MATHS M.C.Q (1 Marks)

2. $-3\leq\text{x}\leq3$

Solution:

Given that |x| < 3

⇒ -3 < x < 3 | x | < a

⇒ -a < x < a.

1. $\big[\frac{11}{3},5\big]$

4. $\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). $

1. {0,1, 2, 3, 4, 5}

Solution:

$10\text{x}\leq50$

Dividing by 10 on both sides, $\text{x}\leq\Big(\frac{50}{10}\Big)=\Rightarrow\text{x}\leq5$

Since x is a whole number so x = 0, 1, 2, 3, 4, 5.

3. breadth x ≥ 20cm

Solution:

Let x be the breadth of a rectangle.

So, length = 3x

Given that the minimum perimeter of a rectangle is 160cm.

Thus, $2(3\text{x}+\text{x})\geq160$

$\Rightarrow4\text{x}\geq80$

$\Rightarrow\text{x}\geq20$

3. $\big(-1,\frac{-1}{3}\big)$

3. $\big(-\infty,3\big)$

1. $\text{x}\in(10,\infty)$

Solution:

− 3x + 17 < −13

Subtracting 17 on both sides, we get

⇒ −3x + 17 − 17 < −13 − 17

⇒ −3x < − 30

Dividing −3 on both sides, we get

$\Rightarrow\frac{-3\text{x}}{-3}>\frac{-30}{-3}$

$\Rightarrow\text{x}>10$

$\Rightarrow\text{x}\in(10,\infty)$

2. [-7, 3]

Solution:

$|\text{x}+2|\leq5$

$\Rightarrow-5\leq\text{x}+2\leq5$

$\Rightarrow-5-2\leq\text{x}+2-2\leq5-2$

$\Rightarrow-7\leq\text{x}\leq3$

$\Rightarrow\text{x}\in[-7,3]$

2. $(\infty, –4)$

1. rational

Solution:

The sum of two rational numbers is a rational number.

Ex: Let two rational numbers are $\frac{1}{2}$ and $\frac{1}{3}$Now, $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ which is a rational number:

2. $\text{x}\in(-\infty,-2\big]$

Solution:

In the given figure, the circle is filled with dark colour at -2 which means -2 is included and the highlighted is towards the left of -2.

So, $\text{x}\in(-\infty,-2\big]$

2. $\text{x}\in[-11, 7]$

Solution:

Given that $|\text{x}+2|\leq9$

$\Rightarrow-9\leq\text{x}+2\leq9$ 

$\Rightarrow-9-2\leq\text{x}\leq9-2[|\text{x}\leq\text{a}|]$ 

$\Rightarrow-11\leq\text{x}\leq7$

$\Rightarrow\text{x}\in[-11, 7]$

3. $\frac{1}{4}$

2. R – (0)

Solution:

Given, f(x) = |x| > 0

We know that modulus is non negative quantity.

So, x ∈ R except that x = 0

$\Rightarrow\text{x}\in\text{R}-(0)$

This is the required solution

2. $(4,2)$

Solution:

We test whether the inequalities are satishfied or not

$(0,8),3(0)+8\geq 83(0)+8\geq8 8\geq8$ is true.

$2(0)+2(8)=16\geq12$ is true.

$0+2(8)=16\geq10$ is true.

$\therefore(0,8)$ is in the feasible region.

$(4,2), 3(4)+2=14\geq 83(4)+2=14\geq8$

$2(4)+2(2)=16\geq12$

$4+2(2)=8\geq10$ is not true.

$\therefore(4,2)$ is not a point in the feasible region.

$\therefore(2)$ is correct.

3. $\text{x}\in(-\infty,-4)\cup(6,\infty)$

Solution:

$|\text{x} – 1| > 5$

$\text{x} – 1 < - 5 $ and $\text{x} – 1 > - 5 $

$\text{x} < -4 $ and $\text{x} > 6 $

$\therefore\text{x}\in(-\infty,-4)\cup(6,\infty)$

1. $|\text{x}|>5$

Solution:

The given graph represents 

x > -5 and x < 5

Combining the two inequalities

|x| > 5.

