MCQ 11 Mark
Choose the correct answer. If $x$ is a real number and $|x| < 3,$ then:
- A
$\text{x}\geq3$
- ✓
$-3<\text{x}<3$
- C
$\text{x}\leq-3$
- D
$-3\leq\text{x}\leq3$
AnswerCorrect option: B. $-3<\text{x}<3$
Given that $|x| < 3$
$\Rightarrow -3 < x < 3 | x | < a$
$\Rightarrow -a < x < a.$
View full question & answer→MCQ 21 Mark
If $−5\leq\frac{5 – 3\text{x}}{2}\leq8, $ then $\text{x}\in$
- ✓
$\big[\frac{11}{3},5\big]$
- B
$\big[-5,5\big]$
- C
$\big[\frac{-11}{3},\infty\big]$
- D
$\big(-\infty,\infty\big)$
AnswerCorrect option: A. $\big[\frac{11}{3},5\big]$
View full question & answer→MCQ 31 Mark
The solution of $\Big|\frac{2}{(\text{x} – 4)}\Big|>1$ where $\text{x}\neq4$ is:
- A
$(2, 6)$
- ✓
$(2, 4)\cup(4, 6)$
- C
$(2, 4)\cup(4,\infty)$
- D
$(-\infty, 4)\cup(4, 6)$
AnswerCorrect option: B. $(2, 4)\cup(4, 6)$
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 $\text{x}\neq4$
$\Rightarrow\text{x}\in(2, 4)\cup(4, 6)$
View full question & answer→MCQ 41 Mark
Write the solution of inequality $\frac{1}{5}\bigg(\frac{3\text{x}}{5}+4\bigg)\geq\frac{1}{3}(\text{x}-6).$
AnswerCorrect option: A. $\text{x}\leq\frac{105}{8}$
$\frac{1}{5}\big(\frac{3\text{x}}{5}+4\big)\geq\frac{1}{3}(\text{x}-6).$
$\Rightarrow3\big(\frac{3\text{x}}{5}+4\big)\geq5\big(\text{x}-6\big)$
$\Rightarrow\big(\frac{9\text{x}}{5}+12\big)\geq5\text{x}-6$
$\Rightarrow(30+12)\geq-\frac{9\text{x}}{5}+5\text{x}$
$\Rightarrow42\geq\frac{-9\text{x}+25\text{x}}{5}$
$\Rightarrow42\geq\frac{16\text{x}}{5}$
$\Rightarrow\frac{42\times5}{16}\geq\text{x}$
$\text{x}\leq\frac{105}{8}$
Therefore option $(1)$ is the correct answere.
View full question & answer→MCQ 51 Mark
If the roots of the equation $x^2- bx + c = 0$ be two consecutive integers, then $b^2 - 4c$ equals:
View full question & answer→MCQ 61 Mark
The length of a rectangle is three times the breadth.If the minimum perimeter of the rectangle is $160\ cm,$ then:
- A
breadth $> 20\ cm$
- B
length $< 20\ cm$
- ✓
breadth $x \geq 20\ cm$
- D
length $\leq 20\ cm$
AnswerCorrect option: C. breadth $x \geq 20\ cm$
Let $x$ be the breadth of a rectangle.
So, length $= 3x$
Given that the minimum perimeter of a rectangle is $160\ cm.$
Thus, $2(3\text{x}+\text{x})\geq160$
$\Rightarrow4\text{x}\geq80$
$\Rightarrow\text{x}\geq20$
View full question & answer→MCQ 71 Mark
Which of the following points lie in the solution set?
- A
$(1, 1)$
- ✓
$(1, 2)$
- C
$(2, 1)$
- D
$(3, 2)$
AnswerCorrect option: B. $(1, 2)$
View full question & answer→MCQ 81 Mark
The value of a for which the sum of the squares of the roots of the equation $x^2- (a - 2)x - a - 1 = 0$ the least value is:
View full question & answer→MCQ 91 Mark
If $x$ is a whole number and $10\text{x}\leq50$ then find solution set of $x.$
- ✓
$\{0,1, 2, 3, 4, 5\}$
- B
$\{1, 2, 3, 4, 5\}$
- C
$\{1, 2, 3, 4\}$
- D
$\{0, 1, 2, 3, 4\}$
AnswerCorrect option: A. $\{0,1, 2, 3, 4, 5\}$
$10\text{x}\leq50$
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, $x = 0, 1, 2, 3, 4, 5.$
View full question & answer→MCQ 101 Mark
If the sum of the roots of the quadratic equation $ \text{ax}^2+\text{bx}+\text{c}=0$ is equal to the sum of the squares of their reciprocals, then $\frac{\text{a}}{\text{c}}, \frac{\text{b}}{\text{a}}$ and $\frac{\text{c}}{\text{b}}$ are in:
- A
- B
- ✓
- D
arithmetico-geometric progression
View full question & answer→MCQ 111 Mark
The quadratic equations $x^2- 6x + a = 0$ and $x^2- cx + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4 : 3.$Then, the common root is:
View full question & answer→MCQ 121 Mark
Solution of $|3\text{x}+2| <1$ is:
- A
$\big[-1,\frac{-1}{3}\big]$
- B
$\big(\frac{-1}{3},-1\big)$
- ✓
$\big(-1,\frac{-1}{3}\big)$
- D
$\text{None of these}$
AnswerCorrect option: C. $\big(-1,\frac{-1}{3}\big)$
View full question & answer→MCQ 131 Mark
A solution is to be kept between $77^\circ F$ and $86^\circ F.$ What is the range in temperature in degree Celsius $(C)$ if the $\frac{\text{Celsius}}{\text{Fahrenheit}}(F) $conversion formula is given by $\text{F}=\frac{9}{5\text{c}}+32^\circ$
- A
$\big[15^\circ, 20^\circ\big]$
- B
$\big[20^\circ, 25^\circ\big]$
- ✓
$\big[25^\circ, 30^\circ\big]$
- D
$\big[30^\circ, 35^\circ\big]$
AnswerCorrect option: C. $\big[25^\circ, 30^\circ\big]$
$\text{F}=\frac{9}{5\text{c}}+32^\circ$
$\text{C}=\text{F}-32^\circ\times\frac{5}{9}$
$77^\circ\leq\text{F}\leq86^\circ$
$\Rightarrow77^\circ-32^\circ\leq\text{F}-32^\circ\leq86^\circ -32^\circ$
$\Rightarrow45^\circ\leq\text{F}-32^\circ\leq54^\circ$
$\Rightarrow45^\circ\times\frac{5}{9}\leq(\text{F}-32^\circ)\times\frac{5}{9}\leq54^\circ\times\frac{5}{9}$
$\Rightarrow25^\circ\leq\text{C}\leq30^\circ$
View full question & answer→MCQ 141 Mark
The graph of the inequations $x ≤ 0, y ≤ 0,$ and $2x + y + 6 ≥ 0$ is:
AnswerCorrect option: B. a triangular region in the $3^{rd}$ quadrant
Given inequalities $x ≥ 0, y ≥ 0, 2x + y + 6 ≥ 0$
Now take $x = 0, y = 0$ and $2x + y + 6 = 0$
when $x = 0, y = -6$
when $y = 0, x = -3$
So, the points are $A(0, 0), B(0, -6)$ and $C(-3, 0)$
So, the graph of the inequations $x ≤ 0, y ≤ 0,$ and $2x + y + 6 ≥ 0$ is a triangular region in the $3^{rd}$ quadrant. View full question & answer→MCQ 151 Mark
If $4\text{x} + 3<6\text{x} + 7$, then$\text{ x}\in$
AnswerCorrect option: B. $\big(-2,\infty\big)$
View full question & answer→MCQ 161 Mark
If $|x−1| x - 1 > 5,$ then:
- A
$\text{x}\in(-4,6)$
- B
$\text{x}\in[-4,6]$
- ✓
$\text{x}\in(-\infty,-4)\cup(6,\infty)$
- D
$\text{x}\in(-\infty,-4)\cup[6,\infty)$
AnswerCorrect option: C. $\text{x}\in(-\infty,-4)\cup(6,\infty)$
$|x−1| > 5$
$\Rightarrow x − 1 > 5$ or $x − 1 < −5$
$\Rightarrow x > 5 + 1$ or $x < −5 + 1$
$\Rightarrow x > 6$ or $x < −4$
$\Rightarrow\text{x}\in(-\infty,-4)\cup(6,\infty)$
View full question & answer→MCQ 171 Mark
If $x$ is a natural number and $20\text{x}\leq100$ then find solution set of $x.$
- A
$\{0, 1, 2, 3, 4, 5\}$
- ✓
$\{1, 2, 3, 4, 5\}$
- C
$\{1, 2, 3, 4\}$
- D
$\{0, 1, 2, 3, 4\}$
AnswerCorrect option: B. $\{1, 2, 3, 4, 5\}$
$20\text{x}\leq100$
Dividing by $20$ on both sides,
$\text{x}\leq\frac{100}{20}$
$\Rightarrow\text{x}\leq5$
Since $x$ is a natural number.
