88506
If \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\), then the minimum and maximum values of ' \(x\) ' are
1 \(5 / 3\) and 7 respectively
2 0 and \(16 / 3\) respectively
3 3 and 9 respectively
4 5 and 21 respectively
Explanation:
(A) : Given, \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\) Then we have to find minimum, maximum value of \(x\). \(0 \leq 3 x-5 \leq 16\) \(5 \leq 3 x \leq 21\) \(\frac{5}{3} \leq x \leq 7\) Therefore minimum and maximum value of \(x\) are \(\frac{5}{3}\) and 7 respectively.
J&K CET-2019
Linear Inequalities and Linear Programming
88507
Solve the inequality \(2 x-5 \leq \frac{(4 x-7)}{3}\).
1 \(x \in(-\infty, 4)\)
2 \(x \in(-\infty, 4)\)
3 \(x \in(-\infty, 8)\)
4 \(x \in(-\infty,-4)\)
Explanation:
(B) : Given inequality, \(2 x-5 \leq \frac{4 x-7}{3}\) On multiplying by 3 in both sides. \(3(2 x-5) \leq(4 x-7)\) \(6 x-15 \leq 4 x-7\) \(6 x-4 x \leq 15-7\) \(2 x \leq 8\) \(x \leq \frac{8}{2}\) Therefore, \(\mathrm{x} \in(-\infty, 4)\)
J&K CET-2014
Linear Inequalities and Linear Programming
88508
Solve the inequality \(3 x+2>-16,2 x-3 \leq 11\)
1 \((-6,7)\)
2 \([-6,7)\)
3 \((-6,7]\)
4 \([-6,7]\)
Explanation:
(C) : Given, inequality is \(3 x+2>-16, \quad 2 x-3 \leq 11\) Consider first inequality, \(3 x+2>-16\) \(3 x>-16-2\) \(3 x>-18\) \(x>\frac{-18}{3}\) \(x>-6\) Consider second inequality \(2 x-3 \leq 11\) \(2 x \leq 11+3\) \(2 x \leq 14\) \(x \leq \frac{14}{2}, \quad x \leq 7\) Hence, \(x \in(-6,7]\)
J&K CET-2014
Linear Inequalities and Linear Programming
88509
The number of integral solution of \(\frac{x+1}{x^{2}+2}>\frac{1}{4} \text { is }\)
1 1
2 2
3 5
4 None of these
Explanation:
(C) : Given, \(\frac{x+1}{x^{2}+2}>\frac{1}{4}\) \(x^{2}+2-4 x-4\lt 0\) \(x^{2}-4 x+4-6\lt 0\) \((x-2)^{2}-(\sqrt{6})^{2}\lt 0\) \(\{(x-2)-\sqrt{6}\}\{(x-2)+\sqrt{6}\}\lt 0\) \({[x-(2+\sqrt{6})][x-(2-\sqrt{6})]\lt 0}\) \(x \in(2-\sqrt{6}, 2+\sqrt{6})\) \(x \in[-0.4,4.4]\) \(x=0,1,2,3,4\) \(\therefore\) There are 5 integral solution of in quality.
Manipal-2019
Linear Inequalities and Linear Programming
88521
If \(7 x-2\lt 4-3 x\) and \(3 x-1\lt 2+5 x\), then \(x\) lies in the interval
1 \(\left(\frac{3}{5}, \frac{3}{2}\right)\)
2 \(\left(\frac{-3}{2}, \frac{3}{5}\right)\)
3 \(\left[-\frac{3}{2}, \frac{3}{5}\right)\)
4 \(\left[-\frac{3}{2}, \frac{3}{5}\right]\)
5 \(\left(-\frac{3}{5}, \frac{3}{2}\right)\)
Explanation:
(B) : We have, \(7 x-2\lt 4-3 x \text { and } 3 x-1\lt 2+5 x\) Therefore, \(7 \mathrm{x}+3 \mathrm{x}\lt 4+2\) And \(3 x-5 x\lt 2+1\) \(10 x\lt 6 \text { and }-2 x\lt 3\) \(x\lt \frac{3}{5} \text { and } x>-\frac{3}{2}\) Therefore, \(x \in\left(-\frac{3}{2}, \frac{3}{5}\right)\)
88506
If \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\), then the minimum and maximum values of ' \(x\) ' are
1 \(5 / 3\) and 7 respectively
2 0 and \(16 / 3\) respectively
3 3 and 9 respectively
4 5 and 21 respectively
Explanation:
(A) : Given, \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\) Then we have to find minimum, maximum value of \(x\). \(0 \leq 3 x-5 \leq 16\) \(5 \leq 3 x \leq 21\) \(\frac{5}{3} \leq x \leq 7\) Therefore minimum and maximum value of \(x\) are \(\frac{5}{3}\) and 7 respectively.