2. $(-2,\infty)$

Solution:

$4\text{x}+3<6\text{x}+7$

Subtracting 3 from both sides,

$4\text{x}+3<6\text{x}+7-3$

$\Rightarrow4\text{x}<6\text{x}+4$

Subtracting 6x from both sides,

$4\text{x} – 6\text{x} <6\text{x} + 4 – 6\text{x}$

$\Rightarrow– 2\text{x}<4$ or

$\Rightarrow\text{x}>-2\text{ i.e..,}$ all the real numbers greater than –2, are the solutions of the given inequality.

Hence, the solution set is $(–2,\infty), \text{i}.\text{e}.\text{x}\in(-2,\infty)$

4. $\text{No Solution}$

Solution:

Given, |x| < -5

Now, LHS ≥ 0 and RHS < 0

Since LHS is non-negative and RHS is negative

So, |x| < -5 does not posses any solution

3. $\big(-\infty,1)\cup\big(2, 3\big)$

1. $\frac{2}{3}$

3. $\text{x}\in[-11,7]$

Solution:

As according to the graph,

x lies between $(-\infty,-5]$ and $[5,\infty)$

$\Rightarrow\text{x}\geq5$ or $\text{x}\leq-5$

$\Rightarrow|\text{x}|\geq5$

1. $|\text{x}|<5$

3. $\big[-2,0\big]\cup\big[2,4\big]$

Solution:

Given, $ 1\leq|\text{x} – 1|\leq3$

$\Rightarrow-3\leq(\text{x} – 1)\leq-1 $ or $1\leq(\text{x}-1)\leq3$

i.e. the distance covered is between 1 unit to 3 unit.

$\Rightarrow-2\leq \text{x}\leq0$ or $2\leq\text{x}\leq4$

Hence, the solution set of the given inequality is.

$\text{x}\in\big[-2, 0\big]\cup\big[2, 4\big]$

1. $\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$

Solution:

Given that x, y and b are real numbers and x < y, b < 0.

Consider, x < y

Divide both sides of the inequality by “b”

$\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}(\text{since, b<0})$

3. $(-13, –3)$

2. $\big(-\infty,4\big)$

1. $\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$

Solution:

Given that x, y and b are real numbers and x < y, b > 0.

Both sides of an inequality can be multiplied or divided by the same positive number.

$\therefore\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$

1. $\text{x}\in(10,\infty)$

Solution:

Given,

-3x + 17 < -13

Subtracting 17 from both sides,

-3x + 17 – 17 < -13 – 17

$\Rightarrow$ -3x < -30

$\Rightarrow$ x > 10 (since the division by negative number inverts the inequality sign)

$\Rightarrow\text{x}\in(10,\infty)$

2. $\text{x}+\text{y}\geq10$

Solution:

The coffee powder is greater than choco.

hence, $\text{x}>\text{y}$

each pack is at least 10gm.

4. $\text{x}\leq3$

2. $\big[–1, 2\big)$

4. $\text{None of the above }$

3. $\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) $

3. $ –1\leq\text{x}<2$

Solution:

Given,

$-8\leq5\text{x}-3$ and $5\text{x}-3<7$

Let us solve these two inequalities simultaneously.

$-8\leq5\text{x}-3$ and $5\text{x}-3<7$ can be written as:

$– 8\leq5\text{x} –3 < 7$

Adding 3, we get

$– 8 + 3\leq5\text{x}-3+3<7+3$

$–5\leq5\text{x}<10$

Dividing by 5, we get

$–1\leq\text{x}<2$

3. (1, 2, 3, 4)

Solution:

$20\text{x}<100$

Dividing by 20 on both sides, $\text{x}<\Big(\frac{100}{20}\Big)\Rightarrow\text{x}<5$

Since x is a positive integer so x = 1, 2, 3, 4.

