NEET Test Series from KOTA - 10 Papers In MS WORD
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Linear Inequalities and Linear Programming
88526
The solution set of the in equation \(\frac{x+11}{x-3}>0\) is
1 \((-\infty,-11) \cup(3, \infty)\)
2 \((-\infty,-10) \cup(2, \infty)\)
3 \((-100,-11) \cup(1, \infty)\)
4 \((0,5) \cup(-1,0)\)
5 \((-5,0) \cup(3,7)\)
Explanation:
(A) : Given, \(\frac{x+11}{x-3}>0\) \((x+11)(x-3)>0\) \(x+11>0 \text { and }(x-3)>0\) Or \(\quad \mathrm{x}+11\lt 0\) and \(\mathrm{x}-3\lt 0\) \(\mathrm{x}>-11\) and \(\mathrm{x}>3\) Or \(\quad \mathrm{x}\lt -11\) and \(\mathrm{x}\lt 3\) \(\mathrm{x}\lt -11\) and \(\mathrm{x}>3\) \(x \in(-\infty,-11) \cup(3, \infty)\)
Kerala CEE-2010
Linear Inequalities and Linear Programming
88527
The set of admissible values of \(x\) such that \(\frac{2 x+3}{2 x-9}\lt 0\) is
(E) : We have, \(\frac{2 x+3}{2 x-9}\lt 0\) \(2 x+3\lt 0 \text { and } 2 x-9>0\) or \(\quad 2 x+3>0\) and \(2 x-9\lt 0\) and \(x \neq \frac{9}{2}\) \(x\lt -\frac{3}{2}\) and \(x>\frac{9}{2}\) or \(x>-\frac{3}{2}\) and \(x\lt \frac{9}{2}\) and \(x\) \(\neq \frac{9}{2}\) \(\left(-\frac{3}{2}, \frac{9}{2}\right)\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88528
Suppose \(a, b\) and \(c\) are real numbers such that \(\frac{a}{b}>1\) and \(\frac{a}{c}\lt 0\). Which one of the following is true?
(C) : We have, \(\frac{\mathrm{a}}{\mathrm{b}}>1 \text { and } \frac{\mathrm{a}}{\mathrm{c}}\lt 0\) We can write to conditions from these two condition as: (i) \(\quad \mathrm{a}>0\) if \(\mathrm{c}\lt 0\) and also \(\mathrm{b}>0\) \(a>c\) \((a-c)>0\) (ii) \(\quad \mathrm{a}\lt 0\) if \(\mathrm{c}>0\) and also \(\mathrm{b}\lt 0\) \(c>b\) \((c-b)>0\) \((b-c)>0\) Therefore, \((a-c)(b-c)>0\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88529
The set of all \(x\) satisfying the inequality \(\frac{4 x-1}{3 x+1} \geq 1\) is :
88526
The solution set of the in equation \(\frac{x+11}{x-3}>0\) is
1 \((-\infty,-11) \cup(3, \infty)\)
2 \((-\infty,-10) \cup(2, \infty)\)
3 \((-100,-11) \cup(1, \infty)\)
4 \((0,5) \cup(-1,0)\)
5 \((-5,0) \cup(3,7)\)
Explanation:
(A) : Given, \(\frac{x+11}{x-3}>0\) \((x+11)(x-3)>0\) \(x+11>0 \text { and }(x-3)>0\) Or \(\quad \mathrm{x}+11\lt 0\) and \(\mathrm{x}-3\lt 0\) \(\mathrm{x}>-11\) and \(\mathrm{x}>3\) Or \(\quad \mathrm{x}\lt -11\) and \(\mathrm{x}\lt 3\) \(\mathrm{x}\lt -11\) and \(\mathrm{x}>3\) \(x \in(-\infty,-11) \cup(3, \infty)\)
Kerala CEE-2010
Linear Inequalities and Linear Programming
88527
The set of admissible values of \(x\) such that \(\frac{2 x+3}{2 x-9}\lt 0\) is
(E) : We have, \(\frac{2 x+3}{2 x-9}\lt 0\) \(2 x+3\lt 0 \text { and } 2 x-9>0\) or \(\quad 2 x+3>0\) and \(2 x-9\lt 0\) and \(x \neq \frac{9}{2}\) \(x\lt -\frac{3}{2}\) and \(x>\frac{9}{2}\) or \(x>-\frac{3}{2}\) and \(x\lt \frac{9}{2}\) and \(x\) \(\neq \frac{9}{2}\) \(\left(-\frac{3}{2}, \frac{9}{2}\right)\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88528
Suppose \(a, b\) and \(c\) are real numbers such that \(\frac{a}{b}>1\) and \(\frac{a}{c}\lt 0\). Which one of the following is true?
