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

\displaystyle \textbf{Question 1: }\text{Solve }7x<24\text{ when }x\in N.
\displaystyle \text{(a) }\{1,2,3,4\}\qquad\text{(b) }\{0,2,3,4\}
\displaystyle \text{(c) }\{1,2,3\}\qquad\text{(d) }\{0,2,3,4\}
\displaystyle \text{Answer:}
\displaystyle 7x<24.
\displaystyle \therefore x<\frac{24}{7}.
\displaystyle \frac{24}{7}=3\frac{3}{7}.
\displaystyle \text{Since }x\in N,\text{ the possible values are }1,2,3.
\displaystyle \therefore \text{Solution set}=\{1,2,3\}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve }3-2x<9\text{ when }x\in R.\text{ Express the solution in the form of interval.}
\displaystyle \text{(a) }[-3,\infty)\qquad\text{(b) }(-3,\infty)
\displaystyle \text{(c) }[-3,\infty]\qquad\text{(d) }[-2,\infty)
\displaystyle \text{Answer:}
\displaystyle 3-2x<9.
\displaystyle -2x<6.
\displaystyle \text{Dividing by }-2,\text{ the inequality sign reverses.}
\displaystyle x>-3.
\displaystyle \therefore \text{Solution interval}=(-3,\infty).
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\frac{x-3}{x-2}>0\text{ then }x\text{ belongs to}
\displaystyle \text{(a) }(-\infty,2)\cup(3,\infty)
\displaystyle \text{(b) }(-\infty,-3)\cup(-5,\infty)
\displaystyle \text{(c) }(-\infty,2]\cup[5,\infty)
\displaystyle \text{(d) }(2,3)
\displaystyle \text{Answer:}
\displaystyle \frac{x-3}{x-2}>0.
\displaystyle \text{The critical points are }x=2\text{ and }x=3.
\displaystyle \text{For }x<2,\text{ both }x-3\text{ and }x-2\text{ are negative, so the quotient is positive.}
\displaystyle \text{For }2<x<3,\text{ the numerator is negative and denominator is positive, so the quotient is negative.}
\displaystyle \text{For }x>3,\text{ both numerator and denominator are positive, so the quotient is positive.}
\displaystyle \text{Also, }x=2\text{ is not defined and }x=3\text{ does not satisfy the strict inequality.}
\displaystyle \therefore x\in(-\infty,2)\cup(3,\infty).
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }x>0\text{ and }y<0\text{ then }(x,y)\text{ lies in}
\displaystyle \text{(a) I quadrant}\qquad\text{(b) II quadrant}
\displaystyle \text{(c) III quadrant}\qquad\text{(d) IV quadrant}
\displaystyle \text{Answer:}
\displaystyle x>0\text{ implies that the }x\text{-coordinate is positive.}
\displaystyle y<0\text{ implies that the }y\text{-coordinate is negative.}
\displaystyle \text{A point with }x>0\text{ and }y<0\text{ lies in the IV quadrant.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }-2<2x-1<2\text{ then the value of }x\text{ lies in the interval}
\displaystyle \text{(a) }\left(\frac{1}{2},\frac{3}{2}\right)\qquad\text{(b) }\left(-\frac{1}{2},\frac{3}{2}\right)
\displaystyle \text{(c) }\left(\frac{3}{2},\frac{1}{2}\right)\qquad\text{(d) }\left(\frac{3}{2},\frac{1}{2}\right)
\displaystyle \text{Answer:}
\displaystyle -2<2x-1<2.
\displaystyle \text{Adding }1\text{ throughout, we get}
\displaystyle -1<2x<3.
\displaystyle \text{Dividing throughout by }2,\text{ we get}
\displaystyle -\frac{1}{2}<x<\frac{3}{2}.
\displaystyle \therefore x\in\left(-\frac{1}{2},\frac{3}{2}\right).
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The solution of }\left|\frac{2}{x-4}\right|>1\text{ where }x\ne4\text{ is}
\displaystyle \text{(a) }(2,6)
\displaystyle \text{(b) }(2,4)\cup(4,6)
\displaystyle \text{(c) }(2,4)\cup(4,\infty)
\displaystyle \text{(d) }(-\infty,4)\cup(4,6)
\displaystyle \text{Answer:}
\displaystyle \left|\frac{2}{x-4}\right|>1.
\displaystyle \therefore \frac{2}{|x-4|}>1.
\displaystyle \therefore |x-4|<2.
\displaystyle -2<x-4<2.
\displaystyle \text{Adding }4\text{ throughout, we get}
\displaystyle 2<x<6.
\displaystyle \text{But }x\ne4.
\displaystyle \therefore x\in(2,4)\cup(4,6).
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The solution of the }0<\frac{3(x-2)}{5}<15\text{ is}
\displaystyle \text{(a) }2<x<27\qquad\text{(b) }27<x<-2
\displaystyle \text{(c) }-27<x<2\qquad\text{(d) }-27<x<-2
\displaystyle \text{Answer:}
\displaystyle 0<\frac{3(x-2)}{5}<15.
\displaystyle \text{Multiplying throughout by }5,\text{ we get}
\displaystyle 0<3(x-2)<75.
\displaystyle \text{Dividing throughout by }3,\text{ we get}
\displaystyle 0<x-2<25.
\displaystyle \text{Adding }2\text{ throughout, we get}
\displaystyle 2<x<27.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Solve: }f(x)=\frac{(x-1)(2-x)}{x-3}\geq0
\displaystyle \text{(a) }(-\infty,1]\cup(2,\infty)
\displaystyle \text{(b) }(-\infty,1]\cup[2,3)
\displaystyle \text{(c) }(-\infty,1]\cup(3,\infty)
\displaystyle \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \frac{(x-1)(2-x)}{x-3}\geq0.
\displaystyle \text{Since }2-x=-(x-2),\text{ we get}
\displaystyle -\frac{(x-1)(x-2)}{x-3}\geq0.
\displaystyle \therefore \frac{(x-1)(x-2)}{x-3}\leq0.
\displaystyle \text{The critical points are }x=1,\ 2,\ 3.
\displaystyle \text{Using the sign of the expression in the intervals determined by these points,}
\displaystyle \frac{(x-1)(x-2)}{x-3}\leq0\text{ for }x\in(-\infty,1]\cup[2,3).
\displaystyle \text{Also, }x=3\text{ is excluded because the expression is not defined there.}
\displaystyle \therefore x\in(-\infty,1]\cup[2,3).
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The solution of the inequality }3(2-x)\geq2(1-x)\text{ for real }x\text{ is:}
\displaystyle \text{(a) }x<4\qquad\text{(b) }x>4\qquad\text{(c) }x\leq4\qquad\text{(d) }x\geq4
\displaystyle \text{Answer:}
\displaystyle 3(2-x)\geq2(1-x).
\displaystyle 6-3x\geq2-2x.
\displaystyle 4\geq x.
\displaystyle \therefore x\leq4.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The solution to }|3x-1|+1<3\text{ is}
