\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{If }x<7,\text{ then}
\displaystyle \text{(a) }-x<-7\qquad \text{(b) }-x\leq-7\qquad \text{(c) }-x>-7\qquad \text{(d) }-x\geq-7
\displaystyle \text{Answer: (c) }-x>-7
\displaystyle x<7.
\displaystyle \text{Multiplying both sides by }-1\text{ reverses the sign of inequality.}
\displaystyle \therefore -x>-7.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }-3x+17<-13,\text{ then}
\displaystyle \text{(a) }x\in(10,\infty)\qquad \text{(b) }x\in[10,\infty)
\displaystyle \text{(c) }x\in(-\infty,10]\qquad \text{(d) }x\in[-10,10)
\displaystyle \text{Answer: (a) }x\in(10,\infty)
\displaystyle -3x+17<-13.
\displaystyle -3x<-30.
\displaystyle \text{Dividing both sides by }-3\text{ reverses the sign of inequality.}
\displaystyle x>10.
\displaystyle \therefore x\in(10,\infty).
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Given that }x,\ y\text{ and }b\text{ are real numbers and }x<y,\ b>0,\text{ then}
\displaystyle \text{(a) }\frac{x}{b}<\frac{y}{b}\qquad \text{(b) }\frac{x}{b}\leq\frac{y}{b}\qquad \text{(c) }\frac{x}{b}>\frac{y}{b}\qquad \text{(d) }\frac{x}{b}\geq\frac{y}{b}
\displaystyle \text{Answer: (a) }\frac{x}{b}<\frac{y}{b}
\displaystyle x<y,\qquad b>0.
\displaystyle \text{Dividing both sides by the positive number }b \text{ does not change the sign of inequality.}
\displaystyle \therefore \frac{x}{b}<\frac{y}{b}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }x\text{ is a real number and }|x|<5,\text{ then}
\displaystyle \text{(a) }x\geq5\qquad \text{(b) }-5<x<5\qquad \text{(c) }x\leq-5\qquad \text{(d) }-5\leq x\leq5
\displaystyle \text{Answer: (b) }-5<x<5
\displaystyle \text{We know that, for }a>0,
\displaystyle |x|<a\Longleftrightarrow -a<x<a.
\displaystyle \therefore |x|<5\Longleftrightarrow -5<x<5.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }x\text{ and }a\text{ are real numbers such that }a>0\text{ and }|x|>a,\text{ then}
\displaystyle \text{(a) }x\in(-a,\infty)\qquad \text{(b) }x\in[-\infty,a]
\displaystyle \text{(c) }x\in(-a,a)\qquad \text{(d) }x\in(-\infty,-a)\cup(a,\infty)
\displaystyle \text{Answer: (d) }x\in(-\infty,-a)\cup(a,\infty)
\displaystyle |x|>a.
\displaystyle \text{Since }a>0,\quad |x|>a\Longleftrightarrow x<-a\text{ or }x>a.
\displaystyle \therefore x\in(-\infty,-a)\cup(a,\infty).
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }|x-1|>5,\text{ then}
\displaystyle \text{(a) }x\in(-4,6)\qquad \text{(b) }x\in[-4,6]
\displaystyle \text{(c) }x\in(-\infty,-4)\cup(6,\infty)\qquad \text{(d) }x\in(-\infty,-4)\cup[6,\infty)
\displaystyle \text{Answer: (c) }x\in(-\infty,-4)\cup(6,\infty)
\displaystyle |x-1|>5.
\displaystyle x-1<-5\quad\text{or}\quad x-1>5.
\displaystyle x<-4\quad\text{or}\quad x>6.
\displaystyle \therefore x\in(-\infty,-4)\cup(6,\infty).
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }|x+2|\leq9,\text{ then}
\displaystyle \text{(a) }x\in(-7,11)\qquad \text{(b) }x\in[-11,7]
\displaystyle \text{(c) }x\in(-\infty,-7)\cup(11,\infty)\qquad \text{(d) }x\in(-\infty,-7)\cup[11,\infty)
\displaystyle \text{Answer: (b) }x\in[-11,7]
\displaystyle |x+2|\leq9.
\displaystyle -9\leq x+2\leq9.
\displaystyle -11\leq x\leq7.
\displaystyle \therefore x\in[-11,7].
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The linear inequality representing the solution set given in Fig.  is}
\displaystyle \text{(a) }|x|<5\qquad \text{(b) }|x|>5\qquad \text{(c) }|x|\geq5\qquad \text{(d) }|x|\leq5 \displaystyle \text{Answer: (c) }|x|\geq5
\displaystyle \text{Fig. represents the region outside the interval }[-5,5].
\displaystyle \text{The endpoints }-5\text{ and }5\text{ are included.}
\displaystyle \therefore x\leq-5\quad\text{or}\quad x\geq5.
\displaystyle \therefore |x|\geq5.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The solution set of the inequation }|x+2|\leq5\text{ is}
\displaystyle \text{(a) }(-7,5)\qquad \text{(b) }[-7,3]\qquad \text{(c) }[-5,5]\qquad \text{(d) }(-7,3)
\displaystyle \text{Answer: (b) }[-7,3]
\displaystyle |x+2|\leq5.
\displaystyle -5\leq x+2\leq5.
\displaystyle -7\leq x\leq3.
\displaystyle \therefore x\in[-7,3].
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\frac{|x-2|}{x-2}\geq0,\text{ then}
\displaystyle \text{(a) }x\in[2,\infty)\qquad \text{(b) }x\in(2,\infty)\qquad \text{(c) }x\in(-\infty,2)\qquad \text{(d) }x\in(-\infty,2]
\displaystyle \text{Answer: (b) }x\in(2,\infty)
\displaystyle \frac{|x-2|}{x-2}\geq0,\qquad x\ne2.
\displaystyle \text{For }x>2,\quad |x-2|=x-2.
\displaystyle \therefore \frac{|x-2|}{x-2}=1\geq0.
\displaystyle \text{For }x<2,\quad |x-2|=-(x-2).
\displaystyle \therefore \frac{|x-2|}{x-2}=-1<0.
\displaystyle \text{Also, }x=2\text{ is excluded since the expression is undefined.}
\displaystyle \therefore x\in(2,\infty).
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }|x+3|\geq10,\text{ then}
\displaystyle \text{(a) }x\in(-13,7]\qquad \text{(b) }x\in(-13,7)
\displaystyle \text{(c) }x\in(-\infty,-13)\cup(7,\infty)\qquad \text{(d) }x\in(-\infty,-13]\cup[7,\infty)
\displaystyle \text{Answer: (d) }x\in(-\infty,-13]\cup[7,\infty)
\displaystyle |x+3|\geq10.
\displaystyle x+3\leq-10\quad\text{or}\quad x+3\geq10.
\displaystyle x\leq-13\quad\text{or}\quad x\geq7.
\displaystyle \therefore x\in(-\infty,-13]\cup[7,\infty).
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\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 \text{The critical points are }x=0\text{ and }x=2.
\displaystyle x^2>0\text{ for }x\ne0.
\displaystyle \text{Hence, the sign of }\frac{x^2}{x-2} \text{ depends on the sign of }x-2.
\displaystyle \frac{x^2}{x-2}>0\quad\text{when}\quad x>2.
\displaystyle x=2\text{ is not defined and }x=0 \text{ does not satisfy the strict inequation.}
\displaystyle \therefore x\in(2,\infty).
\displaystyle \text{Hence, the solution set is }(2,\infty).
\displaystyle \\