(C) : We have, \(\frac{\mathrm{a}}{\mathrm{b}}>1 \text { and } \frac{\mathrm{a}}{\mathrm{c}}\lt 0\) We can write to conditions from these two condition as: (i) \(\quad \mathrm{a}>0\) if \(\mathrm{c}\lt 0\) and also \(\mathrm{b}>0\) \(a>c\) \((a-c)>0\) (ii) \(\quad \mathrm{a}\lt 0\) if \(\mathrm{c}>0\) and also \(\mathrm{b}\lt 0\) \(c>b\) \((c-b)>0\) \((b-c)>0\) Therefore, \((a-c)(b-c)>0\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88529
The set of all \(x\) satisfying the inequality \(\frac{4 x-1}{3 x+1} \geq 1\) is :
88526
The solution set of the in equation \(\frac{x+11}{x-3}>0\) is
1 \((-\infty,-11) \cup(3, \infty)\)
2 \((-\infty,-10) \cup(2, \infty)\)
3 \((-100,-11) \cup(1, \infty)\)
4 \((0,5) \cup(-1,0)\)
5 \((-5,0) \cup(3,7)\)
Explanation:
(A) : Given, \(\frac{x+11}{x-3}>0\) \((x+11)(x-3)>0\) \(x+11>0 \text { and }(x-3)>0\) Or \(\quad \mathrm{x}+11\lt 0\) and \(\mathrm{x}-3\lt 0\) \(\mathrm{x}>-11\) and \(\mathrm{x}>3\) Or \(\quad \mathrm{x}\lt -11\) and \(\mathrm{x}\lt 3\) \(\mathrm{x}\lt -11\) and \(\mathrm{x}>3\) \(x \in(-\infty,-11) \cup(3, \infty)\)
Kerala CEE-2010
Linear Inequalities and Linear Programming
88527
The set of admissible values of \(x\) such that \(\frac{2 x+3}{2 x-9}\lt 0\) is
(E) : We have, \(\frac{2 x+3}{2 x-9}\lt 0\) \(2 x+3\lt 0 \text { and } 2 x-9>0\) or \(\quad 2 x+3>0\) and \(2 x-9\lt 0\) and \(x \neq \frac{9}{2}\) \(x\lt -\frac{3}{2}\) and \(x>\frac{9}{2}\) or \(x>-\frac{3}{2}\) and \(x\lt \frac{9}{2}\) and \(x\) \(\neq \frac{9}{2}\) \(\left(-\frac{3}{2}, \frac{9}{2}\right)\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88528
Suppose \(a, b\) and \(c\) are real numbers such that \(\frac{a}{b}>1\) and \(\frac{a}{c}\lt 0\). Which one of the following is true?
(C) : We have, \(\frac{\mathrm{a}}{\mathrm{b}}>1 \text { and } \frac{\mathrm{a}}{\mathrm{c}}\lt 0\) We can write to conditions from these two condition as: (i) \(\quad \mathrm{a}>0\) if \(\mathrm{c}\lt 0\) and also \(\mathrm{b}>0\) \(a>c\) \((a-c)>0\) (ii) \(\quad \mathrm{a}\lt 0\) if \(\mathrm{c}>0\) and also \(\mathrm{b}\lt 0\) \(c>b\) \((c-b)>0\) \((b-c)>0\) Therefore, \((a-c)(b-c)>0\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88529
The set of all \(x\) satisfying the inequality \(\frac{4 x-1}{3 x+1} \geq 1\) is :
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Linear Inequalities and Linear Programming
88526
The solution set of the in equation \(\frac{x+11}{x-3}>0\) is
1 \((-\infty,-11) \cup(3, \infty)\)
2 \((-\infty,-10) \cup(2, \infty)\)
3 \((-100,-11) \cup(1, \infty)\)
4 \((0,5) \cup(-1,0)\)
5 \((-5,0) \cup(3,7)\)
Explanation:
(A) : Given, \(\frac{x+11}{x-3}>0\) \((x+11)(x-3)>0\) \(x+11>0 \text { and }(x-3)>0\) Or \(\quad \mathrm{x}+11\lt 0\) and \(\mathrm{x}-3\lt 0\) \(\mathrm{x}>-11\) and \(\mathrm{x}>3\) Or \(\quad \mathrm{x}\lt -11\) and \(\mathrm{x}\lt 3\) \(\mathrm{x}\lt -11\) and \(\mathrm{x}>3\) \(x \in(-\infty,-11) \cup(3, \infty)\)
Kerala CEE-2010
Linear Inequalities and Linear Programming
88527
The set of admissible values of \(x\) such that \(\frac{2 x+3}{2 x-9}\lt 0\) is
(E) : We have, \(\frac{2 x+3}{2 x-9}\lt 0\) \(2 x+3\lt 0 \text { and } 2 x-9>0\) or \(\quad 2 x+3>0\) and \(2 x-9\lt 0\) and \(x \neq \frac{9}{2}\) \(x\lt -\frac{3}{2}\) and \(x>\frac{9}{2}\) or \(x>-\frac{3}{2}\) and \(x\lt \frac{9}{2}\) and \(x\) \(\neq \frac{9}{2}\) \(\left(-\frac{3}{2}, \frac{9}{2}\right)\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88528
Suppose \(a, b\) and \(c\) are real numbers such that \(\frac{a}{b}>1\) and \(\frac{a}{c}\lt 0\). Which one of the following is true?
(C) : We have, \(\frac{\mathrm{a}}{\mathrm{b}}>1 \text { and } \frac{\mathrm{a}}{\mathrm{c}}\lt 0\) We can write to conditions from these two condition as: (i) \(\quad \mathrm{a}>0\) if \(\mathrm{c}\lt 0\) and also \(\mathrm{b}>0\) \(a>c\) \((a-c)>0\) (ii) \(\quad \mathrm{a}\lt 0\) if \(\mathrm{c}>0\) and also \(\mathrm{b}\lt 0\) \(c>b\) \((c-b)>0\) \((b-c)>0\) Therefore, \((a-c)(b-c)>0\)
Kerala CEE-2008
Linear Inequalities and Linear Programming
88529
The set of all \(x\) satisfying the inequality \(\frac{4 x-1}{3 x+1} \geq 1\) is :