\displaystyle \text{(a) }2<x<\frac{3}{4}\qquad\text{(b) }-\frac{1}{3}<x<1
\displaystyle \text{(c) }-\frac{1}{3}<x<\frac{1}{4}\qquad\text{(d) }-3<x<3
\displaystyle \text{Answer:}
\displaystyle |3x-1|+1<3.
\displaystyle \therefore |3x-1|<2.
\displaystyle \therefore -2<3x-1<2.
\displaystyle \text{Adding }1\text{ throughout, we get}
\displaystyle -1<3x<3.
\displaystyle \text{Dividing throughout by }3,\text{ we get}
\displaystyle -\frac{1}{3}<x<1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Solve: }3x+5<x-7,\text{ when }x\text{ is a real number.}
\displaystyle \text{(a) }x<-12\qquad\text{(b) }x>-6
\displaystyle \text{(c) }x<-6\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle 3x+5<x-7.
\displaystyle 3x-x<-7-5.
\displaystyle 2x<-12.
\displaystyle \therefore x<-6.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }|x-3|<7\text{ and }x\text{ is a real number then }x\text{ belongs to}
\displaystyle \text{(a) }(4,10)\qquad\text{(b) }(-10,4)
\displaystyle \text{(c) }(4,-10)\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle |x-3|<7.
\displaystyle \therefore -7<x-3<7.
\displaystyle \text{Adding }3\text{ throughout, we get}
\displaystyle -4<x<10.
\displaystyle \therefore x\in(-4,10).
\displaystyle \text{This interval is not given in any of the options.}
\displaystyle \therefore \text{The correct option is (d) None of these.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The solution set of }-x^2>16,\ x\text{ is a real number, is..}
\displaystyle \text{(a) }(-4,4)\qquad\text{(b) }(0,4)
\displaystyle \text{(c) }(-4,0)\qquad\text{(d) Empty set}
\displaystyle \text{Answer:}
\displaystyle -x^2>16.
\displaystyle \text{Multiplying both sides by }-1,\text{ the inequality sign reverses.}
\displaystyle x^2<-16.
\displaystyle \text{But for every real number }x,\quad x^2\geq0.
\displaystyle \therefore x^2<-16\text{ is impossible for real }x.
\displaystyle \therefore \text{the solution set is the empty set.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Set of points in the second quadrant is represented by...}
\displaystyle \text{(a) }x>0\text{ and }y<0\qquad\text{(b) }x<0\text{ and }y<0
\displaystyle \text{(c) }x>0\text{ and }y>0\qquad\text{(d) }x<0\text{ and }y>0
\displaystyle \text{Answer:}
\displaystyle \text{In the second quadrant, the }x\text{-coordinate is negative and the }y\text{-coordinate is positive.}
\displaystyle \therefore x<0\text{ and }y>0.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The graph of }x<y
\displaystyle \text{(a) Contains origin}\qquad\text{(b) Contains }(-2,3)
\displaystyle \text{(c) Contains }(2,2)\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{For }(-2,3),\text{ we have }-2<3,\text{ which is true.}
\displaystyle \text{For }(0,0),\quad0<0\text{ is false.}
\displaystyle \text{For }(2,2),\quad2<2\text{ is false.}
\displaystyle \therefore (-2,3)\text{ lies in the region represented by }x<y.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }-\frac{1}{x}<\frac{2}{3}\text{ then}
\displaystyle \text{(a) }x\text{ can be }2\qquad\text{(b) }x\text{ can be }-1
\displaystyle \text{(c) }x\text{ can be }0\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{For }x=2,\quad-\frac{1}{2}<\frac{2}{3},\text{ which is true.}
\displaystyle \text{For }x=-1,\quad-\frac{1}{-1}=1<\frac{2}{3},\text{ which is false.}
\displaystyle \text{For }x=0,\text{ the expression }-\frac{1}{x}\text{ is not defined.}
\displaystyle \therefore x\text{ can be }2.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }x^2-1\geq8\text{ then }x\text{ belongs to ...}
\displaystyle \text{(a) }R-(-3,3)\qquad\text{(b) }(-3,3)
\displaystyle \text{(c) }(0,3)\qquad\text{(d) }(-3,0)
\displaystyle \text{Answer:}
\displaystyle x^2-1\geq8.
\displaystyle \therefore x^2\geq9.
\displaystyle \therefore |x|\geq3.
\displaystyle \therefore x\leq-3\text{ or }x\geq3.
\displaystyle \therefore x\in(-\infty,-3]\cup[3,\infty).
\displaystyle \text{This can also be written as }R-(-3,3).
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The open upper half plane of }x\text{-axis can be expressed as....}
\displaystyle \text{(a) }x>0\text{ and }y<0\qquad\text{(b) }x>0
\displaystyle \text{(c) }y>0\qquad\text{(d) }x<0
\displaystyle \text{Answer:}
\displaystyle \text{The upper half plane consists of all points lying above the }x\text{-axis.}
\displaystyle \text{For every such point, the }y\text{-coordinate is positive.}
\displaystyle \therefore y>0.
\displaystyle \text{Since the half plane is open, the }x\text{-axis itself is not included.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The word ``linear'' stands for...}
\displaystyle \text{(a) One term}\qquad\text{(b) One power}
\displaystyle \text{(c) One degree}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{A linear expression or equation has the highest power of the variable equal to }1.
\displaystyle \therefore \text{its degree is }1.
\displaystyle \therefore \text{The word ``linear'' stands for one degree.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }|x-3|<0\text{ then solution set of }x\text{ is.....}
\displaystyle \text{(a) }(0,3)\qquad\text{(b) }(-3,0)
\displaystyle \text{(c) }(-3,3)\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle |x-3|\geq0\text{ for every real value of }x.
\displaystyle \therefore |x-3|<0\text{ is not possible for any real }x.
\displaystyle \therefore \text{the solution set is }\varnothing.
\displaystyle \text{Since the empty set is not given among the options,}
\displaystyle \therefore \text{The correct option is (d) None of these.}
\displaystyle \\