3. $(-2, -1)\cup(1, 2)$

Solution:

Given, $\frac{-1}{(|\text{x}| – 2)}\geq1$

$\Rightarrow\frac{-1}{(|\text{x}|-2) -1}\geq0$

$\Rightarrow\frac{-1 – (|\text{x}| – 2)}{(|\text{x}| – 2)}\geq0$

$\Rightarrow\frac{1 – |\text{x}|}{(|\text{x}| – 2)}\geq0$

$\Rightarrow\frac{-(|\text{x}| – 1)}{(|\text{x}| – 2)}\geq0$

Using number line rule:

$1\leq |\text{x}|<2$

$\Rightarrow\text{x}\in(-2, -1)\cup(1, 2)$

2. $\text{x}\in\big[\infty,-2\big]$ 

Solution:

The given graph has all real values of x greater than and equal to -2.

So, $\text{x}\in\big[\infty,-2\big]$

4. 45

Solution:

Average is at least 30 marks.

Let x be the marks in $3^\text{rd}$ test.

Average $=\frac{(20+25+\text{x})}{3}\geq30$

$\Rightarrow45+\text{x}\geq90\Rightarrow\text{x}\geq90-45\Rightarrow\text{x}\geq45.$

Minimum marks in $3^\text{rd}$ test should be 45.

4. $\text{No Solution}$

Solution:

Given, |x| = -5

Since |x| is always positive or zero.

So, it can not be negative.

Hence, given inequality has no solution.

3. $\big(-\infty, 3\big)$

2. $\Big(\frac{-1}{2},\frac{3}{2}\Big)$

Solution:

Given, $-2<2\text{x}-1<2$

$\Rightarrow2+1< 2\text{x}<2+1$

$\Rightarrow-1<2\text{x}<3$

$\Rightarrow\frac{-1}{2}<\text{x}<\frac{3}{2}$

$\Rightarrow\text{x}\in\Big(\frac{-1}{2},\frac{3}{2}\Big)$

2. $\text{|x|}\leq3$

Solution:

As according to the graph,

x lies between −3 and 3

$\Rightarrow-3\leq\text{x}\leq3$

$\Rightarrow|\text{x}|\leq3$

3. $\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$

Solution:

Given that x < y, b < 0

$\Rightarrow\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}},\text{b}<0$

2. $\big(2,\infty\big)$

1. 8

Solution:

Let the four terms be a − 3d, a − d, a + d, and a + 3d with common difference 2d.sum = 4a = 20a = 5a = 5

$\frac{\text{(a-3d)}\text{(a+3d)}}{\text{(a-d)}\text{(a+d)}}=\frac{2}{3}$

$\frac{25-9\text{d}^2}{25-\text{d}^2}=\frac{2}{3}$

$\text{d}^2=1\Rightarrow\text{d}=1$

Largest term = a + 3d = 5 + 3(1) = 8

4. $(-\infty, 1)\cup(2, 3)$

2. $\text{x}\in(7,\infty)$

Solution:

Given,

$\frac{|\text{x}-7|}{(\text{x}-7)}\geq0$

This is possible when $\text{x}-7\geq0,$ and $\text{x}-7\neq0.$

Here, $\text{x}\geq7$ but $\text{x}\neq7$

Therefore, $\text{x}> 7, \text{i}.\text{e}. \text{x}\in(7,\infty).$

1. $\text{x}\in(10, \infty)$

Solution:

Given that - 3x + 17 < - 13

⇒ - 3x < - 17 - 13

⇒ -3x < - 30

⇒ 3x > 30

⇒ x > 10

$\Rightarrow\text{x}\in(10, \infty)$

2. $\big(1,\infty\big)$

Solution:

Given, $2\text{x}+1>3$

$\Rightarrow2\text{x}>3-1$

$\Rightarrow2\text{x}>2$

$\Rightarrow\text{x}>1$

$\Rightarrow\text{x}\in\big(1,\infty\big)$

3. $(-1)^\frac{1}{2}$

Solution:

Explanation: Iota is used to denote complex number.