J&K CET-2019
Linear Inequalities and Linear Programming
88507
Solve the inequality \(2 x-5 \leq \frac{(4 x-7)}{3}\).
1 \(x \in(-\infty, 4)\)
2 \(x \in(-\infty, 4)\)
3 \(x \in(-\infty, 8)\)
4 \(x \in(-\infty,-4)\)
Explanation:
(B) : Given inequality, \(2 x-5 \leq \frac{4 x-7}{3}\) On multiplying by 3 in both sides. \(3(2 x-5) \leq(4 x-7)\) \(6 x-15 \leq 4 x-7\) \(6 x-4 x \leq 15-7\) \(2 x \leq 8\) \(x \leq \frac{8}{2}\) Therefore, \(\mathrm{x} \in(-\infty, 4)\)
J&K CET-2014
Linear Inequalities and Linear Programming
88508
Solve the inequality \(3 x+2>-16,2 x-3 \leq 11\)
1 \((-6,7)\)
2 \([-6,7)\)
3 \((-6,7]\)
4 \([-6,7]\)
Explanation:
(C) : Given, inequality is \(3 x+2>-16, \quad 2 x-3 \leq 11\) Consider first inequality, \(3 x+2>-16\) \(3 x>-16-2\) \(3 x>-18\) \(x>\frac{-18}{3}\) \(x>-6\) Consider second inequality \(2 x-3 \leq 11\) \(2 x \leq 11+3\) \(2 x \leq 14\) \(x \leq \frac{14}{2}, \quad x \leq 7\) Hence, \(x \in(-6,7]\)
J&K CET-2014
Linear Inequalities and Linear Programming
88509
The number of integral solution of \(\frac{x+1}{x^{2}+2}>\frac{1}{4} \text { is }\)
1 1
2 2
3 5
4 None of these
Explanation:
(C) : Given, \(\frac{x+1}{x^{2}+2}>\frac{1}{4}\) \(x^{2}+2-4 x-4\lt 0\) \(x^{2}-4 x+4-6\lt 0\) \((x-2)^{2}-(\sqrt{6})^{2}\lt 0\) \(\{(x-2)-\sqrt{6}\}\{(x-2)+\sqrt{6}\}\lt 0\) \({[x-(2+\sqrt{6})][x-(2-\sqrt{6})]\lt 0}\) \(x \in(2-\sqrt{6}, 2+\sqrt{6})\) \(x \in[-0.4,4.4]\) \(x=0,1,2,3,4\) \(\therefore\) There are 5 integral solution of in quality.
Manipal-2019
Linear Inequalities and Linear Programming
88521
If \(7 x-2\lt 4-3 x\) and \(3 x-1\lt 2+5 x\), then \(x\) lies in the interval
1 \(\left(\frac{3}{5}, \frac{3}{2}\right)\)
2 \(\left(\frac{-3}{2}, \frac{3}{5}\right)\)
3 \(\left[-\frac{3}{2}, \frac{3}{5}\right)\)
4 \(\left[-\frac{3}{2}, \frac{3}{5}\right]\)
5 \(\left(-\frac{3}{5}, \frac{3}{2}\right)\)
Explanation:
(B) : We have, \(7 x-2\lt 4-3 x \text { and } 3 x-1\lt 2+5 x\) Therefore, \(7 \mathrm{x}+3 \mathrm{x}\lt 4+2\) And \(3 x-5 x\lt 2+1\) \(10 x\lt 6 \text { and }-2 x\lt 3\) \(x\lt \frac{3}{5} \text { and } x>-\frac{3}{2}\) Therefore, \(x \in\left(-\frac{3}{2}, \frac{3}{5}\right)\)
88506
If \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\), then the minimum and maximum values of ' \(x\) ' are
1 \(5 / 3\) and 7 respectively
2 0 and \(16 / 3\) respectively
3 3 and 9 respectively
4 5 and 21 respectively
Explanation:
(A) : Given, \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\) Then we have to find minimum, maximum value of \(x\). \(0 \leq 3 x-5 \leq 16\) \(5 \leq 3 x \leq 21\) \(\frac{5}{3} \leq x \leq 7\) Therefore minimum and maximum value of \(x\) are \(\frac{5}{3}\) and 7 respectively.