\displaystyle \text{SOURCE BASED / CASE STUDY QUESTIONS}

\displaystyle \textbf{Question 21: }\text{A beaker contains }640\text{ litres of }8\%\text{ solution of boric acid. This is to be}
\displaystyle \text{diluted by adding }2\%\text{ solution of boric acid to it. Based on this information answer the}
\displaystyle \text{following questions.}
\displaystyle \text{1. If }x\text{ litres of }2\%\text{ boric acid solution is added to the beaker, find the quantity of acid}
\displaystyle \text{in the resulting solution.}
\displaystyle \text{2. Find the initial quantity of water in the beaker.}
\displaystyle \text{3. Find the initial quantity of Boric acid in the beaker.}
\displaystyle \text{Answer:}
\displaystyle \text{Initially, the beaker contains }640\text{ litres of }8\%\text{ boric acid solution.}
\displaystyle \text{Initial quantity of boric acid}=\frac{8}{100}\times640=51.2\text{ litres}.
\displaystyle \text{Initial quantity of water}=640-51.2=588.8\text{ litres}.
\displaystyle \text{1. Boric acid in }x\text{ litres of }2\%\text{ solution}=\frac{2}{100}x=0.02x\text{ litres}.
\displaystyle \therefore \text{quantity of acid in the resulting solution}=51.2+0.02x\text{ litres}.
\displaystyle \text{2. The initial quantity of water in the beaker is }588.8\text{ litres}.
\displaystyle \text{3. The initial quantity of Boric acid in the beaker is }51.2\text{ litres}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Kelvin (K), degree Celsius }(^\circ C)\text{ and degree Fahrenheit }(^\circ F)\text{ are three}
\displaystyle \text{units of temperature. The conversion formula for them is as follows:}
\displaystyle F=\frac{9}{5}C+32\qquad\text{and}\qquad K=C+273.15
\displaystyle \text{Based on the above information, answer the following questions.}

\displaystyle \text{1. To maintain the Celsius temperature of a system at least }5^\circ C,\text{ what minimum}
\displaystyle \text{Fahrenheit temperature should be maintained?}
\displaystyle \text{Answer:}
\displaystyle C\geq5.
\displaystyle F=\frac{9}{5}C+32.
\displaystyle \therefore F\geq\frac{9}{5}(5)+32=9+32=41.
\displaystyle \therefore \text{the minimum Fahrenheit temperature is }41^\circ F.
\displaystyle \\

\displaystyle \text{2. To maintain Kelvin temperature of a system maximum }100\text{ K, what maximum Celsius}
\displaystyle \text{temperature should be maintained?}
\displaystyle \text{Answer:}
\displaystyle K\leq100.
\displaystyle K=C+273.15.
\displaystyle \therefore C+273.15\leq100.
\displaystyle \therefore C\leq100-273.15=-173.15.
\displaystyle \therefore \text{the maximum Celsius temperature is }-173.15^\circ C.
\displaystyle \\

\displaystyle \text{3. Find the Celsius temperature (up to }1^{st}\text{ place after the decimal) for which Kelvin}
\displaystyle \text{and Fahrenheit temperatures are equal.}
\displaystyle \text{Answer:}
\displaystyle K=F.
\displaystyle C+273.15=\frac{9}{5}C+32.
\displaystyle 273.15-32=\frac{9}{5}C-C.
\displaystyle 241.15=\frac{4}{5}C.
\displaystyle C=241.15\times\frac{5}{4}=301.4375.
\displaystyle \therefore C\approx301.4^\circ C.
\displaystyle \text{Hence, Kelvin and Fahrenheit temperatures are equal at approximately }301.4^\circ C.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In three examinations (each of }100\text{ marks), Amit and Arti scored the}
\displaystyle \text{following marks.}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Name}&\text{Test-1}&\text{Test-2}&\text{Test-3}\\ \hline \text{Amit}&60&54&x\\ \hline \text{Arti}&50&84&y\\ \hline \end{array}
\displaystyle \text{Study the above table and answer the following answers.}

\displaystyle \text{1. To attain an average of at least }65\text{ marks, how marks Amit must score in the third test?}
\displaystyle \text{Answer:}
\displaystyle \text{Let Amit score }x\text{ marks in the third test.}
\displaystyle \frac{60+54+x}{3}\geq65.
\displaystyle 114+x\geq195.
\displaystyle \therefore x\geq81.
\displaystyle \therefore \text{Amit must score at least }81\text{ marks in the third test.}
\displaystyle \\

\displaystyle \text{2. The average of Amit is greater than the average of Arti, establish relationship between }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Average marks of Amit}=\frac{60+54+x}{3}=\frac{114+x}{3}.
\displaystyle \text{Average marks of Arti}=\frac{50+84+y}{3}=\frac{134+y}{3}.
\displaystyle \text{Since Amit's average is greater than Arti's average,}
\displaystyle \frac{114+x}{3}>\frac{134+y}{3}.
\displaystyle 114+x>134+y.
\displaystyle \therefore x-y>20.
\displaystyle \text{Thus, the required relationship is }x-y>20\text{ or }x>y+20.
\displaystyle \\

\displaystyle \text{3. To get A grade, one must attain an average of at least }80\text{ marks. Find the minimum marks}
\displaystyle \text{Arti should score to get A grade.}
\displaystyle \text{Answer:}
\displaystyle \text{Let Arti score }y\text{ marks in the third test.}
\displaystyle \frac{50+84+y}{3}\geq80.
\displaystyle 134+y\geq240.
\displaystyle \therefore y\geq106.
\displaystyle \text{But the third test is out of }100\text{ marks, so }y\leq100.
\displaystyle \text{Therefore, it is not possible for Arti to attain an average of at least }80\text{ marks}
\displaystyle \text{with the given scores. Even if she scores }100,\text{ her average will be}
\displaystyle \frac{50+84+100}{3}=78.
\displaystyle \therefore \text{Arti cannot get A grade on the basis of these three test scores.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{IQ of a person is calculated by the formula }\mathrm{IQ}=\frac{M}{C}\times100
\displaystyle \text{where }M\text{ is the mental age and }C\text{ is the chronological age of a person. Answer the following questions.}