so $x = 1, 2, 3, 4, 5.$
View full question & answer→MCQ 181 Mark
If $− 3x + 17 < -13,$ then:
AnswerCorrect option: A. $\text{x}\in(10,\infty)$
$− 3x + 17 < −13$
Subtracting $17$ on both sides, we get
$\Rightarrow −3x + 17 − 17 < −13 − 17$
$\Rightarrow −3x < − 30$
Dividing $−3$ on both sides, we get
$\Rightarrow\frac{-3\text{x}}{-3}>\frac{-30}{-3}$
$\Rightarrow\text{x}>10$
$\Rightarrow\text{x}\in(10,\infty)$
View full question & answer→MCQ 191 Mark
The graph of the inequalities $x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0$ is:
AnswerCorrect option: C. $\{\}$
Given inequalities $x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0$
Now take $x = 0, y = 0$ and $2x + y + 6 = 0$
when $x = 0, y = -6$
when $y = 0, x = -3$
So, the points are $A(0, 0), B(0, -6)$ and $C(-3, 0)$
Since region is outside from the line $2x + y + 6 = 0$
So, it does not represent any figure. View full question & answer→MCQ 201 Mark
If $x$ is a real number and $|x| < 5,$ then:
- A
$\text{x}\geq5$
- ✓
$-5<\text{x}<5$
- C
$\text{x}\leq-5$
- D
$-5\leq\text{x}\leq5$
AnswerCorrect option: B. $-5<\text{x}<5$
If $x$ is a real number.
$|x| < 5$
$\Rightarrow -5 < x < 5$
View full question & answer→MCQ 211 Mark
If $|\text{x} + 3|\geq10,$ then:
- A
$\text{x}\in\big(-13, 7\big] $
- B
$\text{x}\in\big(–13, 7\big)$
- ✓
$\text{x}\in\big(-\infty,-13\big]\cup\big[7,\infty\big)$
- D
$\text{x}\in\big(-\infty,-13\big)\cup\big[7,\infty\big)$
AnswerCorrect option: C. $\text{x}\in\big(-\infty,-13\big]\cup\big[7,\infty\big)$
View full question & answer→MCQ 221 Mark
The solution of the inequality $\frac{3(\text{x}-2)}{5}\geq\frac{5(2-\text{x})}{3}$ is:
- A
$\text{x}\in(2,\infty)$
- B
$\text{x}\in\big[-2,\infty)$
- C
$\text{x}\in\big[\infty,2)$
- ✓
$\text{x}\in\big[2,\infty)$
AnswerCorrect option: D. $\text{x}\in\big[2,\infty)$
Given, $\frac{3(\text{x}-2)}{5}\geq\frac{5(2-\text{x})}{3}$
$\Rightarrow3(\text{x} – 2)\times3\geq5(2 – \text{x})\times5$
$\Rightarrow9(\text{x} – 2)\geq25(2 – \text{x})$
$\Rightarrow9\text{x} – 18\geq50 – 25\text{x}$
$\Rightarrow9\text{x} – 18 + 25\text{x}\geq50$
$\Rightarrow34\text{x}-18\geq50$
$\Rightarrow34\text{x}\geq50+18$
$\Rightarrow34\geq68$
$\Rightarrow\text{x}\geq\frac{68}{34}$
$\Rightarrow\text{x}\geq2$
$\Rightarrow\text{x}\in\big[2,\infty)$
View full question & answer→MCQ 231 Mark
If the equstion $\text{a}_\text{n}\text{x}^\text{n}+\text{a}_\text{n-1}\text{x}^\text{n-1}+...+\text{a}_1\text{x}=0,$
$\text{a}_1\neq0,\text{n}\geq2,$ has positive root $\text{x}=\alpha$ then the eqestions
$\text{na}_\text{n}\text{x}^\text{n-1}+(\text{n-1})\text{a}_\text{n-1}\text{x}^\text{n-2}+...+\text{a}_1=0$ has a positive root, which is:
AnswerCorrect option: C. smaller than $\alpha$
View full question & answer→MCQ 241 Mark
Question: If one root of the equation $\text{x}^2+\text{px}+12=0$ is $4,$ while the equation $\text{x}^2+\text{px}+\text{q}=0$ has equal roots, then the value of $q$ is:
- ✓
$\frac{49}{4}$
- B
$12$
- C
$3$
- D
$4$
AnswerCorrect option: A. $\frac{49}{4}$
View full question & answer→MCQ 251 Mark
Solution of $\bigg|\text{x}+\frac{1}{\text{x}}\bigg|<4$ is:
- ✓
$\big(2-3, 2+3\big)\cup\big(-2-3-2+3\big)$
- B
$\text{R}\big(2-3, 2+3\big)$
- C
$\text{R}-\big(-2-3-2 + 3\big)$
- D
$\text{None of these}$
AnswerCorrect option: A. $\big(2-3, 2+3\big)\cup\big(-2-3-2+3\big)$
View full question & answer→MCQ 261 Mark
The number of the real solutions of the equation $x^2-3 |x| + 2 = 0$ is:
View full question & answer→MCQ 271 Mark
If the difference between the roots of the equation $\text{x}^2 +\text{ax}+1=0$ is less than $\sqrt{5},$ then the set of possible values of $a$ is:
- ✓
$(-3,3)$
- B
$(-3,\infty)$
- C
$(3,\infty)$
- D
$(-\infty,-3)$
AnswerCorrect option: A. $(-3,3)$
View full question & answer→MCQ 281 Mark
All the values of $m$ for which both roots of the equation $x^2-2mx + m^2-1 = 0$ are greater than $-2$
but less than $4$ lie in the interval:
- A
$m > 3$
- ✓
$-1 < m < 3$
- C
$1 < m < 4$
- D
$– 2 < m < 0$
AnswerCorrect option: B. $-1 < m < 3$
View full question & answer→MCQ 291 Mark
If $7\text{x} + 3 < 5\text{x} + 9$ then $\text{x}\in$
AnswerCorrect option: C. $\big(-\infty,3\big)$
View full question & answer→MCQ 301 Mark
The inequality representing the following graph is:
- ✓
$|\text{x}|<5$
- B
$|\text{x}|\leq5$
- C
$|\text{x}|>5$
- D
$|\text{x}|\geq5$
AnswerCorrect option: A. $|\text{x}|<5$
View full question & answer→MCQ 311 Mark
$ax + b > 0$ is, $.............?$
AnswerSince it has highest power of $x\ '1\ '$ and has inequality sign
so, it is called linear inequality.
It is not numerical inequality as it does not have numbers on both sides of inequality.
It does not have two inequality signs so it is not double inequality.
View full question & answer→MCQ 321 Mark
If $\text{|x}+2|\leq9,$ then:
- A
$\text{x}\in(-7,11)$
- ✓
$\text{x}\in[-11,7]$
- C
$\text{x}\in(-\infty,-7)\cup(11,\infty)$
- D
$\text{x}\in(-\infty,-7)\cup[11,\infty)$
AnswerCorrect option: B. $\text{x}\in[-11,7]$
$|\text{x}+2|\leq9$
$\Rightarrow-9\leq\text{x}+2\leq9$
$\Rightarrow-9-2\leq\text{x}+2-2\leq9-2$
$\Rightarrow-11\leq\text{x}\leq7$
$\Rightarrow\text{x}\in[-11,7]$
View full question & answer→MCQ 331 Mark
$ax^2+ bx + c > 0$ is, $............?$
AnswerSince it has highest power of $x\ '2\ '$ and has inequality sign
so, it is called quadratic inequality.
It is not numerical inequality as it does not have numbers on both sides of inequality.
It does not have two inequality signs.
so it is not double inequality.