J&K CET-2019
Linear Inequalities and Linear Programming
88507
Solve the inequality \(2 x-5 \leq \frac{(4 x-7)}{3}\).
1 \(x \in(-\infty, 4)\)
2 \(x \in(-\infty, 4)\)
3 \(x \in(-\infty, 8)\)
4 \(x \in(-\infty,-4)\)
Explanation:
(B) : Given inequality, \(2 x-5 \leq \frac{4 x-7}{3}\) On multiplying by 3 in both sides. \(3(2 x-5) \leq(4 x-7)\) \(6 x-15 \leq 4 x-7\) \(6 x-4 x \leq 15-7\) \(2 x \leq 8\) \(x \leq \frac{8}{2}\) Therefore, \(\mathrm{x} \in(-\infty, 4)\)
J&K CET-2014
Linear Inequalities and Linear Programming
88508
Solve the inequality \(3 x+2>-16,2 x-3 \leq 11\)
1 \((-6,7)\)
2 \([-6,7)\)
3 \((-6,7]\)
4 \([-6,7]\)
Explanation:
(C) : Given, inequality is \(3 x+2>-16, \quad 2 x-3 \leq 11\) Consider first inequality, \(3 x+2>-16\) \(3 x>-16-2\) \(3 x>-18\) \(x>\frac{-18}{3}\) \(x>-6\) Consider second inequality \(2 x-3 \leq 11\) \(2 x \leq 11+3\) \(2 x \leq 14\) \(x \leq \frac{14}{2}, \quad x \leq 7\) Hence, \(x \in(-6,7]\)
J&K CET-2014
Linear Inequalities and Linear Programming
88509
The number of integral solution of \(\frac{x+1}{x^{2}+2}>\frac{1}{4} \text { is }\)
1 1
2 2
3 5
4 None of these
Explanation:
(C) : Given, \(\frac{x+1}{x^{2}+2}>\frac{1}{4}\) \(x^{2}+2-4 x-4\lt 0\) \(x^{2}-4 x+4-6\lt 0\) \((x-2)^{2}-(\sqrt{6})^{2}\lt 0\) \(\{(x-2)-\sqrt{6}\}\{(x-2)+\sqrt{6}\}\lt 0\) \({[x-(2+\sqrt{6})][x-(2-\sqrt{6})]\lt 0}\) \(x \in(2-\sqrt{6}, 2+\sqrt{6})\) \(x \in[-0.4,4.4]\) \(x=0,1,2,3,4\) \(\therefore\) There are 5 integral solution of in quality.
Manipal-2019
Linear Inequalities and Linear Programming
88521
If \(7 x-2\lt 4-3 x\) and \(3 x-1\lt 2+5 x\), then \(x\) lies in the interval
1 \(\left(\frac{3}{5}, \frac{3}{2}\right)\)
2 \(\left(\frac{-3}{2}, \frac{3}{5}\right)\)
3 \(\left[-\frac{3}{2}, \frac{3}{5}\right)\)
4 \(\left[-\frac{3}{2}, \frac{3}{5}\right]\)
5 \(\left(-\frac{3}{5}, \frac{3}{2}\right)\)
Explanation:
(B) : We have, \(7 x-2\lt 4-3 x \text { and } 3 x-1\lt 2+5 x\) Therefore, \(7 \mathrm{x}+3 \mathrm{x}\lt 4+2\) And \(3 x-5 x\lt 2+1\) \(10 x\lt 6 \text { and }-2 x\lt 3\) \(x\lt \frac{3}{5} \text { and } x>-\frac{3}{2}\) Therefore, \(x \in\left(-\frac{3}{2}, \frac{3}{5}\right)\)
88506
If \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\), then the minimum and maximum values of ' \(x\) ' are
1 \(5 / 3\) and 7 respectively
2 0 and \(16 / 3\) respectively
3 3 and 9 respectively
4 5 and 21 respectively
Explanation:
(A) : Given, \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\) Then we have to find minimum, maximum value of \(x\). \(0 \leq 3 x-5 \leq 16\) \(5 \leq 3 x \leq 21\) \(\frac{5}{3} \leq x \leq 7\) Therefore minimum and maximum value of \(x\) are \(\frac{5}{3}\) and 7 respectively.