The value of i (iota) is $\sqrt{-1}\text{ i.e.}(-1)^\frac{1}{2}$

4. Infinite

Solution:

$\frac{2\text{x}}{5}+1<-3.$

$\frac{2\text{x}}{5}<-3-1$

$2\text{x}<-4\times5$

$\text{x}<-\frac{20}{2}$

$\text{x}<-10$

As we know the number of rational numbers on either side of a number on the number line is infinite.

$\therefore$ set x = (infinite number of integers which are < - 10)

4. $\text{x}\in\big[2,\infty)$

Solution:

Given, $\frac{3(\text{x}-2)}{5}\geq\frac{5(2-\text{x})}{3}$

$\Rightarrow3(\text{x} – 2)\times3\geq5(2 – \text{x})\times5$

$\Rightarrow9(\text{x} – 2)\geq25(2 – \text{x})$

$\Rightarrow9\text{x} – 18\geq50 – 25\text{x}$

$\Rightarrow9\text{x} – 18 + 25\text{x}\geq50$

$\Rightarrow34\text{x}-18\geq50$

$\Rightarrow34\text{x}\geq50+18$

$\Rightarrow34\geq68$

$\Rightarrow\text{x}\geq\frac{68}{34}$

$\Rightarrow\text{x}\geq2$

$\Rightarrow\text{x}\in\big[2,\infty)$

1. $\big(2-3, 2+3\big)\cup\big(-2-3-2+3\big)$

3. $\text{x}\in[-11,7]$

Solution:

$|\text{x}+2|\leq9$

$\Rightarrow-9\leq\text{x}+2\leq9$

$\Rightarrow-9-2\leq\text{x}+2-2\leq9-2$

$\Rightarrow-11\leq\text{x}\leq7$

$\Rightarrow\text{x}\in[-11,7]$

1. $\text{x}\in\big(10,\infty\big)$

2. $\text{x}\geq2$

2. $-1,1+2\omega,1-2\omega^2$

4. $\text{None of the above }$

3. $\big(-\infty ,0\big)\cup\big(\frac{2}{3},\infty\big) $

4. $\text{x}\in(-\infty,-\text{b})\cup(\text{b,}\infty)$

Solution:

Given that |x| > b, b > 0

⇒ x < -b or x > b

$\Rightarrow\text{x}\in(-\infty,-\text{b})\cup(\text{b,}\infty)$

3. $\bigg(\frac{-4}{9},\frac{2}{9}\bigg)-\bigg(\frac{-1}{3}\bigg)$

3. $\text{x}\in(-\infty,-4)\cup(6,\infty)$

Solution:

|x−1| > 5

⇒ x − 1 > 5 or x − 1 < −5

⇒ x > 5 + 1 or x < −5 + 1

⇒ x > 6 or x < −4

$\Rightarrow\text{x}\in(-\infty,-4)\cup(6,\infty)$

3. smaller than $\alpha$

4. $\text{x}\in(-\infty,-13]\cup[7,\infty)$

Solution:

$|\text{x}+3|\geq10$

$\Rightarrow\text{x}+3\geq10\text{ or }\text{x}+3;\leq-10$

$\Rightarrow\text{x}\geq10-3\text{ or }\text{x}\leq-10-3$

$\Rightarrow\text{x}\geq7\ \text{or}\ \text{x}\leq-13$

$\Rightarrow\text{x}\in(-\infty,-13)\cup[7,\infty)$

3. $\text{x}\in\big(-\infty, –4\big)\cup\big(6,\infty\big)$

4. $\big(-\infty,\frac{-3}{2}\big]\cup\big[3,\infty\big)$

3. right side with complete line x = 2

Solution:

$3\text{x}-6\geq0\Rightarrow\text{x}\geq2.$

(0, 0) does not satisfy te equation so region is right side of x = 2 with complete line x = 2 due to presence of equality sign along with inequality sign.