View full question & answer→MCQ 341 Mark
Solution of $|\text{x} – 1|\geq|\text{x}-3| $ is:
- A
$\text{x}\leq2$
- ✓
$\text{x}\geq2$
- C
$\big[1, 3\big]$
- D
$\text{None of these}$
AnswerCorrect option: B. $\text{x}\geq2$
View full question & answer→MCQ 351 Mark
If $(1 – p)$ is a root of quadratic equation $x^2+ px + (1-p) = 0$, then its roots are:
- A
$0, 1$
- B
$– 1, 1$
- ✓
$0, -1$
- D
$– 1, 2$
AnswerCorrect option: C. $0, -1$
View full question & answer→MCQ 361 Mark
If $–3\text{x}+17<-13, $ then:
- ✓
$\text{x}\in\big(10,\infty\big)$
- B
$\text{x}\in\big[10,\infty\big)$
- C
$\text{x}\in\big(-\infty,10\big)$
- D
$\text{x}\in\big[-10,10\big)$
AnswerCorrect option: A. $\text{x}\in\big(10,\infty\big)$
View full question & answer→MCQ 371 Mark
If $|\text{x}+3|\geq10,$ then:
- A
$\text{x}\in(-12,7]$
- B
$\text{x}\in(-13,7)$
- C
$\text{x}\in(\infty,-13)\cup(7,\infty)$
- ✓
$\text{x}\in(-\infty,-13]\cup[7,\infty)$
AnswerCorrect option: D. $\text{x}\in(-\infty,-13]\cup[7,\infty)$
$|\text{x}+3|\geq10$
$\Rightarrow\text{x}+3\geq10$ or $\text{x}+3;\leq-10$
$\Rightarrow\text{x}\geq10-3$ or $\text{x}\leq-10-3$
$\Rightarrow\text{x}\geq7$ or $\text{x}\leq-13$
$\Rightarrow\text{x}\in(-\infty,-13)\cup[7,\infty)$
View full question & answer→MCQ 381 Mark
Choose the correct answer. If $x < 5,$ then.
- A
$-\text{x} < – 5$
- B
$-\text{x}\leq-5$
- ✓
$-\text{x} > – 5$
- D
$-\text{x}\leq-5$
AnswerCorrect option: C. $-\text{x} > – 5$
If $x > 5$ then $- x > - 5.$
View full question & answer→MCQ 391 Mark
If $|\text{x} – 1|>5,$ then
- A
$\text{x}\in\big(-4, 6\big)$
- B
$\text{x}\in\big[–4, 6\big]$
- ✓
$\text{x}\in\big(-\infty, –4\big)\cup\big(6,\infty\big)$
- D
$\text{x}\in\big(-\infty, –4\big)\cup\big[6,\infty\big)$
AnswerCorrect option: C. $\text{x}\in\big(-\infty, –4\big)\cup\big(6,\infty\big)$
View full question & answer→MCQ 401 Mark
The solution set of the inequation $|\text{x}+2|\leq5$ is:
- A
$(-7, 5)$
- ✓
$[-7, 3]$
- C
$[-5, 5]$
- D
$(-7, 3)$
AnswerCorrect option: B. $[-7, 3]$
$|\text{x}+2|\leq5$
$\Rightarrow-5\leq\text{x}+2\leq5$
$\Rightarrow-5-2\leq\text{x}+2-2\leq5-2$
$\Rightarrow-7\leq\text{x}\leq3$
$\Rightarrow\text{x}\in[-7,3]$
View full question & answer→MCQ 411 Mark
The inequality representing the following graph is:
- ✓
$|x| < 5$
- B
$|x| ≤ 5$
- C
$|x| > 5$
- D
$|x| ≥ 5$
AnswerCorrect option: A. $|x| < 5$
The given graph shows the shaded region corresponding to $x > – 5$ and $x < 5.$
Therefore, by combining the above two inequalities, we get $|x| < 5.$
View full question & answer→MCQ 421 Mark
Choose correct option which suitably represents value of $\text{x.x}<5,\text{x}\in\text{N}$
- A
$\{0, 1, 2, 3, 4\}$
- ✓
$\{1, 2, 3, 4\}$
- C
$\{1, 2, 3, 4, 5\}$
- D
AnswerCorrect option: B. $\{1, 2, 3, 4\}$
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 $\text{x}<5$
View full question & answer→MCQ 431 Mark
Solutions of the inequalities comprising a system in variable $x$ are represented on number lines as given below, then $36$
- A
$\text{x}\in\big(-\infty,-4\big]\cup\big[3,\infty\big)$
- ✓
$\text{x}\in\big[–3, 1\big]$
- C
$\text{x}\in\big(-\infty, -4\big)\cup\big[3,\infty\big)$
- D
$\text{x}\in\big[–4, 3\big]$
AnswerCorrect option: B. $\text{x}\in\big[–3, 1\big]$
View full question & answer→MCQ 441 Mark
Solution of $(\text{x}-1) 2 (\text{x}+4)<0$ is:
- A
$(-\infty, 1)$
- ✓
$(\infty, –4)$
- C
$(– 1, 4)$
- D
$(1, 4)$
AnswerCorrect option: B. $(\infty, –4)$
View full question & answer→MCQ 451 Mark
If $x < 7,$ then:
- A
$-\text{x}<-7$
- B
$-\text{x}\leq-7$
- ✓
$-\text{x}>-7$
- D
$-\text{x}\geq-7$
AnswerCorrect option: C. $-\text{x}>-7$
subtracting $x$ on both sides, we get
$\Rightarrow x − x < 7 − x$
$\Rightarrow 0 < 7 − x$
subtracting $7$ on both sides, we get
$\Rightarrow 0 − 7 < 7 − x − 7$
$\Rightarrow −7 < − x$
$\Rightarrow − x > −7$
View full question & answer→MCQ 461 Mark
If the roots of the equation $bx^2+ cx + a = 0$ be imaginary, then for all real values of $x,$ the expression $3b^2x^2+ 6bcx + 2c^2$
- A
greater than $4ab$
- B
less than $4ab$
- ✓
greater than $-4ab$
- D
less than $-4ab$
AnswerCorrect option: C. greater than $-4ab$
View full question & answer→MCQ 471 Mark
$\text{IQ}$ of a person is given by the formula.
$\text{IQ}=\Big(\frac{\text{MA}}{\text{CA}}\Big)\times100$ where $\text{MA}$ is mental age and $\text{CA}$ is chronological age.
If $40\leq\text{IQ}\leq120$ for a group of $10$ years old children, find the range of their mental age.
- A
$\big(9,16\big)$
- B
$\big[9,16\big]$
- C
$\big(4,12\big)$
- ✓
$\big[4,12\big]$
AnswerCorrect option: D. $\big[4,12\big]$
$\text{IQ}=\Big(\frac{\text{MA}}{\text{CA}}\Big)\times100$
$\Rightarrow\text{MA}=\text{IQ}\times\frac{\text{CA}}{100}$ Given, $\text{CA}=10$ years
$40\leq\text{IQ}\leq120$
$\Rightarrow40\times\text{CA}\leq\text{IQ}\times\text{CA}\leq120\times\text{CA}$
$\Rightarrow40\times10\leq\text{IQ}\times\text{CA}\leq120\times10$
$\Rightarrow40\times10100\leq\text{Q}\times\text{CA}100\leq120\times10100$
$\Rightarrow40\leq\text{MA}\leq120$
View full question & answer→MCQ 481 Mark
If $|3 – 4\text{x}|\geq9,$ then $\text{x}\in$
- A
$\big(-\infty –3\big)\cup\big(3,\infty\big)$
- B
$\big(\infty,\frac{ -3}{2}\big]\cup\big(3 ,\infty\big)$
- C
$\big(−\infty ,\frac{-3}{2}\big]\cup\big(0,\infty\big)$
- ✓
$\big(-\infty,\frac{-3}{2}\big]\cup\big[3,\infty\big)$
AnswerCorrect option: D. $\big(-\infty,\frac{-3}{2}\big]\cup\big[3,\infty\big)$
View full question & answer→MCQ 491 Mark
The solution of the $15<\frac{3(\text{x} – 2)}{5}<0$ is:
- ✓
$27 < x < 2$
- B
$27 < x < -2$
- C
$-27 < x < 2$
- D
$27 < x < -2$
AnswerCorrect option: A. $27 < x < 2$
Given inequality is:
$15<\frac{3(\text{x} – 2)}{5}<0$
$\Rightarrow15\times5<3(\text{x} – 2)<0 × 5$
$\Rightarrow75<3(\text{x} – 2)<0$
$\Rightarrow\frac{75}{3}<\text{x} – 2<0$
$\Rightarrow25<\text{x}-2<0$
$\Rightarrow25 + 2 <\text{x}<0 + 2$
$\Rightarrow27 < x < 2$
View full question & answer→MCQ 501 Mark
The longest side of a triangle is $2$ times the shortest side and the third side is $4\ cm$ shorter than the longest side.If the perimeter of the triangle is at least $61\ cm,$ find the minimum length of the shortest side.