J&K CET-2019
Linear Inequalities and Linear Programming
88507
Solve the inequality \(2 x-5 \leq \frac{(4 x-7)}{3}\).
1 \(x \in(-\infty, 4)\)
2 \(x \in(-\infty, 4)\)
3 \(x \in(-\infty, 8)\)
4 \(x \in(-\infty,-4)\)
Explanation:
(B) : Given inequality, \(2 x-5 \leq \frac{4 x-7}{3}\) On multiplying by 3 in both sides. \(3(2 x-5) \leq(4 x-7)\) \(6 x-15 \leq 4 x-7\) \(6 x-4 x \leq 15-7\) \(2 x \leq 8\) \(x \leq \frac{8}{2}\) Therefore, \(\mathrm{x} \in(-\infty, 4)\)
J&K CET-2014
Linear Inequalities and Linear Programming
88508
Solve the inequality \(3 x+2>-16,2 x-3 \leq 11\)
1 \((-6,7)\)
2 \([-6,7)\)
3 \((-6,7]\)
4 \([-6,7]\)
Explanation:
(C) : Given, inequality is \(3 x+2>-16, \quad 2 x-3 \leq 11\) Consider first inequality, \(3 x+2>-16\) \(3 x>-16-2\) \(3 x>-18\) \(x>\frac{-18}{3}\) \(x>-6\) Consider second inequality \(2 x-3 \leq 11\) \(2 x \leq 11+3\) \(2 x \leq 14\) \(x \leq \frac{14}{2}, \quad x \leq 7\) Hence, \(x \in(-6,7]\)
J&K CET-2014
Linear Inequalities and Linear Programming
88509
The number of integral solution of \(\frac{x+1}{x^{2}+2}>\frac{1}{4} \text { is }\)
1 1
2 2
3 5
4 None of these
Explanation:
(C) : Given, \(\frac{x+1}{x^{2}+2}>\frac{1}{4}\) \(x^{2}+2-4 x-4\lt 0\) \(x^{2}-4 x+4-6\lt 0\) \((x-2)^{2}-(\sqrt{6})^{2}\lt 0\) \(\{(x-2)-\sqrt{6}\}\{(x-2)+\sqrt{6}\}\lt 0\) \({[x-(2+\sqrt{6})][x-(2-\sqrt{6})]\lt 0}\) \(x \in(2-\sqrt{6}, 2+\sqrt{6})\) \(x \in[-0.4,4.4]\) \(x=0,1,2,3,4\) \(\therefore\) There are 5 integral solution of in quality.
Manipal-2019
Linear Inequalities and Linear Programming
88521
If \(7 x-2\lt 4-3 x\) and \(3 x-1\lt 2+5 x\), then \(x\) lies in the interval
1 \(\left(\frac{3}{5}, \frac{3}{2}\right)\)
2 \(\left(\frac{-3}{2}, \frac{3}{5}\right)\)
3 \(\left[-\frac{3}{2}, \frac{3}{5}\right)\)
4 \(\left[-\frac{3}{2}, \frac{3}{5}\right]\)
5 \(\left(-\frac{3}{5}, \frac{3}{2}\right)\)
Explanation:
(B) : We have, \(7 x-2\lt 4-3 x \text { and } 3 x-1\lt 2+5 x\) Therefore, \(7 \mathrm{x}+3 \mathrm{x}\lt 4+2\) And \(3 x-5 x\lt 2+1\) \(10 x\lt 6 \text { and }-2 x\lt 3\) \(x\lt \frac{3}{5} \text { and } x>-\frac{3}{2}\) Therefore, \(x \in\left(-\frac{3}{2}, \frac{3}{5}\right)\)
88506
If \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\), then the minimum and maximum values of ' \(x\) ' are
1 \(5 / 3\) and 7 respectively
2 0 and \(16 / 3\) respectively
3 3 and 9 respectively
4 5 and 21 respectively
Explanation:
(A) : Given, \(x \in R\) and \(0 \leq(3 x-5) / 2 \leq 8\) Then we have to find minimum, maximum value of \(x\). \(0 \leq 3 x-5 \leq 16\) \(5 \leq 3 x \leq 21\) \(\frac{5}{3} \leq x \leq 7\) Therefore minimum and maximum value of \(x\) are \(\frac{5}{3}\) and 7 respectively.