\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,\qquad x\ne0.
\displaystyle x+\frac{1}{x}-2\geq0.
\displaystyle \frac{x^2-2x+1}{x}\geq0.
\displaystyle \frac{(x-1)^2}{x}\geq0.
\displaystyle \text{The critical points are }x=0\text{ and }x=1.
\displaystyle (x-1)^2\geq0\text{ for all real }x.
\displaystyle \text{For }x<0,\quad \frac{(x-1)^2}{x}<0.
\displaystyle \text{For }x>0,\quad \frac{(x-1)^2}{x}\geq0.
\displaystyle x=0\text{ is excluded since the expression is not defined.}
\displaystyle \therefore x\in(0,\infty).
\displaystyle \text{Hence, the solution set is }(0,\infty).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Write the set of values of }x\text{ satisfying the inequation}
\displaystyle (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 \text{The critical points are }x=1\text{ and }x=4.
\displaystyle (x-1)^2\geq0\text{ for all real }x.
\displaystyle \text{At }x=1,\text{ the expression is equal to }0.
\displaystyle \text{For }x\ne1,\text{ the sign depends on }x-4.
\displaystyle \text{Thus, the expression is positive for }x>4.
\displaystyle \text{Also, equality holds at }x=1\text{ and }x=4.
\displaystyle \therefore x\in\{1\}\cup[4,\infty).
\displaystyle \text{Hence, the solution set is }\{1\}\cup[4,\infty).
\displaystyle \\