AnswerLet shortest side be $x.$
Then longest side $=2\text{x}.$
Third side $=2\text{x}-4.$
Given, perimeter of triangle is at least $61\ cm$
$\Rightarrow\text{x}+2\text{x}+2\text{x} - 4\geq61$
$\Rightarrow5\text{x}\geq65=\text{x}\geq13.$
Minimum length of the shortest side is $13\ cm.$
View full question & answer→MCQ 511 Mark
$\text{x} + 2\text{y}\leq8$
$\text{x}\geq0, \text{y}\geq0$
$\text{x}\leq0,\text{y}\leq0$
$ 2\text{x} + \text{y}\leq8$
$4\text{x}+5\text{y}\geq40$ - A
$(i), (iii)$ and $(v)$
- B
$(i), (iv)$ and $(v)$
- C
$(i), (iii)$ and $(iv)$
- ✓
$(i), (ii)$ and $(iv)$
AnswerCorrect option: D. $(i), (ii)$ and $(iv)$
View full question & answer→MCQ 521 Mark
The solution of the inequality $\frac{\text{x}}{4}>\frac{\text{x}}{2}+1$ will be:
- A
$\text{x}>4$
- B
$\text{x}>-4$
- ✓
$\text{x}<-4$
- D
$-4 <\text{ x} >4$
AnswerCorrect option: C. $\text{x}<-4$
Given: $\frac{\text{x}}{4}>\frac{\text{x}}{2}+1$
$\Rightarrow\frac{\text{x}}{4}-\frac{\text{x}}{2}>1$
$\Rightarrow\frac{\text{x}-2\text{x}}{4}>1$
$\Rightarrow\frac{\text{-x}}{4}>1$
$\Rightarrow-\text{x}>4$
$\Rightarrow\text{x}<-4$
View full question & answer→MCQ 531 Mark
$3\text{x}-6\geq0$ are, $...........?$
- A
right side with dotted $x = 2$
- B
left side with dotted $x = 2$
- ✓
right side with complete line $x = 2$
- D
left side with complete line $x = 2$
AnswerCorrect option: C. right side with complete line $x = 2$
$3\text{x}-6\geq0$
$\Rightarrow\text{x}\geq2.$
$(0, 0)$ 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.
View full question & answer→MCQ 541 Mark
The value of $x$ for which $|\text{x} + 1|+\sqrt{(\text{x} – 1)} = 0$
- A
$0$
- B
$1$
- C
$-1$
- ✓
No value of $x$
AnswerCorrect option: D. No value of $x$
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.
View full question & answer→MCQ 551 Mark
Solution of $ |\text{x}^2 – 10|\leq6$ is:
- A
$\big(2, 4\big)$
- B
$\big(– 4, – 2\big)$
- C
$\big(– 4, – 2\big)\cup\big(2, 4\big)$
- ✓
$\big[– 4, – 2\big]\cup\big[2, 4\big]$
AnswerCorrect option: D. $\big[– 4, – 2\big]\cup\big[2, 4\big]$
View full question & answer→MCQ 561 Mark
Sum of two rational numbers is, $.............$ number:
AnswerThe 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:
View full question & answer→MCQ 571 Mark
Solution of $2\text{x}-\frac{3}{3\text{x}}-5\geq3$ is:
- A
$\big[1,\frac{12}{7}\big]$
- ✓
$\big(\frac{5}{3},\frac{12}{7}\big]$
- C
$\big(-\infty,\frac{5}{3}\big)$
- D
$\big[\frac{2}{7},\infty\big)$
AnswerCorrect option: B. $\big(\frac{5}{3},\frac{12}{7}\big]$
View full question & answer→MCQ 581 Mark
If $|2\text{x} – 3|<|\text{x} + 5|,$ then $x$ belongs to:
- A
$\big(–3, 5\big)$
- B
$\big(5, 9\big)$
- ✓
$\big(\frac{-2}{3} ,8\big)$
- D
$\big(\frac{-8}{2} ,3\big)$
AnswerCorrect option: C. $\big(\frac{-2}{3} ,8\big)$
View full question & answer→MCQ 591 Mark
If $\alpha $ and $\beta$ are the roots of the equation $\text{x}^2-\text{x}+1=0,$ then a $2009+\beta^{2009}$ is equal to:
View full question & answer→MCQ 601 Mark
Solution of $\bigg|\text{x}+\frac{1}{\text{x}}\bigg|>2$ is:
- A
$R – (0)$
- ✓
$R – (–1, 0, 1)$
- C
$R – (1)$
- D
$R – (–1, 1)$
AnswerCorrect option: B. $R – (–1, 0, 1)$
View full question & answer→MCQ 611 Mark
If the cube roots of unity are $1,\omega$ and $\omega^2$, then the roots of the equation $(\text{x} -1)3+8=0,$ are:
- A
$-1,1+2\omega,1+2\omega^2$
- ✓
$-1,1+2\omega,1-2\omega^2$
- C
$-1,-1,-1$
- D
$-1,-1+2\omega-1-2\omega^2$
AnswerCorrect option: B. $-1,1+2\omega,1-2\omega^2$
View full question & answer→MCQ 621 Mark
Solve: $|\text{x}-1|\leq 5, |\text{x}|\geq2$
- A
$[2, 6]$
- B
$[-4, -2]$
- ✓
$[-4, -2]\cup[2, 6]$
- D
$\text{None of these}$
AnswerCorrect option: C. $[-4, -2]\cup[2, 6]$
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.
$\text{x}\in[-4, -2]\cup[2, 6]$
View full question & answer→MCQ 631 Mark
Ordered pair that satisfy the equation $x + y + 1 < 0$ is:
- A
$(0, -1)$
- B
$(-2,0)$
- ✓
$(2, -4)$
- D
Both $(B)$ and $(C)$
AnswerCorrect option: C. $(2, -4)$
Given inequation is $x + y + 1 < 0$
From option $A, 0 + (-1) + 1 < 0$
$\Rightarrow 0 < 0$ which is false
Hence, $(0, -1)$ is not a solution.
From option $B, -2 + 0 + 1 < 0$
$\Rightarrow -1 < 0$ which is true
Hence, $(-2,0)$ there is a solution.
From option $C, 2 - 4 + 1 < 0$
$\Rightarrow -1 < 0$ which is true.
Hence, $(2,-4)$ is a solution.
View full question & answer→MCQ 641 Mark
Find all pairs of consecutive odd natural numbers, both of which are larger than $10,$ such that their sum is less than $40.$
- ✓
$(11, 13), (13, 15), (15, 17), (17, 19)$
- B
$(11, 13), (13, 15), (15, 17)$
- C
$(21, 23), (23, 25), (25, 27), (27, 29)$
- D
$(15, 17), (17, 19), (19, 21), (21, 23)$
AnswerCorrect option: A. $(11, 13), (13, 15), (15, 17), (17, 19)$
View full question & answer→MCQ 651 Mark
Observe the figure given below:
The interval at which the value of $x$ lies is: - A
$\text{x}\in(-\infty,-2)$
- ✓
$\text{x}\in(-\infty,-2\big]$
- C
$\text{x}\in(-2,\infty\big]$
- D
$\text{x}\in\big[-2,\infty)$
AnswerCorrect option: B. $\text{x}\in(-\infty,-2\big]$
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, $\text{x}\in(-\infty,-2\big]$
View full question & answer→MCQ 661 Mark
If $a$ is an irrational number which is divisible by $b$ then the number $b:$
- A
- ✓
- C
May be rational or irrational
- D
AnswerIf $a$ 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, $\sqrt{2}\sqrt{3}=\sqrt{\Big(\frac{2}{3}}\Big)$
View full question & answer→MCQ 671 Mark
If $x > 7$ then which is impossible?
- A
$x > 4$
- ✓
$x < 6$
- C
$x > 9$
- D
$x < 14$
AnswerCorrect option: B. $x < 6$
$x > 7$ and $7 > 4$
$\Rightarrow x > 7 > 4$
$\Rightarrow x > 4.$
If $x > 7$ then $x$ cannot be less than $6.$
If $x = 11$ then $x > 7$ and $x > 9.$
If $x = 11$ then $x > 7$ and $x < 14.$
View full question & answer→MCQ 681 Mark
Choose the correct answer. If $|\text{x}+2|\leq9,$ then:
- A
$\text{x}\in(-7,11)$
- ✓
$\text{x}\in[-11, 7]$
- C
$\text{x}\in[-\infty,-7)\cup(11,\infty) $
- D
$\text{x}\in(-\infty,-7)\cup[11,\infty) $
AnswerCorrect option: B. $\text{x}\in[-11, 7]$
Given that $|\text{x}+2|\leq9$
$\Rightarrow-9\leq\text{x}+2\leq9$
$\Rightarrow-9-2\leq\text{x}\leq9-2[|\text{x}\leq\text{a}|]$
$\Rightarrow-11\leq\text{x}\leq7$
$\Rightarrow\text{x}\in[-11, 7]$
View full question & answer→MCQ 691 Mark
Inequations involved in the given region are$...........?$
- A
$2\text{x}+\text{y}\geq6,\text{x}\geq0,\text{y}\geq0$
- B
$2\text{x}+\text{y}>,\text{x}\geq0,\text{y}\geq0$
- C
$2\text{x}+\text{y}<6,\text{x}\geq0,\text{y}\geq0$
- ✓
$2\text{x}+\text{y}\leq6,\text{x}\geq0,\text{y}\geq0$
AnswerCorrect option: D. $2\text{x}+\text{y}\leq6,\text{x}\geq0,\text{y}\geq0$
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.$
View full question & answer→MCQ 701 Mark
If the expression $\big(\text{mx} – 1 +\frac{1}{\text{x}}\big)$ is always nonnegative, then the minimum value of $m$ must be:
- A
$\frac{-1}{2}$
- B
$0$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{4}$
View full question & answer→MCQ 711 Mark
The solution of the function $f(x) = |x| > 0$ is:
- A
$R$
- ✓
$R – (0)$
- C
$R – (1)$
- D
$R – (-1)$
AnswerCorrect option: B. $R – (0)$
Given, $f(x) = |x| > 0$
We know that modulus is non negative quantity.