2. Must be irrational

Solution:

If a is an irrational number which is divisible by b then the number b must be irrational.

Ex: Let the two irrational numbers are $\sqrt{2}$ and $\sqrt{3}$ Now, $\sqrt{2}\sqrt{3}=\sqrt{\Big(\frac{2}{3}}\Big)$

4. $2\text{x}+\text{y}\leq6,\text{x}\geq0,\text{y}\geq0$

Solution:

Since region involves 1^st quadrant so $\text{x}\geq0,\text{y}\geq0.$

Two points on line are $(0,6)$ and$(3,0)$

$\frac{(\text{y}-6)}{(0-6)} =\frac{(\text{x}-0)}{(3-0)}$

$\Rightarrow\frac{(\text{y}-6)}{(-6)}=\frac{\text{x}}{3}$

$\Rightarrow\text{y}-6=2\text{x}\Rightarrow2\text{x}+\text{y}=6$

$2\text{x}+\text{y}\leq6$ since $(0,0) $ should also satisfy.

So, $ 2\text{x}+\text{y}\leq6, \text{x}\geq0, \text{y}\geq0.$

4. $\text{x}\in(-\infty,-\text{a})\cup(\text{a},\infty)$

Solution:

If x and a are real numbers such that a > 0.

|x| > a

⇒ x > a or x < −a

$\Rightarrow\text{x}\in(-\infty,-\text{a})\cup(\text{a},\infty)$

2. $\text{x}\in\big[–11, 7\big]$

4. $\text{x}\in(-\infty, –\text{b}\big)\cup\big(\text{b},\infty\big)$

4. $\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$

Solution:

$|\text{x} + 3|\geq10$

$\Rightarrow\text{x} + 3\leq-10$ or $\text{x}+3\geq10$

$\Rightarrow\text{x}\leq -13 $ or $\text{x}\geq7$

$\Rightarrow\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$

2. $15\text{x}+5\text{y}\leq600:24\text{x}+14\text{y}\geq1000$

1. $\text{x}\leq\frac{105}{8}$

Solution:

$\frac{1}{5}\big(\frac{3\text{x}}{5}+4\big)\geq\frac{1}{3}(\text{x}-6).$

$​​\Rightarrow3\big(\frac{3\text{x}}{5}+4\big)\geq5\big(\text{x}-6\big)$

$​​\Rightarrow\big(\frac{9\text{x}}{5}+12\big)\geq5\text{x}-6$

$\Rightarrow(30+12)\geq-\frac{9\text{x}}{5}+5\text{x}$

$\Rightarrow42\geq\frac{-9\text{x}+25\text{x}}{5}$

$\Rightarrow42\geq\frac{16\text{x}}{5}$

$\Rightarrow\frac{42\times5}{16}\geq\text{x}$

$\text{x}\leq\frac{105}{8}$

Therefore option (1) is the correct answere.

3. $\text{x}<-4$

Solution:

Given: $\frac{\text{x}}{4}>\frac{\text{x}}{2}+1$

$\Rightarrow\frac{\text{x}}{4}-\frac{\text{x}}{2}>1$

$\Rightarrow\frac{\text{x}-2\text{x}}{4}>1$

$\Rightarrow\frac{\text{-x}}{4}>1$

$\Rightarrow-\text{x}>4$

$\Rightarrow\text{x}<-4$

2. $(−6, 0)$

Solution

Given: $0<−\frac{\text{x}}{2} < 3$

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

$\Rightarrow0 <-\text{x}<6$

$\Rightarrow−6 <\text{x}<0$

$\therefore\text{x}$ lies in $(−6, 0)$

3. $\big(1,3\big]$

Solution:

$\frac{2\text{x}+4}{\text{x}-1}\geq5$

$\Rightarrow\frac{2\text{x}+4}{\text{x}-1}-5\geq0$

$\Rightarrow\frac{2\text{x}+4-5(\text{x-1})}{\text{x-1}}\geq0$

$\Rightarrow\frac{2\text{x}+4-\text{5x+5}}{\text{x-1}}\geq0$

$\Rightarrow\frac{9-3\text{x}}{\text{x-1}}\geq0$

$\Rightarrow\frac{-3\text{(x-3)}}{\text{x}-1}\geq0$

Multiplying each side of an inequality by a negative number $\big(\frac{-1}{3}\big)$ reverses the direction of the inequality symbol.