\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 \text{Case I: }x\geq2.
\displaystyle |2-x|=x-2.
\displaystyle \therefore |2-x|=x-2\text{ is satisfied for every }x\geq2.
\displaystyle \text{Case II: }x<2.
\displaystyle |2-x|=2-x.
\displaystyle 2-x=x-2.
\displaystyle 2x=4.
\displaystyle x=2.
\displaystyle \text{But }x=2\text{ does not satisfy }x<2,\text{ so there is no solution in this case.}
\displaystyle \therefore x\in[2,\infty).
\displaystyle \text{Hence, the solution set is }[2,\infty).
\displaystyle \\

\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 \therefore x\in[-2,4].
\displaystyle \text{Also, }|x-1|\leq1.
\displaystyle -1\leq x-1\leq1.
\displaystyle 0\leq x\leq2.
\displaystyle \therefore x\in[0,2].
\displaystyle \text{Since both inequations must be satisfied, we take their intersection.}
\displaystyle [-2,4]\cap[0,2]=[0,2].
\displaystyle \therefore \text{The required set of values of }x\text{ is }[0,2].
\displaystyle \\

\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{Thus, }\frac{1}{x}>-2\text{ and }\frac{1}{x}<6.
\displaystyle \frac{2x+1}{x}>0\Rightarrow x\in\left(-\infty,-\frac{1}{2}\right)\cup(0,\infty).
\displaystyle \frac{1-6x}{x}<0\Rightarrow x\in(-\infty,0)\cup\left(\frac{1}{6},\infty\right).
\displaystyle \text{Taking the intersection of the two solution sets,}
\displaystyle x\in\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{1}{6},\infty\right).
\displaystyle \therefore \text{The solution set is } \left(-\infty,-\frac{1}{2}\right)\cup\left(\frac{1}{6},\infty\right).
\displaystyle \\

\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 \text{Since }x^2+1>0\text{ for all real }x,
\displaystyle 2(x+2)>x^2+1.
\displaystyle 2x+4>x^2+1.
\displaystyle x^2-2x-3<0.
\displaystyle (x-3)(x+1)<0.
\displaystyle \therefore -1<x<3.
\displaystyle \text{The integral solutions are }x=0,1,2.
\displaystyle \therefore \text{The number of integral solutions is }3.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the set of values of }x\text{ satisfying the inequations}
\displaystyle 5x+2<3x+8\text{ and }\frac{x+2}{x-1}<4.
\displaystyle \text{Answer:}
\displaystyle 5x+2<3x+8.
\displaystyle 2x<6.
\displaystyle \therefore x<3.\qquad ...(1)
\displaystyle \frac{x+2}{x-1}<4.
\displaystyle \frac{x+2-4(x-1)}{x-1}<0.
\displaystyle \frac{-3x+6}{x-1}<0.
\displaystyle \frac{x-2}{x-1}>0.
\displaystyle \text{The critical points are }x=1\text{ and }x=2.
\displaystyle \therefore x\in(-\infty,1)\cup(2,\infty).\qquad ...(2)
\displaystyle \text{Taking the intersection of (1) and (2),}
\displaystyle (-\infty,3)\cap\left[(-\infty,1)\cup(2,\infty)\right]
\displaystyle =(-\infty,1)\cup(2,3).
\displaystyle \therefore \text{The required set of values of }x\text{ is } (-\infty,1)\cup(2,3).
\displaystyle \\

\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,\qquad x\ne0.
\displaystyle x+\frac{1}{x}>2\quad\text{or}\quad x+\frac{1}{x}<-2.
\displaystyle \text{For }x+\frac{1}{x}>2,
\displaystyle \frac{x^2-2x+1}{x}>0.
\displaystyle \frac{(x-1)^2}{x}>0.
\displaystyle \therefore x\in(0,1)\cup(1,\infty).
\displaystyle \text{For }x+\frac{1}{x}<-2,
\displaystyle \frac{x^2+2x+1}{x}<0.
\displaystyle \frac{(x+1)^2}{x}<0.
\displaystyle \therefore x\in(-\infty,-1)\cup(-1,0).
\displaystyle \text{Combining the two solution sets,}
\displaystyle x\in(-\infty,-1)\cup(-1,0)\cup(0,1)\cup(1,\infty).
\displaystyle \therefore x\in\mathbb{R}-\{-1,0,1\}.
\displaystyle \text{Hence, the solution set is }\mathbb{R}-\{-1,0,1\}.
\displaystyle \\

\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 \text{Since both sides are non-negative, we can square both sides.}
\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 \therefore x\in[2,\infty).
\displaystyle \text{Hence, the solution set is }[2,\infty).
\displaystyle \\

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