So, $x \in R$ except that $x = 0$
$\Rightarrow\text{x}\in\text{R}-(0)$
This is the required solution
View full question & answer→MCQ 721 Mark
The point which does not belong to the feasible region of the $\text{LPP:}$
Minimize: $\text{Z}=60\text{x}+10\text{y}$
subject to $3\text{x}+\text{y}\geq18$
$2\text{x}+2\text{y}\geq12$
$\text{x}+2\text{y}\geq10$
$\text{x,y}\geq0$ is.
- A
$(0,8)$
- ✓
$(4,2)$
- C
$(6,2)$
- D
$(10,0)$
AnswerCorrect option: B. $(4,2)$
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.
View full question & answer→MCQ 731 Mark
If $|\text{x} – 1| > 5$ then:
- A
$\text{x}\in(-4,6)$
- B
$\text{x}\in\big[-4,6\big]$
- ✓
$\text{x}\in(-\infty,-4)\cup(6,\infty)$
- D
$\text{x}\in\big[-\infty,-4)\cup\big[6,\infty)$
AnswerCorrect option: C. $\text{x}\in(-\infty,-4)\cup(6,\infty)$
$|\text{x} – 1| > 5$
$\text{x} – 1 < - 5 $ and $\text{x} – 1 > - 5 $
$\text{x} < -4 $ and $\text{x} > 6 $
$\therefore\text{x}\in(-\infty,-4)\cup(6,\infty)$
View full question & answer→MCQ 741 Mark
Choose the correct answer. The inequality representing the following graph is:
- ✓
$|\text{x}|>5$
- B
$|\text{x}|\leq5$
- C
$|\text{x}|>5$
- D
$|\text{x}|\geq5$
AnswerCorrect option: A. $|\text{x}|>5$
The given graph represents
$x > -5$ and $x < 5$
Combining the two inequalities
$|x| > 5.$
View full question & answer→MCQ 751 Mark
If $4\text{x}+3<6\text{x}+7$ then $x$ belongs to the interval:
- A
$(2,\infty)$
- ✓
$(-2,\infty)$
- C
$(-\infty, 2)$
- D
$(-4,\infty)$
AnswerCorrect option: B. $(-2,\infty)$
$4\text{x}+3<6\text{x}+7$
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 $(–2,\infty),$
$\text{i}.\text{e}.\text{x}\in(-2,\infty)$
View full question & answer→MCQ 761 Mark
If $x$ and $a$ are real numbers such that $a > 0$ and $|x| > a$, then:
- A
$\text{x}\in(-\text{a},\infty)$
- B
$\text{x}\in[-\infty,\text{a}]$
- C
$\text{x}\in(-\text{a},\text{a})$
- ✓
$\text{x}\in(-\infty,-\text{a})\cup(\text{a},\infty)$
AnswerCorrect option: D. $\text{x}\in(-\infty,-\text{a})\cup(\text{a},\infty)$
If $x$ and a are real numbers such that $a > 0.$
$|x| > a$
$\Rightarrow x > a$ or $x < −a$
$\Rightarrow\text{x}\in(-\infty,-\text{a})\cup(\text{a},\infty)$
View full question & answer→MCQ 771 Mark
If $|x| < -5$ then the value of $x$ lies in the interval:
- A
$(-\infty,-5)$
- B
$(\infty,5)$
- C
$(-5,\infty)$
- ✓
$\text{No Solution}$
AnswerCorrect option: D. $\text{No Solution}$
Given, $|x| < -5$
Now, $\text{LHS} ≥ 0$ and $\text{RHS} < 0$
Since $\text{LHS}$ is non$-$negative and $\text{RHS}$ is negative
So, $|x| < -5$ does not posses any solution
View full question & answer→MCQ 781 Mark
If $(\text{x} – 1)(\text{x}^2 – 5\text{x} + 7)<(\text{x} – 1), $ then $x$ belongs to:
- A
$\big(1, 2\big)\cup\big(3,\infty\big)$
- B
$\big(2, 3\big)$
- ✓
$\big(-\infty,1)\cup\big(2, 3\big)$
- D
$\text{None of these}$
AnswerCorrect option: C. $\big(-\infty,1)\cup\big(2, 3\big)$
View full question & answer→MCQ 791 Mark
The value of a for which one root of the quadratic equation $(\text{a}^2-5\text{a}+3) \text{x}^2+3\text{a}-1)\text{x}+2$ is twice as large as the other, is:
- ✓
$\frac{2}{3}$
- B
$\frac{-2}{3}$
- C
$\frac{1}{3}$
- D
$\frac{-1}{3}$
AnswerCorrect option: A. $\frac{2}{3}$
View full question & answer→MCQ 801 Mark
The linear inequality representing the solution set given in Fig. is:
- A
$\text{|x|}<5$
- B
$\text{|x|}>5$
- ✓
$\text{|x|}\geq5$
- D
$\text{|x|}\geq5$
AnswerCorrect option: C. $\text{|x|}\geq5$
As according to the graph,
$x$ lies between $(-\infty,-5]$ and $[5,\infty)$
$\Rightarrow\text{x}\geq5$ or $\text{x}\leq-5$
$\Rightarrow|\text{x}|\geq5$
View full question & answer→MCQ 811 Mark
Solve: $1\leq|\text{x} – 1|\leq3$
AnswerCorrect option: C. $\big[-2,0\big]\cup\big[2,4\big]$
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
$\text{x}\in\big[-2, 0\big]\cup\big[2, 4\big]$
View full question & answer→MCQ 821 Mark
If $|\text{x} + 2|\leq9,$ then:
- A
$\text{x}\in\big(–7, 11\big)$
- ✓
$\text{x}\in\big[–11, 7\big]$
- C
$\text{x}\in\big(-\infty, –7\big)\cup\big(11,\infty\big)$
- D
$\text{x}\in\big(-\infty, -7\big)\cup\big[11,\infty\big)$
AnswerCorrect option: B. $\text{x}\in\big[–11, 7\big]$
View full question & answer→MCQ 831 Mark
Given that $x, y$ and $b$ are real numbers and $x < y, b < 0,$ then:
- ✓
$\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
- B
$\frac{\text{x}}{\text{b}}\leq\frac{\text{y}}{\text{b}}$
- C
$\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
- D
$\frac{\text{x}}{\text{b}}\geq\frac{\text{y}}{\text{b}}$
AnswerCorrect option: A. $\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
Given that $x, y$ and $b$ are real numbers and $x < y, b < 0.$
Consider, $x < y$
Divide both sides of the inequality by $“b”$
$\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}(\text{since, b<0})$
View full question & answer→MCQ 841 Mark
Solution of $\frac{\text{x}-7}{\text{x}+3}>2$ is:
- A
$(– 3,\infty)$
- B
$\big(-\infty, –13\big)$
- ✓
$(-13, –3)$
- D
$(– 13, 3)$
AnswerCorrect option: C. $(-13, –3)$
View full question & answer→MCQ 851 Mark
What is the solution set for $0<−\frac{\text{x}}{2} < 3$
- A
$(−6, 6)$
- ✓
$(−6, 0)$
- C
$(0, 6)$
- D
$(-\infty, -6)$
AnswerCorrect option: B. $(−6, 0)$
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 $(−6, 0)$
View full question & answer→MCQ 861 Mark
If $\text{x}\in\text{l},$ the solution set of the inequation $-2\leq\text{x}<3$ is:
- ✓
$−2, −1, 0, 1, 2$
- B
$−1, 0, 1, 2, 3$
- C
$0, 1, 2, 3$
- D
AnswerCorrect option: A. $−2, −1, 0, 1, 2$
$\text{x}\in\text{l},$ solution set $-2\leq\text{x}<3$
$= −2, −1, 0, 1, 2$
View full question & answer→MCQ 871 Mark
Solve the following in equations $\frac{2\text{x}+4}{\text{x}-1}\geq5$
- A
$\big(1,2\big]$
- B
$\big(1,3\big)$
- ✓
$\big(1,3\big]$
- D
$\big(1,4\big]$
AnswerCorrect option: C. $\big(1,3\big]$
$\frac{2\text{x}+4}{\text{x}-1}\geq5$
$\Rightarrow\frac{2\text{x}+4}{\text{x}-1}-5\geq0$
$\Rightarrow\frac{2\text{x}+4-5(\text{x-1})}{\text{x-1}}\geq0$
$\Rightarrow\frac{2\text{x}+4-\text{5x+5}}{\text{x-1}}\geq0$
$\Rightarrow\frac{9-3\text{x}}{\text{x-1}}\geq0$
$\Rightarrow\frac{-3\text{(x-3)}}{\text{x}-1}\geq0$
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 $\big(1, 3\big]$ View full question & answer→MCQ 881 Mark
Solution of $2x – 1 = |x + 7|$ is:
View full question & answer→MCQ 891 Mark
Let two numbers have arithmetic mean $9$ and geometric mean $4$.Then, these numbers are the roots of the quadratic equation:
- A
$x^2+ 18x + 16 = 0$
- ✓
$x^2- 18x + 16 = 0$
- C
$x^2+ 18x - 16 = 0$
- D
$x^2- 18x - 16 = 0$
AnswerCorrect option: B. $x^2- 18x + 16 = 0$
View full question & answer→MCQ 901 Mark
Find all pairs of consecutive odd positive integers both of which are smaller than $8$ such that their sum is more than $10.$
- ✓