\displaystyle \text{1. If IQ of a person of Mental age }30\text{ years is }60,\text{ find the chronological age of the person.}
\displaystyle \text{Answer:}
\displaystyle \mathrm{IQ}=\frac{M}{C}\times100.
\displaystyle 60=\frac{30}{C}\times100.
\displaystyle 60C=3000.
\displaystyle \therefore C=50.
\displaystyle \therefore \text{the chronological age of the person is }50\text{ years.}
\displaystyle \\

\displaystyle \text{2. A person's Chronological age is }50\text{ years, to have IQ at least }70,\text{ what minimum }
\displaystyle \text{Mental age he should have in completed years.}
\displaystyle \text{Answer:}
\displaystyle C=50\text{ and }\mathrm{IQ}\geq70.
\displaystyle \frac{M}{50}\times100\geq70.
\displaystyle 2M\geq70.
\displaystyle \therefore M\geq35.
\displaystyle \therefore \text{the minimum mental age should be }35\text{ completed years.}
\displaystyle \\

\displaystyle \text{3. Mental age of a person is }30\text{ years. To have IQ between }50\text{ and }60,\text{ what his}
\displaystyle \text{chronological age should be?}
\displaystyle \text{Answer:}
\displaystyle M=30\text{ and }50<\mathrm{IQ}<60.
\displaystyle 50<\frac{30}{C}\times100<60.
\displaystyle 50<\frac{3000}{C}<60.
\displaystyle \text{Since }C>0,\text{ multiplying throughout by }C,\text{ we get}
\displaystyle 50C<3000<60C.
\displaystyle \text{From }50C<3000,\quad C<60.
\displaystyle \text{From }3000<60C,\quad C>50.
\displaystyle \therefore 50<C<60.
\displaystyle \therefore \text{the chronological age should be between }50\text{ and }60\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Consider the following system of linear inequalities }
\displaystyle x-2y\leq0,\ 2x+y\leq4, \ y\leq2,\ x\geq0.\text{ Based on the above information,} \\ \text{answer the following questions:}

\displaystyle \text{1. The graph of }x-2y\leq0\text{ contains the origin?}
\displaystyle \text{Answer:}
\displaystyle \text{Substituting }(0,0)\text{ in }x-2y\leq0,\text{ we get}
\displaystyle 0-2(0)\leq0.
\displaystyle 0\leq0,\text{ which is true.}
\displaystyle \therefore \text{the graph of }x-2y\leq0\text{ contains the origin.}
\displaystyle \\

\displaystyle \text{2. In which quadrants the solution region cannot lie.}
\displaystyle \text{Answer:}
\displaystyle x-2y\leq0\implies x\leq2y\implies y\geq\frac{x}{2}.
\displaystyle \text{Also, }x\geq0.
\displaystyle \text{Since }x\geq0\text{ and }y\geq\frac{x}{2}\geq0,\text{ we have }y\geq0.
\displaystyle \therefore \text{the solution region can lie only in the first quadrant and on the axes.}
\displaystyle \therefore \text{it cannot lie in the second, third and fourth quadrants.}
\displaystyle \\

\displaystyle \text{3. Show the region }2x+y\leq4.
\displaystyle \text{Answer:}
\displaystyle \text{First draw the boundary line }2x+y=4.
\displaystyle \text{Putting }x=0,\text{ we get }y=4,\text{ so the line passes through }(0,4).
\displaystyle \text{Putting }y=0,\text{ we get }x=2,\text{ so the line passes through }(2,0).
\displaystyle \text{Now test the point }(0,0).
\displaystyle 2(0)+0\leq4\implies0\leq4,\text{ which is true.}
\displaystyle \therefore \text{the required region is the half-plane containing the origin, including the line }2x+y=4.
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER TYPE QUESTIONS}

\displaystyle \textbf{Question 26: }\text{Minimum value of }x-3\text{ is }7,\text{ find minimum value of }x.
\displaystyle \text{Answer:}
\displaystyle x-3\geq7.
\displaystyle \therefore x\geq10.
\displaystyle \therefore \text{the minimum value of }x\text{ is }10.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Maximum value of }2m-3\text{ is }7;\text{ find the maximum value of }m.
\displaystyle \text{Answer:}
\displaystyle 2m-3\leq7.
\displaystyle 2m\leq10.
\displaystyle \therefore m\leq5.
\displaystyle \therefore \text{the maximum value of }m\text{ is }5.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }-3<m<7,\text{ find the minimum value of }|2m+10|.
\displaystyle \text{Answer:}
\displaystyle -3<m<7.
\displaystyle \text{Multiplying throughout by }2,\text{ we get}
\displaystyle -6<2m<14.
\displaystyle \text{Adding }10\text{ throughout, we get}
\displaystyle 4<2m+10<24.
\displaystyle \therefore |2m+10|>4.
\displaystyle \text{Since }m=-3\text{ is not included, the value }4\text{ is not attained.}
\displaystyle \therefore \text{there is no minimum value; the infimum is }4.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }|x+2|\leq9,\text{ find interval of }x.
\displaystyle \text{Answer:}
\displaystyle |x+2|\leq9.
\displaystyle \therefore -9\leq x+2\leq9.
\displaystyle \text{Subtracting }2\text{ throughout, we get}
\displaystyle -11\leq x\leq7.
\displaystyle \therefore x\in[-11,7].
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }x\geq-3,\text{ find the minimum value of }x+5.
\displaystyle \text{Answer:}
\displaystyle x\geq-3.
\displaystyle \text{Adding }5\text{ on both sides, we get}
\displaystyle x+5\geq2.
\displaystyle \therefore \text{the minimum value of }x+5\text{ is }2.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{If }x^2\leq9,\text{ find the interval of }x.
\displaystyle \text{Answer:}
\displaystyle x^2\leq9.
\displaystyle \therefore |x|\leq3.
\displaystyle \therefore -3\leq x\leq3.
\displaystyle \therefore x\in[-3,3].
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If }|x|-3<8,\text{ find the interval of }x.
\displaystyle \text{Answer:}
\displaystyle |x|-3<8.
\displaystyle \therefore |x|<11.
\displaystyle \therefore -11<x<11.
\displaystyle \therefore x\in(-11,11).
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If }-11<4x-3\leq13,\text{ find interval of }x.
\displaystyle \text{Answer:}
\displaystyle -11<4x-3\leq13.
\displaystyle \text{Adding }3\text{ throughout, we get}
\displaystyle -8<4x\leq16.
\displaystyle \text{Dividing throughout by }4,\text{ we get}
\displaystyle -2<x\leq4.
\displaystyle \therefore x\in(-2,4].
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Represent solution set of }x-3<6\text{ on number line.}
\displaystyle \text{Answer:}
\displaystyle x-3<6.
\displaystyle \text{Adding }3\text{ on both sides, we get}
\displaystyle x<9.
\displaystyle \therefore \text{Solution set}=(-\infty,9).
\displaystyle \text{On the number line, mark an open circle at }9\text{ and shade the region to its left.}\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Represent solution set of }\frac{x}{-2}<3\text{ on number line.}
\displaystyle \text{Answer:}
\displaystyle \frac{x}{-2}<3.
\displaystyle \text{Multiplying both sides by }-2,\text{ the inequality sign reverses.}
\displaystyle x>-6.
\displaystyle \therefore \text{Solution set}=(-6,\infty).
\displaystyle \text{On the number line, mark an open circle at }-6\text{ and shade the region to its right.}\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Represent solution set of }x^2\leq4\text{ on number line.}
\displaystyle \text{Answer:}
\displaystyle x^2\leq4.
\displaystyle \therefore |x|\leq2.
\displaystyle \therefore -2\leq x\leq2.
\displaystyle \therefore \text{Solution set}=[-2,2].
\displaystyle \text{On the number line, mark closed circles at }-2\text{ and }2\text{ and shade the region between them.}\displaystyle \\

