\text{x}-4\geq0$ $\Rightarrow\text{x}\geq2$ But $1\leq\text{x}\leq3$ Therefore in case the solution set of the given inequation is [2, 3) Case 3: When $3\leq\text{x}<\infty$ |x - 1| = -(x - 1) and |x - 3| = -(x - 3) $\therefore|\text{x}-1|-|\text{x}-3|\geq0$ ⇒ (x - 1) + (x - 3) $\geq$ 0 $\Rightarrow2\geq0$ The solution set of the given inequation is $[3,\infty)$ Combining 1 and 3 we obtain that the solution set of the given inequation is $[2,3)\cup[3,\infty)=[2,\infty)$

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Question 31 Mark

Write the solution set of the inequation $\text{x}+\frac{1}{\text{x}}\geq2$

Answer

We have
$\text{x}+\frac{1}{\text{x}}\geq2$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}\geq2$
This is only possible if $\text{x}\in(0,\infty)$
$\therefore\text{x}\in[0,\infty)$

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Question 41 Mark

Write the solution set of the inequation |2 - x| = x - 2.

Answer

|2 - x| = x - 2 ...(i)
$2-\text{x}\geq0$ for $\text{x}\leq2$
⇒ |2 - x| = 2 - x
2 - x < 0 for x > 2
⇒ |2 - x|= - (2 - x)
for x > 2 from (i)
|2 - x| = x - 2
⇒ x = 2
Which is not true as x < 2 and x = 2 cannot happen at same time.
For $\text{x}\geq2$ from (i)
|2 - x| = -(2 - x)
From (i)
$\therefore$ -(2 - x) = x - 2x
⇒ x - 2x = x - 2x which is not true.
Hence the solution on set of the given inequation is $(2,\infty)$

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Question 51 Mark

Write the solution set of x satisfying the inequation $|\text{x}-1|\leq3$ and $|\text{x}-1|\leq1.$

Answer

$|\text{x}-1|\leq3$ $\Leftrightarrow1-3\leq\text{x}\leq1+3$ $\Leftrightarrow-2\leq\text{x}\leq4$ $\Leftrightarrow\text{x}\in[-2,4]$ $|\text{x}-1|\geq1$ $\Leftrightarrow\text{x}\leq1-1$ or $\text{x}\geq1+1$ $\Leftrightarrow\text{x}\leq0$ or $\text{x}\geq2$ $\Leftrightarrow\text{x}\in(-\infty,0)$ or $\text{x}\in[2,\infty)$ $\Leftrightarrow\text{x}\in(-\infty,0]\cup[2,\infty)$ Hence, the solution set of the given system of inequation is $[-2,\infty]\cup[2,4]$

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Question 61 Mark

The longest side of a triangle is three times the shortest side and third side is 2cm shorter than the longest side if the perimeter of the triangles at least 61cm, find the minimum length of the shortest-side.

Answer

$\Big|\text{x}+\frac{1}{\text{x}}\Big|>2$
$\Rightarrow\text{x}+\frac{1}{\text{x}}<-2\ \text{or }\text{x}+\frac{1}{\text{x}}>2$
Case 1: When $\text{x}+\frac{1}{\text{x}}<-2$
$\text{x}+\frac{1}{\text{x}}<-2$
$\Rightarrow\frac{\text{x}^2+2\text{x}+1}{\text{x}}<0$
$\Rightarrow\frac{(\text{x}+1)^2}{\text{x}}<0$
(x + 1) positive for x > -1, (x + 1) Negative for x < -1, (x + 1) = 0 for x = -1 x positive for x > 0, x > 0, x Negative for x < 0, x zero for x = 0
$\Rightarrow\frac{(\text{x}+1)^2}{\text{x}}<0$ for x < -1 or $\frac{(\text{x}+1)^2}{\text{x}}<0$ for - 1 x < 0
case 2: When $\text{x}+\frac{1}{\text{x}}>2$
$\text{x}+\frac{1}{\text{x}}>2$
$\Rightarrow\frac{(\text{x}-1)^2}{\text{x}}>0$
(x - 1) Positive for x > 1, (x - 1) negative for x < 1, (x - 1) = 0 for x = 1 x postive for x > 0, x Negative for x < 0, x zero for x = 0
$\Rightarrow\frac{(\text{x}-1)^2}{\text{x}}>0$ For 0 < x < 1 or $\frac{(\text{x}-1)^2}{\text{x}}>0$ For x > 1
Combing 1 and 2, we obtain that the solution on set of the given inequation is
x < -1 or -1 < x < 0 or 0 < x < 1 or x > 1
$\frac{\Rightarrow\text{x}\in\text{R}}{(-1,0,1)}$

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Question 71 Mark

Write the number of intregals solution of $\frac{\text{x}+2}{\text{x}^2+1}>\frac{1}{2}$

Answer

$\frac{\text{x}+2}{\text{x}^2+1}>\frac{1}{2}$ $\Rightarrow 2x + 4 > x^2 + 1 \Rightarrow 4 > x^2 - 2x + 1 \Rightarrow 4 > (x - 1)^2 \Rightarrow 2 > |x - 1| \Rightarrow 1 - 2 < x < 1 + 2 \Rightarrow -1 < x < 3$ $\Rightarrow\text{x}\in(-1,3)$ x can take value 0 , 1 , 2 as the integrals solution for $\frac{\text{x}+2}{\text{x}^2+1}>\frac{1}{2}$ Hence, the number of intergrals solution of $\frac{\text{x}+2}{\text{x}^2+1}>\frac{1}{2}$ is 3

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Question 81 Mark

A company manufactures cassettes and its cost and revenue functions for a week are $\text{C}=300+\frac{3}{2}\text{x}$ R = 2x respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold for the company to realize a profit?

Answer

5x + 2 < 3x + 8
⇒ 2x < 6
⇒ x < 3
$\therefore$ Solution set for inequation on (i) is $(-\infty,3)$
$\frac{\text{x}+2}{\text{x}-1}<4$
$\Rightarrow\frac{\text{x}+2}{\text{X}-1}-4<0$
$\Rightarrow\frac{\text{x}+2-4\text{x}+4}{\text{x}-1}<0$
$\Rightarrow\frac{-3\text{x}+6}{\text{x}-1}<0$
$\Rightarrow\frac{\text{x}-2}{\text{x}-1}>0$
$\therefore$ Solution set for inequation (ii) is $(-\infty,1)\cup(2,\infty)$

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Question 91 Mark

Write the solution set of $x$ satisfying the inequation $\left(x^2-2 x+1\right)(x-4)>0$.

Answer

$\left(x^2-2 x+1\right)(x-4)<0$
$\Rightarrow(x-1)^2(x-4)<0$
$(x-1)^2$ is positive for all $x \neq 1$
$(x-1)^2=0 \text { for } x=0$
$(x-4)$ is positive for all $x>4$
$(x-4)$ is negative for all $x<4$
$(x-4)=0$ for $x=0$
$(x-1)^2(x-4)<0$ for all $x<4$.
Hence the solution set of the given inequation is $(-\infty, 4)$

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Question 101 Mark

Write the solution set of the inequation $\frac{\text{x}^2}{\text{x}-2}>0$

Answer

We have
$\frac{\text{x}^2}{\text{x}-2}>0$
$\Rightarrow\text{x}-2\geq0$ $\Big[\because\frac{\text{a}}{\text{b}}>0$ and $\text{a}>0\Rightarrow\text{b}\geq0\Big]$
$\Rightarrow\text{x}\geq2 $
$\Rightarrow\text{x}\in[2,\infty)$
Hence, the solution set of the given inequation is $[2,\infty)$

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