$(5, 7)$
- B
$(3, 5), (5, 7)$
- C
$(3, 5), (5, 7), (7, 9)$
- D
$(5, 7), (7, 9)$
AnswerCorrect option: A. $(5, 7)$
Let two numbers be $x$ and $x + 2.$
$x + x + 2 > 10$
$\Rightarrow 2x > 8$
$\Rightarrow x > 4$ and $x < 8$ and $x + 2 < 8$
$\Rightarrow x < 6.$
$4 < x < 6$
$\Rightarrow x$ can be $5.$
For $x = 5, x + 2 = 7$
So, Pairs of odd consecutive positive integers are $(5, 7).$
View full question & answer→MCQ 911 Mark
The quantity of $A$ and $B$ in one day for which profit will be maximum is:
- A
$25, 30$
- B
$30, 35$
- C
$25, 25$
- ✓
$30, 30$
AnswerCorrect option: D. $30, 30$
View full question & answer→MCQ 921 Mark
If both the roots of the quadratic equation $\text{x}^2 -2\text{kx}+\text{k}^2-5=0 $ are less than $5,$ than $k$ lies in the interval:
- A
$\big[4,5\big]$
- ✓
$\big(-\infty,4\big)$
- C
$\big(6,\infty\big)$
- D
$\big(5,6\big]$
AnswerCorrect option: B. $\big(-\infty,4\big)$
View full question & answer→MCQ 931 Mark
Given that $x, y$ and $b$ are real numbers and $x < y, b > 0,$ then:
- ✓
$\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
- B
$\frac{\text{x}}{\text{b}}\leq\frac{\text{y}}{\text{b}}$
- C
$\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
- D
$\frac{\text{x}}{\text{b}}\geq\frac{\text{y}}{\text{b}}$
AnswerCorrect option: A. $\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
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.
$\therefore\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
View full question & answer→MCQ 941 Mark
If $– 3x + 17 < – 13,$ then:
- ✓
$\text{x}\in(10,\infty)$
- B
$\text{x}\in\big[10,\infty)$
- C
$\text{x}\in(-\infty,10\big]$
- D
$\text{x}\in\big[-10,10)$
AnswerCorrect option: A. $\text{x}\in(10,\infty)$
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$)$
$\Rightarrow\text{x}\in(10,\infty)$
View full question & answer→MCQ 951 Mark
All the values of $m$ for which both roots of the equation $x^2-2mx + m^2-1 = 0$ are greater than $-2$
but less than $4$ lie in the interval:
- A
$m > 3$
- ✓
$-1 < m < 3$
- C
$1 < m < 4$
- D
$– 2 < m < 0$
AnswerCorrect option: B. $-1 < m < 3$
View full question & answer→MCQ 961 Mark
A pack of coffee powder contains a mixture of $x$ gms of coffee and y gms of choco. The amount of coffee powder is greater than that of chocolate and each pack weights at least $10g$. Which of the followinginequalities describe the given condition?
AnswerCorrect option: B. $\text{x}+\text{y}\geq10$
The coffee powder is greater than choco.
hence, $\text{x}>\text{y}$
each pack is at least $10gm.$
View full question & answer→MCQ 971 Mark
Solution of $|3-\text{x}| = 3-\text{x} $ is:
- A
$\text{x}<3$
- B
$\text{x}>3$
- C
$\text{x}\geq3$
- ✓
$\text{x}\leq3$
AnswerCorrect option: D. $\text{x}\leq3$
View full question & answer→MCQ 981 Mark
If $x - 1 > -x + 7$ then which is true?
- ✓
$x > 4$
- B
$x < 4$
- C
$x > 2$
- D
$x < 2$
AnswerCorrect option: A. $x > 4$
$x - 1 > -x + 7$
$\Rightarrow 2x > 8$
$\Rightarrow x > 4.$
View full question & answer→MCQ 991 Mark
$x$ and $b$ are real numbers.If $\text{b}>0$ and $|\text{x}|>\text{b},$ then:
- A
$\text{x}\in\big(–\text{b},\infty\big)$
- B
$\text{x}\in\big(-\infty,\text{b}\big)$
- C
$\text{x}\in\big(–\text{b}, \text{b}\big)$
- ✓
$\text{x}\in(-\infty, –\text{b}\big)\cup\big(\text{b},\infty\big)$
AnswerCorrect option: D. $\text{x}\in(-\infty, –\text{b}\big)\cup\big(\text{b},\infty\big)$
View full question & answer→MCQ 1001 Mark
If $–8\leq5\text{x} – 3<7,$ then $\text{x}\in$
- A
$\big(–1, 2\big)$
- ✓
$\big[–1, 2\big)$
- C
$\big[–2,\infty\big)$
- D
$\big[-2,0\big)$
AnswerCorrect option: B. $\big[–1, 2\big)$
View full question & answer→MCQ 1011 Mark
If Ram has $x$ rupees and he pay $40$ rupees to shopkeeper then find range of $x$ if amount of money left with Ram is at least $10$ rupees is given by inequation, $.........?$
- A
$x ≥ 10$
- B
$x ≤ 10$
- C
$x ≤ 50$
- ✓
$x ≥ 50$
AnswerCorrect option: D. $x ≥ 50$
Amount left is at least $10$ rupees i.e.amount left $≥ 10.$
$x - 40 \geq 10$
$\Rightarrow x \geq 50.$
View full question & answer→MCQ 1021 Mark
Consider the linear inequations and solve them graphically: $3\text{x−y}-2>0,\text{x+y}\leq4:\text{x}>0\text{y}\geq0$ Which of the following points belong to the feasible solution region?
- A
$\Big(\frac{1}{2},0\Big)$
- B
$\Big(\frac{1}{2},\frac{1}{2}\Big)$
- C
$\Big(\frac{3}{2},\frac{5}{2}\Big)$
- ✓
View full question & answer→MCQ 1031 Mark
Choose the correct answer. If $|x - 1| > 5,$ then:
- A
$\text{x}\in(-4, 6)$
- B
$\text{x}\in[-4,6]$
- ✓
$\text{x}\in[-\infty,-4)\cup(6,\infty) $
- D
$\text{x}\in[-\infty,-4)\cup[6,\infty) $
AnswerCorrect option: C. $\text{x}\in[-\infty,-4)\cup(6,\infty) $
Given that $|x - 1| > 5$
$\Rightarrow (x - 1) < -5$ or $(x - 1) > 5$
$\Rightarrow x < -5 + 1$ or $x > 5 + 1$
$\Rightarrow x < -4$ or $x > 6$
$\Rightarrow\text{x}\in[-\infty,-4)\cup(6,\infty) $
View full question & answer→MCQ 1041 Mark
The inequality $\frac{2}{\text{x}}<3$ is true, when $x$ belongs to:
- A
$\big[\frac{2}{3},\infty\big) $
- B
$\big(-\infty,\frac{2}{3}\big)$
- ✓
$\big(-\infty ,0\big)\cup\big(\frac{2}{3},\infty\big) $
- D
$\text{None of these}$
AnswerCorrect option: C. $\big(-\infty ,0\big)\cup\big(\frac{2}{3},\infty\big) $
View full question & answer→MCQ 1051 Mark
If $7\text{x}+3<5\text{x}+9$ then $\text{x}\in$
AnswerCorrect option: C. $\big(-\infty, 3\big)$
View full question & answer→MCQ 1061 Mark
If $\frac{\text{x} – 2}{\text{x} + 5}> 2 ,$ then $\text{x}\in$
- A
$(–12, 5)$
- ✓
$(–12, –5)$
- C
$(–5, 12)$
- D
$(5, 12)$
AnswerCorrect option: B. $(–12, –5)$
View full question & answer→MCQ 1071 Mark
Solution of $11 +\frac{3}{\text{xl}}> 2 $ is:
AnswerCorrect option: C. $\big(-1, 0\big)\cup\big(0, 3\big)$
View full question & answer→MCQ 1081 Mark
Solving $ – 8 \leq 5\text{x} – 3<7, $ we get:
AnswerCorrect option: C. $ –1\leq\text{x}<2$
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
$–1\leq\text{x}<2$
View full question & answer→MCQ 1091 Mark
The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is $160\ cm,$ then:
- A
breadth $> 20\ cm$
- B
length $< 20\ cm$
- ✓
breadth $\geq 20\ cm$
- D
length $\leq 20\ cm$
AnswerCorrect option: C. breadth $\geq 20\ cm$
View full question & answer→MCQ 1101 Mark
Number of integral solutions satisfy inequality $|\text{x-3}|-|2\text{x}+5|\geq|\text{x}+8|$
View full question & answer→MCQ 1111 Mark
If $x > 7$ then $-x > -7$ is, .$............?$
AnswerIf we multiply by negative number on both sides of inequality then,
sign of inequality will change
i.e. if $x > 7$ then $(-1) x < (-1)^7$
$\Rightarrow -x < -7.$
View full question & answer→MCQ 1121 Mark
If the equations $\text{x}^2+2\text{x}+3=0 $ and $\text{ax}^2+\text{bx}+\text{c}=0,\text{abc}\in\text{R},$ I have a common root, then $a : b : c$ is.