\displaystyle \textbf{Question 37: }\text{How many points are there in the solution set of }x^2<0,\ x\in R?
\displaystyle \text{Answer:}
\displaystyle x^2\geq0\text{ for every real number }x.
\displaystyle \therefore x^2<0\text{ has no real solution.}
\displaystyle \therefore \text{Solution set}=\varnothing.
\displaystyle \therefore \text{the number of points in the solution set is }0.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{How many integral solutions are there for the inequation }-5<3x<6\text{?}
\displaystyle \text{Answer:}
\displaystyle -5<3x<6.
\displaystyle \text{Dividing throughout by }3,\text{ we get}
\displaystyle -\frac{5}{3}<x<2.
\displaystyle \text{The integral values of }x\text{ are }-1,0,1.
\displaystyle \therefore \text{the number of integral solutions is }3.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Write any one solution of }x-2y<-6.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=0\text{ and }y=4.
\displaystyle x-2y=0-2(4)=-8<-6.
\displaystyle \therefore (0,4)\text{ is one solution of the given inequality.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Write one solution of }|x-3|<y.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=3\text{ and }y=1.
\displaystyle |x-3|=|3-3|=0<1.
\displaystyle \therefore (3,1)\text{ is one solution of the given inequality.}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Represent first quadrant in terms of inequalities of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{In the first quadrant, both }x\text{-coordinate and }y\text{-coordinate are positive.}
\displaystyle \therefore x>0,\qquad y>0.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Represent the first and fourth quadrants together in terms of inequalities}
\displaystyle \text{of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{In both the first and fourth quadrants, the }x\text{-coordinate is positive.}
\displaystyle \text{The }y\text{-coordinate may be positive or negative.}
\displaystyle \therefore x>0,\qquad y\in R,\ y\ne0.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{Represent third quadrant in terms of inequalities of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{In the third quadrant, both }x\text{-coordinate and }y\text{-coordinate are negative.}
\displaystyle \therefore x<0,\qquad y<0.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Write the interval in which }\sin^2x\text{ lies for any real }x.
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin x\leq1.
\displaystyle \therefore 0\leq\sin^2x\leq1.
\displaystyle \therefore \sin^2x\in[0,1].
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Write the interval in which }\sin x+3\text{ lies for any real }x.
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin x\leq1.
\displaystyle \text{Adding }3\text{ throughout, we get}
\displaystyle 2\leq\sin x+3\leq4.
\displaystyle \therefore \sin x+3\in[2,4].
\displaystyle \\