J&K CET-2019
Linear Inequalities and Linear Programming
88507
Solve the inequality \(2 x-5 \leq \frac{(4 x-7)}{3}\).
1 \(x \in(-\infty, 4)\)
2 \(x \in(-\infty, 4)\)
3 \(x \in(-\infty, 8)\)
4 \(x \in(-\infty,-4)\)
Explanation:
(B) : Given inequality, \(2 x-5 \leq \frac{4 x-7}{3}\) On multiplying by 3 in both sides. \(3(2 x-5) \leq(4 x-7)\) \(6 x-15 \leq 4 x-7\) \(6 x-4 x \leq 15-7\) \(2 x \leq 8\) \(x \leq \frac{8}{2}\) Therefore, \(\mathrm{x} \in(-\infty, 4)\)
J&K CET-2014
Linear Inequalities and Linear Programming
88508
Solve the inequality \(3 x+2>-16,2 x-3 \leq 11\)
1 \((-6,7)\)
2 \([-6,7)\)
3 \((-6,7]\)
4 \([-6,7]\)
Explanation:
(C) : Given, inequality is \(3 x+2>-16, \quad 2 x-3 \leq 11\) Consider first inequality, \(3 x+2>-16\) \(3 x>-16-2\) \(3 x>-18\) \(x>\frac{-18}{3}\) \(x>-6\) Consider second inequality \(2 x-3 \leq 11\) \(2 x \leq 11+3\) \(2 x \leq 14\) \(x \leq \frac{14}{2}, \quad x \leq 7\) Hence, \(x \in(-6,7]\)
J&K CET-2014
Linear Inequalities and Linear Programming
88509
The number of integral solution of \(\frac{x+1}{x^{2}+2}>\frac{1}{4} \text { is }\)
1 1
2 2
3 5
4 None of these
Explanation:
(C) : Given, \(\frac{x+1}{x^{2}+2}>\frac{1}{4}\) \(x^{2}+2-4 x-4\lt 0\) \(x^{2}-4 x+4-6\lt 0\) \((x-2)^{2}-(\sqrt{6})^{2}\lt 0\) \(\{(x-2)-\sqrt{6}\}\{(x-2)+\sqrt{6}\}\lt 0\) \({[x-(2+\sqrt{6})][x-(2-\sqrt{6})]\lt 0}\) \(x \in(2-\sqrt{6}, 2+\sqrt{6})\) \(x \in[-0.4,4.4]\) \(x=0,1,2,3,4\) \(\therefore\) There are 5 integral solution of in quality.
Manipal-2019
Linear Inequalities and Linear Programming
88521
If \(7 x-2\lt 4-3 x\) and \(3 x-1\lt 2+5 x\), then \(x\) lies in the interval
1 \(\left(\frac{3}{5}, \frac{3}{2}\right)\)
2 \(\left(\frac{-3}{2}, \frac{3}{5}\right)\)
3 \(\left[-\frac{3}{2}, \frac{3}{5}\right)\)
4 \(\left[-\frac{3}{2}, \frac{3}{5}\right]\)
5 \(\left(-\frac{3}{5}, \frac{3}{2}\right)\)
Explanation:
(B) : We have, \(7 x-2\lt 4-3 x \text { and } 3 x-1\lt 2+5 x\) Therefore, \(7 \mathrm{x}+3 \mathrm{x}\lt 4+2\) And \(3 x-5 x\lt 2+1\) \(10 x\lt 6 \text { and }-2 x\lt 3\) \(x\lt \frac{3}{5} \text { and } x>-\frac{3}{2}\) Therefore, \(x \in\left(-\frac{3}{2}, \frac{3}{5}\right)\)