- ✓
$1 : 2 : 3$
- B
$3 : 2 : 1$
- C
$1 : 3 : 2$
- D
$3 : 1 : 2$
AnswerCorrect option: A. $1 : 2 : 3$
View full question & answer→MCQ 1131 Mark
If $\frac{\text{lx}-2\text{l}}{\text{x}-2}\geq0$ then, $\text{x}\in$
- A
$\big[-2,\infty\big)$
- ✓
$\big(2,\infty\big)$
- C
$\big(\infty, 2\big)$
- D
$\big(-\infty, 2\big]$
AnswerCorrect option: B. $\big(2,\infty\big)$
View full question & answer→MCQ 1141 Mark
If $x$ is a positive integer and $20 x < 100$ then find solution set of $x.$
- A
$(0, 1, 2, 3, 4, 5)$
- B
$(1, 2, 3, 4, 5)$
- ✓
$(1, 2, 3, 4)$
- D
$(0, 1, 2, 3, 4)$
AnswerCorrect option: C. $(1, 2, 3, 4)$
$20\text{x}<100$
Dividing by $20$ on both sides, $\text{x}<\Big(\frac{100}{20}\Big)$
$\Rightarrow\text{x}<5$
Since $x$ is a positive integer so $x = 1, 2, 3, 4.$
View full question & answer→MCQ 1151 Mark
Choose the correct answer: $x$ and $b$ are real numbers. If $b > 0$ and $|x| > b,$ then:
- A
$\text{x}\in(-\text{b},\infty)$
- B
$\text{x}\in(\infty,-\text{b})$
- C
$\text{x}\in(-\text{b},\text{b})$
- ✓
$\text{x}\in(-\infty,-\text{b})\cup(\text{b},\infty)$
AnswerCorrect option: D. $\text{x}\in(-\infty,-\text{b})\cup(\text{b},\infty)$
Given that $|x| > b, b > 0$
$\Rightarrow x < -b$ or $x > b$
$\Rightarrow\text{x}\in(-\infty,-\text{b})\cup(\text{b,}\infty)$
View full question & answer→MCQ 1161 Mark
Solve : $2\text{x}+1>3$
- A
$\big[-1,\infty\big]$
- ✓
$(1,\infty)$
- C
$(\infty,\infty)$
- D
$(\infty,1)$
AnswerCorrect option: B. $(1,\infty)$
Given, $2\text{x}+1>3$
$\Rightarrow2\text{x}>3-1$
$\Rightarrow2\text{x}>2$
$\Rightarrow\text{x}>1$
$\Rightarrow\text{x}\in(1,\infty)$
View full question & answer→MCQ 1171 Mark
If $2x + 1 > 5$ then which is true?
- A
$x > 4$
- B
$x < 4$
- ✓
$x > 2$
- D
$x < 2$
AnswerCorrect option: C. $x > 2$
$2x + 1 > 5$
$\Rightarrow 2x > 5 - 1$
$\Rightarrow 2x > 4$
$\Rightarrow x > 2.$
View full question & answer→MCQ 1181 Mark
If $1\leq|\text{x} -2|\leq3,$ then $\text{x}\in$
- A
$\big[–1, 5\big]$
- ✓
$\big[–1, 1\big]\cup\big[3, 5\big]$
- C
$\big(–1, 0\big)\cup\big(2, 5\big)$
- D
$\big(–1, 5\big)$
AnswerCorrect option: B. $\big[–1, 1\big]\cup\big[3, 5\big]$
View full question & answer→MCQ 1191 Mark
If $x^2 < 4$ then the value of $x$ is:
- A
$(0, 2)$
- ✓
$(-2, 2)$
- C
$(-2, 0)$
- D
AnswerCorrect option: B. $(-2, 2)$
Given, $x^2 < 4$
$\Rightarrow x^2– 4 < 0$
$\Rightarrow (x – 2) \times (x + 2) < 0$
$\Rightarrow -2 < x < 2$
$\Rightarrow x \in (-2, 2)$
View full question & answer→MCQ 1201 Mark
Solve: $\frac{-1}{(|\text{x}| – 2)}\geq1$ where $\text{x}\in\text{R}, \text{x}\neq\pm2$
- A
$(-2, -1)$
- B
$(-2, 2)$
- ✓
$(-2, -1)\cup(1, 2)$
- D
$\text{None of these}$
AnswerCorrect option: C. $(-2, -1)\cup(1, 2)$
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:
$1\leq |\text{x}|<2$
$\Rightarrow\text{x}\in(-2, -1)\cup(1, 2)$ View full question & answer→MCQ 1211 Mark
Which of the following statements is correct:
- ✓
If $x > y$ and $b < 0,$ then $bx < by$
- B
If $x > y,$ then $x > 0$ and $y < 0$
- C
If $xy < 0,$ then $x > 0$ and $y > 0$
- D
AnswerCorrect option: A. If $x > y$ and $b < 0,$ then $bx < by$
View full question & answer→MCQ 1221 Mark
Rahul obtained $20$ and $25$ marks in first two tests. Find the minimum marks he should get in the third test to have an average of at least $30$ marks.
AnswerAverage 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 $45.$
View full question & answer→MCQ 1231 Mark
If $|\text{x}| = -5$ then the value of $x$ lies in the interval:
- A
$(-5,\infty)$
- B
$(5,\infty)$
- C
$(\infty,-5)$
- ✓
$\text{No Solution}$
AnswerCorrect option: D. $\text{No Solution}$
Given, $|x| = -5$
Since $|x|$ is always positive or zero.
So, it can not be negative.
Hence, given inequality has no solution.
View full question & answer→MCQ 1241 Mark
If $-2<2\text{x}-1<2$ then the value of $x$ lies in the interval:
- A
$\Big(\frac{1}{2},\frac{3}{2}\Big)$
- ✓
$\Big(\frac{-1}{2},\frac{3}{2}\Big)$
- C
$\Big(\frac{3}{2},\frac{1}{2}\Big)$
- D
$\Big(\frac{3}{2},\frac{-1}{2}\Big)$
AnswerCorrect option: B. $\Big(\frac{-1}{2},\frac{3}{2}\Big)$
Given, $-2<2\text{x}-1<2$
$\Rightarrow2+1< 2\text{x}<2+1$
$\Rightarrow-1<2\text{x}<3$
$\Rightarrow\frac{-1}{2}<\text{x}<\frac{3}{2}$
$\Rightarrow\text{x}\in\Big(\frac{-1}{2},\frac{3}{2}\Big)$
View full question & answer→MCQ 1251 Mark
If the roots of the quadratic equation $x^2+ px + q = 0$ are $\tan 30^\circ$ and $\tan 15^\circ$ then the value of $2 + q - p$ is:
View full question & answer→MCQ 1261 Mark
Consider the linear inequations and solve them graphically: $3\text{x−y}-2>0,\text{x+y}\leq4:\text{x}>0\text{y}\geq0$ Which of the following points belong to the feasible solution region?