\displaystyle \text{ASSERTION REASONING TYPE QUESTIONS}

\displaystyle \textbf{Question 46: }\text{Assertion: }|3x-5|>9\text{ implies }x\in\left(-\infty,-\frac{4}{3}\right)\cup\left(\frac{14}{3},\infty\right).
\displaystyle \text{Reason: The region containing all the solutions of inequality is called the solution region.}
\displaystyle \text{(a) Assertion and reason both are correct statements and reason is the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(b) Assertion and reason both are correct statements and reason is not the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(c) Assertion is correct statement and reason is wrong statement.}
\displaystyle \text{(d) Reason is correct statement and assertion is wrong statement.}
\displaystyle \text{Answer:}
\displaystyle |3x-5|>9.
\displaystyle \therefore 3x-5>9\quad\text{or}\quad3x-5<-9.
\displaystyle \therefore 3x>14\quad\text{or}\quad3x<-4.
\displaystyle \therefore x>\frac{14}{3}\quad\text{or}\quad x<-\frac{4}{3}.
\displaystyle \therefore x\in\left(-\infty,-\frac{4}{3}\right)\cup\left(\frac{14}{3},\infty\right).
\displaystyle \text{Hence, the Assertion is correct.}
\displaystyle \text{The Reason is also correct, as the region containing all solutions of an inequality is called}
\displaystyle \text{the solution region, but it does not explain the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Assertion: If }11x-9\leq68,\text{ then }x\in(-\infty,7].
\displaystyle \text{Reason: If an inequality consists of signs }\leq\text{ or }\geq,\text{ then the points on the line are also}
\displaystyle \text{included in the solution region.}
\displaystyle \text{(a) Assertion and reason both are correct statements and reason is the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(b) Assertion and reason both are correct statements and reason is not the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(c) Assertion is correct statement and reason is wrong statement.}
\displaystyle \text{(d) Reason is correct statement and assertion is wrong statement.}
\displaystyle \text{Answer:}
\displaystyle 11x-9\leq68.
\displaystyle \therefore 11x\leq77.
\displaystyle \therefore x\leq7.
\displaystyle \therefore x\in(-\infty,7].
\displaystyle \text{Hence, the Assertion is correct.}
\displaystyle \text{The Reason is also correct because the sign }\leq\text{ includes the boundary point } \\ x=7.
\displaystyle \text{However, the Reason explains only why }7\text{ is included, not why the complete solution is } \\ x\leq7.
\displaystyle \therefore \text{the Reason is not the correct explanation of the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{Assertion: If }x\geq-3,\text{ then }x+5\geq2.
\displaystyle \text{Reason: Same number can be added to both sides of the inequality without changing the sign}
\displaystyle \text{of inequality.}
\displaystyle \text{(a) Assertion and reason both are correct statements and reason is the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(b) Assertion and reason both are correct statements and reason is not the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(c) Assertion is correct statement and reason is wrong statement.}
\displaystyle \text{(d) Reason is correct statement and assertion is wrong statement.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x\geq-3.
\displaystyle \text{Adding }5\text{ to both sides, we get}
\displaystyle x+5\geq-3+5.
\displaystyle \therefore x+5\geq2.
\displaystyle \text{Hence, the Assertion is correct.}
\displaystyle \text{The Reason is also correct because adding the same number to both sides does not change}
\displaystyle \text{the sign of an inequality. It correctly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Assertion: If }a<b,\ c<0,\text{ then }\frac{a}{c}>\frac{b}{c}.
\displaystyle \text{Reason: If both sides are divided by the same negative quantity, then the inequality} \\ \text{is reversed.}
\displaystyle \text{(a) Assertion and reason both are correct statements and reason is the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(b) Assertion and reason both are correct statements and reason is not the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(c) Assertion is correct statement and reason is wrong statement.}
\displaystyle \text{(d) Reason is correct statement and assertion is wrong statement.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a<b\text{ and }c<0.
\displaystyle \text{Dividing both sides of }a<b\text{ by the negative number }c,\text{ the inequality sign reverses.}
\displaystyle \therefore \frac{a}{c}>\frac{b}{c}.
\displaystyle \text{Hence, the Assertion is correct.}
\displaystyle \text{The Reason is also correct and it correctly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Assertion: If }-5\leq2x+9\leq2,\text{ then }x\in[-7,-3.5].
\displaystyle \text{Reason: The graphical representation of }-5\leq2x+9\leq2\text{ shows the interval }[-7,-3.5].
\displaystyle \text{(a) Assertion and reason both are correct statements and reason is the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(b) Assertion and reason both are correct statements and reason is not the correct} \\ \text{explanation of the assertion.}
\displaystyle \text{(c) Assertion is correct statement and reason is wrong statement.}
\displaystyle \text{(d) Reason is correct statement and assertion is wrong statement.}
\displaystyle \text{Answer:}
\displaystyle -5\leq2x+9\leq2.
\displaystyle \text{Subtracting }9\text{ throughout, we get}
\displaystyle -14\leq2x\leq-7.
\displaystyle \text{Dividing throughout by }2,\text{ we get}
\displaystyle -7\leq x\leq-\frac{7}{2}.
\displaystyle \therefore x\in[-7,-3.5].
\displaystyle \text{Hence, the Assertion is correct.}
\displaystyle \text{The graphical representation also correctly shows the closed interval from }-7\text{ to }-3.5.\displaystyle \text{Thus, the Reason is correct and correctly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{SHORT ANSWER TYPE QUESTIONS }

\displaystyle \textbf{Question 51: }\text{Solve }2m-3<-8\text{ when }m\text{ is a natural number.}
\displaystyle \text{Answer:}
\displaystyle 2m-3<-8.
\displaystyle \therefore 2m<-5.
\displaystyle \therefore m<-\frac{5}{2}.
\displaystyle \text{But }m\text{ is a natural number, so no natural number satisfies the inequality.}
\displaystyle \therefore \text{Solution set}=\varnothing.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Solve: }(m-3)(m+2)<0\text{ when }m\text{ is a real number.}
\displaystyle \text{Answer:}
\displaystyle (m-3)(m+2)<0.
\displaystyle \text{The critical points are }m=-2\text{ and }m=3.
\displaystyle \text{The product is negative when }m\text{ lies between the two critical points.}
\displaystyle \therefore -2<m<3.
\displaystyle \therefore \text{Solution set}=(-2,3).
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Solve: }|m-5|<6.
\displaystyle \text{Answer:}
\displaystyle |m-5|<6.
\displaystyle \therefore -6<m-5<6.
\displaystyle \text{Adding }5\text{ throughout, we get}
\displaystyle -1<m<11.
\displaystyle \therefore \text{Solution set}=(-1,11).
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{Solve }4x+3<5x+7\text{ if }x\text{ is a natural number.}
\displaystyle \text{Answer:}
\displaystyle 4x+3<5x+7.
\displaystyle 3-7<5x-4x.
\displaystyle -4<x.
\displaystyle \therefore x>-4.
\displaystyle \text{Since }x\text{ is a natural number, all natural numbers satisfy the inequality.}
\displaystyle \therefore \text{Solution set}=N.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Solve: }\frac{x-3}{x+5}<0,\text{ if }x\text{ is a real number.}
\displaystyle \text{Answer:}
\displaystyle \frac{x-3}{x+5}<0.
\displaystyle \text{The critical points are }x=-5\text{ and }x=3.
\displaystyle \text{For }x<-5,\text{ both numerator and denominator are negative, so the quotient is positive.}
\displaystyle \text{For }-5<x<3,\text{ the numerator is negative and denominator is positive, so the quotient is negative.}
\displaystyle \text{For }x>3,\text{ both numerator and denominator are positive, so the quotient is positive.}
\displaystyle \text{Also, }x=-5\text{ is not defined and }x=3\text{ does not satisfy the strict inequality.}
\displaystyle \therefore -5<x<3.
\displaystyle \therefore \text{Solution set}=(-5,3).
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Solve: }x+y<5\text{ graphically. (Not in syllabus).}
\displaystyle \text{Answer:}
\displaystyle \text{First draw the boundary line }x+y=5.
\displaystyle \text{Putting }x=0,\text{ we get }y=5.
\displaystyle \text{Putting }y=0,\text{ we get }x=5.
\displaystyle \therefore \text{the boundary line passes through }(0,5)\text{ and }(5,0).
\displaystyle \text{Since the inequality is strict, the boundary line }x+y=5\text{ is not included.}
\displaystyle \text{Hence, it is represented by a dotted line.}
\displaystyle \text{To determine the required half-plane, consider the point }(0,0).
\displaystyle 0+0<5,
\displaystyle \text{which is true.}
\displaystyle \therefore \text{the half-plane containing the origin is the required solution region.}
\displaystyle \therefore \text{the solution is the region }x+y<5\text{ below the line }x+y=5.\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Solve }3x+2y>6\text{ graphically. (Not in syllabus).}
\displaystyle \text{Answer:}
\displaystyle \text{First draw the boundary line }3x+2y=6.
\displaystyle \text{Putting }x=0,\text{ we get }y=3.
\displaystyle \text{Putting }y=0,\text{ we get }x=2.
\displaystyle \therefore \text{the boundary line passes through }(0,3)\text{ and }(2,0).
\displaystyle \text{Since the inequality is strict, the boundary line }3x+2y=6\text{ is not included.}
\displaystyle \text{Hence, it is represented by a dotted line.}
\displaystyle \text{To determine the required half-plane, consider the point }(0,0).
\displaystyle 3(0)+2(0)>6\implies0>6,
\displaystyle \text{which is false.}
\displaystyle \therefore \text{the half-plane not containing the origin is the required solution region.}
\displaystyle \therefore \text{the solution is the region }3x+2y>6\text{ above the line }3x+2y=6.\displaystyle \\

