Class 8-12 Mathematics Practice Questions, Solutions | ICSE ISC CBSE Math Mathematics Made Easy for School Students

$\displaystyle \textbf{Question 1. }\text{Write the solution set of the inequation }\frac{x^2}{x-2}>0.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x^2}{x-2}>0$
$\displaystyle x^2>0\text{ for all }x\neq0$
$\displaystyle \therefore \frac{x^2}{x-2}>0\text{ when }x-2>0$
$\displaystyle \therefore x>2$
$\displaystyle \text{Solution set }=(2,\infty)$
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$\displaystyle \textbf{Question 2. }\text{Write the solution set of the inequation }x+\frac{1}{x}\geq2.$
$\displaystyle \text{Answer:}$
$\displaystyle x+\frac{1}{x}\geq2$
$\displaystyle \frac{x^2-2x+1}{x}\geq0$
$\displaystyle \frac{(x-1)^2}{x}\geq0$
$\displaystyle (x-1)^2\geq0\text{ for all }x,\text{ and }x\neq0$
$\displaystyle \therefore x>0$
$\displaystyle \text{Solution set }=(0,\infty)$
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$\displaystyle \textbf{Question 3. }\text{Write the set of values of }x\text{ satisfying }(x^2-2x+1)(x-4)\geq0.$
$\displaystyle \text{Answer:}$
$\displaystyle (x^2-2x+1)(x-4)\geq0$
$\displaystyle (x-1)^2(x-4)\geq0$
$\displaystyle (x-1)^2\geq0\text{ for all }x$
$\displaystyle \therefore x=1\text{ or }x-4\geq0$
$\displaystyle \therefore x=1\text{ or }x\geq4$
$\displaystyle \text{Solution set }=\{1\}\cup[4,\infty)$
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$\displaystyle \textbf{Question 4. }\text{Write the solution set of the equation }|2-x|=x-2.$
$\displaystyle \text{Answer:}$
$\displaystyle |2-x|=x-2$
$\displaystyle |x-2|=x-2$
$\displaystyle \text{This is true when }x-2\geq0$
$\displaystyle \therefore x\geq2$
$\displaystyle \therefore \text{Solution set}=[2,\infty)$
$\\$

$\displaystyle \textbf{Question 5. }\text{Write the set of values of }x\text{ satisfying }|x-1|\leq3 \\ \text{ and }|x-1|\leq1.$
$\displaystyle \text{Answer:}$
$\displaystyle |x-1|\leq3$
$\displaystyle -3\leq x-1\leq3$
$\displaystyle -2\leq x\leq4$
$\displaystyle |x-1|\leq1$
$\displaystyle -1\leq x-1\leq1$
$\displaystyle 0\leq x\leq2$
$\displaystyle \text{Taking common values,}$
$\displaystyle 0\leq x\leq2$
$\displaystyle \therefore \text{Solution set}=[0,2]$
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$\displaystyle \textbf{Question 6. }\text{Write the solution set of the inequation }\left|\frac{1}{x}-2\right|<4.$
$\displaystyle \text{Answer:}$
$\displaystyle \left|\frac{1}{x}-2\right|<4$
$\displaystyle -4<\frac{1}{x}-2<4$
$\displaystyle -2<\frac{1}{x}<6$
$\displaystyle \text{Solving, we get }x<-\frac{1}{2}\text{ or }x>\frac{1}{6}$
$\displaystyle \text{Solution set }=\left(-\infty,-\frac{1}{2}\right)\cup\left(\frac{1}{6},\infty\right)$
$\\$

$\displaystyle \textbf{Question 7. }\text{Write the number of integral solutions of }\frac{x+2}{x^2+1}>\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+2}{x^2+1}>\frac{1}{2}$
$\displaystyle 2x+4>x^2+1$
$\displaystyle x^2-2x-3<0$
$\displaystyle (x-3)(x+1)<0$
$\displaystyle -1<x<3$
$\displaystyle \text{Integral solutions are }0,1,2$
$\displaystyle \therefore \text{Number of integral solutions }=3$
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$\displaystyle \textbf{Question 8. }\text{Write the set of values of }x\text{ satisfying }5x+2<3x+8\text{ and } \\ \frac{x+2}{x-1}<4.$
$\displaystyle \text{Answer:}$
$\displaystyle 5x+2<3x+8$
$\displaystyle \therefore x<3$
$\displaystyle \frac{x+2}{x-1}<4$
$\displaystyle \frac{x+2-4x+4}{x-1}<0$
$\displaystyle \frac{-3x+6}{x-1}<0$
$\displaystyle \frac{x-2}{x-1}>0$
$\displaystyle \therefore x<1\text{ or }x>2$
$\displaystyle \text{Combining with }x<3,\text{ we get }x<1\text{ or }2<x<3$
$\displaystyle \text{Solution set }=(-\infty,1)\cup(2,3)$
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$\displaystyle \textbf{Question 9. }\text{Write the solution set of }\left|x+\frac{1}{x}\right|>2.$
$\displaystyle \text{Answer:}$
$\displaystyle \left|x+\frac{1}{x}\right|>2$
$\displaystyle x+\frac{1}{x}>2\text{ or }x+\frac{1}{x}<-2$
$\displaystyle \frac{(x-1)^2}{x}>0\text{ or }\frac{(x+1)^2}{x}<0$
$\displaystyle \therefore x>0,\ x\neq1\text{ or }x<0,\ x\neq-1$
$\displaystyle \therefore \text{Solution set }=(-\infty,-1)\cup(-1,0)\cup(0,1)\cup(1,\infty)$
$\\$

$\displaystyle \textbf{Question 10. }\text{Write the solution set of the inequation }|x-1|\geq|x-3|.$
$\displaystyle \text{Answer:}$
$\displaystyle |x-1|\geq|x-3|$
$\displaystyle (x-1)^2\geq(x-3)^2$
$\displaystyle x^2-2x+1\geq x^2-6x+9$
$\displaystyle 4x\geq8$
$\displaystyle x\geq2$
$\displaystyle \text{Solution set }=[2,\infty)$
$\\$