- A
$\Big(\frac{1}{2},0\Big)$
- B
$\Big(\frac{1}{2},\frac{1}{2}\Big)$
- C
$\Big(\frac{3}{2},\frac{5}{2}\Big)$
- ✓
View full question & answer→MCQ 1271 Mark
If $\frac{\text{lx}-2\text{l}}{\text{x}-2}\geq0$ then, $\text{x}\in$
- A
$\big[2,\infty\big)$
- ✓
$\big(2,\infty\big)$
- C
$\big(\infty, 2\big)$
- D
$\big(-\infty, 2\big]$
AnswerCorrect option: B. $\big(2,\infty\big)$
View full question & answer→MCQ 1281 Mark
The marks obtained by a student of Class $XI$ in first and second terminal examination are $62$ and $48,$ respectively.Find the minimum marks he should get in the annual examination to have an average of at least $60$ marks.
View full question & answer→MCQ 1291 Mark
The inequality representing the following graph is:
- A
$\text{|x|}<3$
- ✓
$\text{|x|}\leq3$
- C
$\text{|x|}>3$
- D
$\text{|x|}\geq3$
AnswerCorrect option: B. $\text{|x|}\leq3$
As according to the graph,
$x$ lies between $−3$ and $3$
$\Rightarrow-3\leq\text{x}\leq3$
$\Rightarrow|\text{x}|\leq3$
View full question & answer→MCQ 1301 Mark
Solution of $0<|3\text{x}+1\big|<\frac{1}{3} $ is:
- A
$(\frac{-4}{9},\frac{2}{9})$
- B
$[\frac{-4}{9},\frac{-2}{9}]$
- ✓
$(\frac{-4}{9},\frac{2}{9})-(\frac{-1}{3})$
- D
$[\frac{-4}{9},\frac{-2}{9}]-(\frac{-1}{3})$
AnswerCorrect option: C. $(\frac{-4}{9},\frac{2}{9})-(\frac{-1}{3})$
View full question & answer→MCQ 1311 Mark
Choose the correct answer. Given that $x, y$ and $b$ are real numbers and $x < y, b < 0,$ then:
- A
$\frac{\text{x}}{\text{b}}<\frac{\text{y}}{\text{b}}$
- B
$\frac{\text{x}}{\text{b}}\leq\frac{\text{y}}{\text{b}}$
- ✓
$\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
- D
$\frac{\text{x}}{\text{b}}\geq\frac{\text{y}}{\text{b}}$
AnswerCorrect option: C. $\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}}$
Given that $x < y, b < 0$
$\Rightarrow\frac{\text{x}}{\text{b}}>\frac{\text{y}}{\text{b}},\text{b}<0$
View full question & answer→MCQ 1321 Mark
The sum of four numbers in $AP$ is $20.$The numbers are such that the ratio of the product of first and fourth is to the product of second and third as $2 : 3.$The greatest number is:
AnswerLet 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 $= a + 3d = 5 + 3(1) = 8$
View full question & answer→MCQ 1331 Mark
The set of values of $x$ which satisfy the inequations $5\text{x}+2<3\text{x}+8$ and $\text{x}+\frac{\text{x}+2}{\text{x}-1}<4$ is:
- A
$(-\infty, 1)$
- B
$(2, 3)$
- C
$\big(-\infty, 3)$
- ✓
$(-\infty, 1)\cup(2, 3)$
AnswerCorrect option: D. $(-\infty, 1)\cup(2, 3)$
View full question & answer→MCQ 1341 Mark
If $\text{x}^2<-4$ then the value of $x$ is:
- A
$(-2,2)$
- B
$(2,\infty)$
- C
$(-2,\infty)$
- ✓
$\text{No solution}$
AnswerCorrect option: D. $\text{No solution}$
Given, $\text{x}^2<-4$
$\Rightarrow\text{x}^2+4<0$
Which is not possible.
So, there is no solution.
View full question & answer→MCQ 1351 Mark
If $\frac{\big[\text{x} – 7\big]}{(\text{x} – 7)\geq 0}$ then:
- A
$\text{x}\in\big[7,\infty)$
- ✓
$\text{x}\in(7,\infty)$
- C
$\text{x}\in(\infty, 7)$
- D
$\text{x}\in(-\infty, 7)$
AnswerCorrect option: B. $\text{x}\in(7,\infty)$
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, $\text{x}> 7,$
i.e $\text{x}\in(7,\infty).$
View full question & answer→MCQ 1361 Mark
If $3\text{x}-7<5+\text{x},11-5\text{x}\leq1,$, then $\text{x}\in$
- A
$\big[2, 6\big]$
- B
$\big[–2, 6\big]$
- ✓
$\big[2,6\big)$
- D
$\big(-2,6\big)$
AnswerCorrect option: C. $\big[2,6\big)$
View full question & answer→MCQ 1371 Mark
If $|\text{x} + 3|\geq10,$ then:
- A
$\text{x}\in(-13,7\big)$
- B
$\text{x}\in(-13,7\big]$
- C
$\text{x}\in(-\infty -13\big]\cup \big[7,\infty)$
- ✓
$\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$
AnswerCorrect option: D. $\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$
$|\text{x} + 3|\geq10$
$\Rightarrow\text{x} + 3\leq-10$ or $\text{x}+3\geq10$
$\Rightarrow\text{x}\leq -13 $ or $\text{x}\geq7$
$\Rightarrow\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$
View full question & answer→MCQ 1381 Mark
The cost and revenue functions of a product are given by $C(x) = 20x + 4000$ and $R(x) = 60x + 2000,$ respectively, where $x$ is the number of items produced and sold. How many items must be sold to realise some profit?
- A
Less than $40$
- ✓
More than $50$
- C
Less than $50$
- D
Exactly $50$
AnswerCorrect option: B. More than $50$
View full question & answer→MCQ 1391 Mark
Choose the correct answer. If $– 3x + 17 < – 13,$ then:
- ✓
$\text{x}\in(10, \infty)$
- B
$\text{x}\in[10, \infty)$
- C
$\text{x}\in(-\infty\text{j},10]$
- D
$\text{x}\in[-10, 10)$
AnswerCorrect option: A. $\text{x}\in(10, \infty)$
Given that $- 3x + 17 < - 13$
$\Rightarrow - 3x < - 17 - 13$
$\Rightarrow -3x < - 30$
$\Rightarrow 3x > 30$
$\Rightarrow x > 10$
$\Rightarrow\text{x}\in(10, \infty)$
View full question & answer→MCQ 1401 Mark
Solve: $2\text{x}+1>3$
AnswerCorrect option: B. $\big(1,\infty\big)$
Given, $2\text{x}+1>3$
$\Rightarrow2\text{x}>3-1$
$\Rightarrow2\text{x}>2$
$\Rightarrow\text{x}>1$
$\Rightarrow\text{x}\in\big(1,\infty\big)$
View full question & answer→MCQ 1411 Mark
$y < -2$ involves region are, $...........?$
- A
above dotted line $y = -2$
- ✓
below dotted line $y = -2$
- C
above complete line $y = -2$
- D
below complete line $y = -2$
AnswerCorrect option: B. below dotted line $y = -2$
$y < -2$ does not satisfy $(0, 0)$
so, region is below $y = -2.$
Since only inequality sign given,
so dotted line $y = -2.$
View full question & answer→MCQ 1421 Mark
Value of $\text{i(iota)}$ is, $...........?$
- A
$-1$
- B
$1$
- ✓
$(-1)^\frac{1}{2}$
- D
$(-1)^\frac{1}{4}$
AnswerCorrect option: C. $(-1)^\frac{1}{2}$
Explanation: Iota is used to denote complex number.
The value of $\text{i (iota)}$ is $\sqrt{-1}$
$\text{ i.e.}(-1)^\frac{1}{2}$
View full question & answer→MCQ 1431 Mark
Find the number of real numbers in the solution set of following $\frac{2\text{x}}{5}+1<-3.$
Answer$\frac{2\text{x}}{5}+1<-3.$
$\frac{2\text{x}}{5}<-3-1$
$2\text{x}<-4\times5$
$\text{x}<-\frac{20}{2}$
$\text{x}<-10$
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 $< - 10)$
View full question & answer→MCQ 1441 Mark
Formulate the equations for the above problem.$(x$ and $y$ are the number of units of $A$ and $B$ manufactured in a day respectively$)$
- A
$15\text{x}+5\text{y}\leq10:24\text{x}+14\text{y}\geq1000$
- ✓
$15\text{x}+5\text{y}\leq600:24\text{x}+14\text{y}\geq1000$
- C
$15\text{x}+15\text{y}\leq600:24\text{x}+14\text{y}\geq1000$
- D
$15\text{x}+15\text{y}\leq10:24\text{x}+14\text{y}\geq1000$
AnswerCorrect option: B. $15\text{x}+5\text{y}\leq600:24\text{x}+14\text{y}\geq1000$
View full question & answer→MCQ 1451 Mark
$x > 5$ is, $............?$
AnswerSince $a$ variable $'x\ '$ is compared with number $'5\ '$ with inequality sign so it is called literal inequality.
View full question & answer→