\displaystyle \textbf{Question 58: }\text{Solve }5x-3>3x-5,\text{ if }x\text{ is a real number.}
\displaystyle \text{Answer:}
\displaystyle 5x-3>3x-5.
\displaystyle 5x-3x>-5+3.
\displaystyle 2x>-2.
\displaystyle \therefore x>-1.
\displaystyle \therefore \text{Solution set}=(-1,\infty).
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{Solve: }\frac{x-3}{x-5}>2,\text{ for real }x.
\displaystyle \text{Answer:}
\displaystyle \frac{x-3}{x-5}>2.
\displaystyle \frac{x-3}{x-5}-2>0.
\displaystyle \frac{x-3-2(x-5)}{x-5}>0.
\displaystyle \frac{7-x}{x-5}>0.
\displaystyle \therefore \frac{x-7}{x-5}<0.
\displaystyle \text{The critical points are }x=5\text{ and }x=7.
\displaystyle \text{The expression is negative when }x\text{ lies between the two critical points.}
\displaystyle \therefore 5<x<7.
\displaystyle \therefore \text{Solution set}=(5,7).
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Solve: }0<-\frac{x}{2}<3\text{ for real }x.
\displaystyle \text{Answer:}
\displaystyle 0<-\frac{x}{2}<3.
\displaystyle \text{Multiplying throughout by }2,\text{ we get}
\displaystyle 0<-x<6.
\displaystyle \text{Multiplying throughout by }-1,\text{ all the inequality signs are reversed.}
\displaystyle 0>x>-6.
\displaystyle \therefore -6<x<0.
\displaystyle \therefore \text{Solution set}=(-6,0).
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{Solve: }|x-3|\geq3\text{ for real }x.
\displaystyle \text{Answer:}
\displaystyle |x-3|\geq3.
\displaystyle \therefore x-3\geq3\quad\text{or}\quad x-3\leq-3.
\displaystyle \therefore x\geq6\quad\text{or}\quad x\leq0.
\displaystyle \therefore \text{Solution set}=(-\infty,0]\cup[6,\infty).
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{We know that }\sin x\in[-1,1],\ \forall x\in R. \\ \text{Find the range of }2\sin x-3.
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin x\leq1.
\displaystyle \text{Multiplying throughout by }2,\text{ we get}
\displaystyle -2\leq2\sin x\leq2.
\displaystyle \text{Subtracting }3\text{ throughout, we get}
\displaystyle -5\leq2\sin x-3\leq-1.
\displaystyle \therefore \text{Range of }2\sin x-3=[-5,-1].
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{Solve: }x+5<3\text{ on number line.}
\displaystyle \text{Answer:}
\displaystyle x+5<3.
\displaystyle \text{Subtracting }5\text{ from both sides, we get}
\displaystyle x<-2.
\displaystyle \therefore \text{Solution set}=(-\infty,-2).
\displaystyle \text{On the number line, mark an open circle at }-2\text{ and shade the region to its left.}\displaystyle \\

\displaystyle \textbf{Question 64: }\text{Represent the solution set of }3x-5\geq7\text{ on real number line.}
\displaystyle \text{Answer:}
\displaystyle 3x-5\geq7.
\displaystyle \text{Adding }5\text{ to both sides, we get}
\displaystyle 3x\geq12.
\displaystyle \text{Dividing both sides by }3,\text{ we get}
\displaystyle x\geq4.
\displaystyle \therefore \text{Solution set}=[4,\infty).
\displaystyle \text{On the number line, mark a closed circle at }4\text{ and shade the region to its right.}\displaystyle \\

\displaystyle \textbf{Question 65: }\text{Represent the solution set of }|x+5|<6\text{ on real number line.}
\displaystyle \text{Answer:}
\displaystyle |x+5|<6.
\displaystyle \therefore -6<x+5<6.
\displaystyle \text{Subtracting }5\text{ throughout, we get}
\displaystyle -11<x<1.
\displaystyle \therefore \text{Solution set}=(-11,1).
\displaystyle \text{On the number line, mark open circles at }-11\text{ and }1\text{ and shade the region between them.}\displaystyle \\