$\Rightarrow\frac{(\text{x}-3)}{\text{x}-1}\leq0$ Here $\text{x}-1\neq0\Rightarrow\text{x}\neq1$

Hence the solution set of the given in equations is $\big(1, 3\big]$

2. $(2, 4)\cup(4, 6)$

Solution:

Given,$\Big|\frac{2}{(\text{x} – 4)}\Big|>1$

$\Rightarrow2 > |\text{x} – 4|$

$\Rightarrow |\text{x} – 4|<2$

$\Rightarrow-2<\text{x}-4<2$

$\Rightarrow-2+4 < \text{x} < 2+4$

$\Rightarrow2<\text{x}<6$

$\Rightarrow\text{x}\in(2, 6), $ where $\text{x}\neq4$

$\Rightarrow\text{x}\in(2, 4)\cup(4, 6)$

3. $\big[25^\circ, 30^\circ\big]$

Solution:

$\text{F}=\frac{9}{5\text{c}}+32^\circ$

$\text{C}=\text{F}-32^\circ\times\frac{5}{9}$

$77^\circ\leq\text{F}\leq86^\circ$

$\Rightarrow77^\circ-32^\circ\leq\text{F}-32^\circ\leq86^\circ -32^\circ$

$\Rightarrow45^\circ\leq\text{F}-32^\circ\leq54^\circ$

$\Rightarrow45^\circ\times\frac{5}{9}\leq(\text{F}-32^\circ)\times\frac{5}{9}\leq54^\circ\times\frac{5}{9}$

$\Rightarrow25^\circ\leq\text{C}\leq30^\circ$

2. $\big(-2,\infty\big)$

2. {1, 2, 3, 4, 5}

Solution:

$20\text{x}\leq100$

Dividing by 20 on both sides, $\text{x}\leq\frac{100}{20}\Rightarrow\text{x}\leq5$

Since x is a natural number so x = 1, 2, 3, 4, 5.

3. $\text{x}\in\big(-\infty,-13\big]\cup\big[7,\infty\big)$

2. [-7, 3]

Solution:

$|\text{x}+2|\leq5$

$\Rightarrow-5\leq\text{x}+2\leq5$

$\Rightarrow-5-2\leq\text{x}+2-2\leq5-2$

$\Rightarrow-7\leq\text{x}\leq3$

$\Rightarrow\text{x}\in[-7,3]$

2. $-5<\text{x}<5$

Solution:

If x is a real number.

|x| < 5

⇒ -5 < x < 5

1. $\frac{49}{4}$

1. $(-3,3)$

3. $-\text{x} > – 5 $

Solution:

If x > 5 then - x > - 5.

3. $-\text{x}>-7$

Solution:

x < 7

subtracting x on both sides, we get

⇒ x − x < 7 − x

⇒ 0 < 7 − x

subtracting 7 on both sides, we get

⇒ 0 − 7 < 7 − x − 7

⇒ −7 < − x

⇒ − x > −7

4. $\big[4,12\big]$

Solution:

$\text{IQ}=\Big(\frac{\text{MA}}{\text{CA}}\Big)\times100$

$\Rightarrow\text{MA}=\text{IQ}\times\frac{\text{CA}}{100}$ Given, $\text{CA}=10$ years

$40\leq\text{IQ}\leq120$

$\Rightarrow40\times\text{CA}\leq\text{IQ}\times\text{CA}\leq120\times\text{CA}$

$\Rightarrow40\times10\leq\text{IQ}\times\text{CA}\leq120\times10$

$\Rightarrow40\times10100\leq\text{Q}\times\text{CA}100\leq120\times10100$

$\Rightarrow40\leq\text{MA}\leq120$

1. $27 <\times< 2$

Solution:

Given inequality is:

$15<\frac{3(\text{x} – 2)}{5}<0$

$\Rightarrow15\times5<3(\text{x} – 2)<0 × 5$

$\Rightarrow75<3(\text{x} – 2)<0$

$\Rightarrow\frac{75}{3}<\text{x} – 2<0$

$\Rightarrow25<\text{x}-2<0$

$\Rightarrow25 + 2 <\text{x}<0 + 2$

$\Rightarrow27 <\times< 2$

4. 13

Solution:

Let shortest side be x.Then longest side $=2\text{x}.$

Third side $=2\text{x}-4.$

Given, perimeter of triangle is at least 61cm

$\Rightarrow\text{x}+2\text{x}+2\text{x} - 4\geq61\Rightarrow5\text{x}\geq65=\text{x}\geq13.$

Minimum length of the shortest side is 13cm.

2. $\text{x}\in\big[\frac{9}{2},\infty\big]$

Solution:

The given graph has all real values of x greater than and equal to $\frac{9}{2}.$

 So, $\text{x}\in\big[\frac{9}{2}\infty\big]$

1. $\text{x}\in\big(-\infty,\frac{7}{2}\big)$

Solution:

The given graph all real values of x greater than and equal $\frac{7}{2}$ on real number line.

So, $\text{x}\in\big(-\infty,\frac{7}{2}\big)$

4. No value of x

Solution:

Given, $|\text{x} + 1| +\sqrt{(\text{x} – 1)}= 0, $where each term is non - negative.

So, $ |\text{x} + 1| = 0 $ and $\sqrt{\text{(x-1})}=0$ should be zero simultaneously.

$\text{i}.\text{e}. \text{x} = -1 $ and $\text{x}=1,$ which is not possible.

So, there is no value of x for which each term is zero simultaneously.

4. $\big[– 4, – 2\big]\cup\big[2, 4\big]$

2. {1, 2, 3, 4}

Solution:

Given, $\text{x}<5$ and $\text{x}\in\text{N}$ Natural numbers are counting numbers whose set is.

N = {1, 2, 3, ..} Therefore, {1, 2, 3, 4} represents $\text{x}<5$

2. $\big(\frac{5}{3},\frac{12}{7}\big]$

2. $\text{x}\in\big[–3, 1\big]$

3. $\big(\frac{-2}{3} ,8\big)$

3. $[-4, -2]\cup[2, 6]$

Solution:

Given, $|\text{x}-1|\leq 5, |\text{x}|\geq2$

$\Rightarrow-(5\leq(\text{x} – 1)\leq5), (\text{x}\leq -2 \text{or} \text{x} \geq 2)$

$\Rightarrow-(4\leq\text{x}\leq6), (\text{x}\leq-2 \text{or}\text{x}\geq 2)$

Now, required solution is.

$\text{x}\in[-4, -2]\cup[2, 6]$

1. −2, −1, 0, 1, 2

Solution:

$\text{x}\in\text{l},$ solution set $-2\leq\text{x}<3$

= −2, −1, 0, 1, 2

2. $\big(2,\infty\big)$

2. $\big[–1, 1\big]\cup\big[3, 5\big]$

4. $\text{No solution}$

Solution:

Given, $\text{x}^2<-4$

$\Rightarrow\text{x}^2+4<0$

Which is not possible.

So, there is no solution.

2. $(1,\infty)$

Solution:

Given, $2\text{x}+1>3$

$\Rightarrow2\text{x}>3-1$

$\Rightarrow2\text{x}>2$

$\Rightarrow\text{x}>1$

$\Rightarrow\text{x}\in(1,\infty)$

3. $\big[2,6\big)$

3. $\big(-1, 0\big)\cup\big(0, 3\big)$