\displaystyle \textbf{Question 66: }\text{The maximum value of }7x-5\text{ is }9.\text{ Find the maximum value of }2x+3.
\displaystyle \text{Answer:}
\displaystyle 7x-5\leq9.
\displaystyle \text{Adding }5\text{ to both sides, we get}
\displaystyle 7x\leq14.
\displaystyle \therefore x\leq2.
\displaystyle \text{Hence, the maximum value of }x\text{ is }2.
\displaystyle \therefore 2x+3\leq2(2)+3=7.
\displaystyle \therefore \text{the maximum value of }2x+3\text{ is }7.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{The minimum value of }x-2\text{ is }12,\text{ find the minimum value of }3x+5.
\displaystyle \text{Answer:}
\displaystyle x-2\geq12.
\displaystyle \text{Adding }2\text{ to both sides, we get}
\displaystyle x\geq14.
\displaystyle \text{Hence, the minimum value of }x\text{ is }14.
\displaystyle \therefore 3x+5\geq3(14)+5=47.
\displaystyle \therefore \text{the minimum value of }3x+5\text{ is }47.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{Find the solutions of }|2-x|=x-2.
\displaystyle \text{Answer:}
\displaystyle |2-x|=|x-2|.
\displaystyle \therefore |x-2|=x-2.
\displaystyle \text{We know that }|x-2|=x-2\text{ when }x-2\geq0.
\displaystyle \therefore x\geq2.
\displaystyle \therefore \text{Solution set}=[2,\infty).
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{Solve: }1\leq|x-2|\leq3\text{ for real }x.
\displaystyle \text{Answer:}
\displaystyle 1\leq|x-2|\leq3.
\displaystyle \text{From }|x-2|\leq3,\text{ we get}
\displaystyle -3\leq x-2\leq3.
\displaystyle \therefore -1\leq x\leq5.
\displaystyle \text{Also, }|x-2|\geq1\text{ gives}
\displaystyle x-2\leq-1\quad\text{or}\quad x-2\geq1.
\displaystyle \therefore x\leq1\quad\text{or}\quad x\geq3.
\displaystyle \text{Combining the two conditions, we get}
\displaystyle -1\leq x\leq1\quad\text{or}\quad3\leq x\leq5.
\displaystyle \therefore \text{Solution set}=[-1,1]\cup[3,5].
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{Solve: }10\leq-5(x-2)<20\text{ for real }x.
\displaystyle \text{Answer:}
\displaystyle 10\leq-5(x-2)<20.
\displaystyle \text{Dividing throughout by }-5,\text{ all the inequality signs are reversed.}
\displaystyle -2\geq x-2>-4.
\displaystyle \therefore -4<x-2\leq-2.
\displaystyle \text{Adding }2\text{ throughout, we get}
\displaystyle -2<x\leq0.
\displaystyle \therefore \text{Solution set}=(-2,0].
\displaystyle \\

\displaystyle \text{LONG ANSWER TYPE QUESTIONS.} \textbf{(Not in syllabus). }

\displaystyle \textbf{Question 71: }\text{Solve the following system of inequalities graphically:}
\displaystyle 2x+y\leq24,\qquad x+y\leq11,\qquad 2x+5y\leq40,\qquad x\geq0,\qquad y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are}
\displaystyle 2x+y=24,\qquad x+y=11,\qquad 2x+5y=40.
\displaystyle \text{For }2x+y=24,\text{ the intercepts are }(12,0)\text{ and }(0,24).
\displaystyle \text{For }x+y=11,\text{ the intercepts are }(11,0)\text{ and }(0,11).
\displaystyle \text{For }2x+5y=40,\text{ the intercepts are }(20,0)\text{ and }(0,8).
\displaystyle \text{Since all three inequalities contain the point }(0,0),\text{ we take the half-planes containing the origin.}
\displaystyle \text{Also, }x\geq0\text{ and }y\geq0,\text{ so the solution region lies in the first quadrant.}
\displaystyle \text{Now, }x+y=11\text{ and }2x+5y=40\text{ intersect at}
\displaystyle x+y=11,\qquad 2x+5y=40.
\displaystyle x=11-y.
\displaystyle 2(11-y)+5y=40.
\displaystyle 22+3y=40\implies y=6,\qquad x=5.
\displaystyle \therefore \text{the point of intersection is }(5,6).
\displaystyle \text{Also, }2x+y=24\text{ and }x+y=11\text{ intersect at}
\displaystyle x=13,\qquad y=-2,
\displaystyle \text{which does not lie in the first quadrant and hence does not bound the feasible region.}
\displaystyle \text{Thus, in the first quadrant, }2x+y\leq24\text{ is automatically satisfied by the region}
\displaystyle \text{determined by the other inequalities.}
\displaystyle \therefore \text{the vertices of the feasible region are }(0,0),\ (11,0),\ (5,6)\text{ and }(0,8).
\displaystyle \therefore \text{the required solution is the closed region bounded by these four vertices.}
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{Solve the following system of inequalities graphically:}
\displaystyle 3x+2y\geq24,\qquad 3x+y\leq15,\qquad x\geq4.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are}
\displaystyle 3x+2y=24,\qquad 3x+y=15,\qquad x=4.
\displaystyle \text{For }3x+2y=24,\text{ the intercepts are }(8,0)\text{ and }(0,12).
\displaystyle \text{For }3x+y=15,\text{ the intercepts are }(5,0)\text{ and }(0,15).
\displaystyle 3x+2y\geq24\implies y\geq12-\frac{3x}{2}.
\displaystyle \therefore \text{we take the region on or above the line }3x+2y=24.
\displaystyle 3x+y\leq15\implies y\leq15-3x.
\displaystyle \therefore \text{we take the region on or below the line }3x+y=15.
\displaystyle \text{Also, }x\geq4\text{ represents the region on or to the right of the line }x=4.
\displaystyle \text{The lines }3x+2y=24\text{ and }3x+y=15\text{ intersect at}
\displaystyle 3x+2y=24,\qquad3x+y=15.
\displaystyle \text{Subtracting the second equation from the first, we get }y=9.
\displaystyle 3x+9=15\implies x=2.
\displaystyle \therefore \text{their point of intersection is }(2,9).
\displaystyle \text{But this point does not satisfy }x\geq4.
\displaystyle \text{For }x\geq4,\text{ we would require simultaneously}
\displaystyle y\geq12-\frac{3x}{2}\qquad\text{and}\qquad y\leq15-3x.
\displaystyle \text{These conditions can hold only if }12-\frac{3x}{2}\leq15-3x.
\displaystyle \therefore \frac{3x}{2}\leq3\implies x\leq2.
\displaystyle \text{This contradicts the given condition }x\geq4.
\displaystyle \therefore \text{there is no common solution region.}
\displaystyle \therefore \text{Solution set}=\varnothing.
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{Solve the following system of inequalities graphically:}
\displaystyle x-2y\leq3,\qquad3x+4y>12,\qquad x\geq0,\qquad y\geq1.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are}
\displaystyle x-2y=3,\qquad3x+4y=12,\qquad x=0,\qquad y=1.
\displaystyle x-2y\leq3\implies y\geq\frac{x-3}{2}.
\displaystyle \therefore \text{we take the region on or above the line }x-2y=3.
\displaystyle 3x+4y>12\implies y>3-\frac{3x}{4}.
\displaystyle \therefore \text{we take the region above the line }3x+4y=12.
\displaystyle \text{Since the inequality is strict, the line }3x+4y=12\text{ is not included.}
\displaystyle \text{Also, }x\geq0\text{ represents the region on or to the right of the }y\text{-axis,}
\displaystyle \text{and }y\geq1\text{ represents the region on or above the line }y=1.
\displaystyle \text{The line }3x+4y=12\text{ intersects }y=1\text{ at}
\displaystyle 3x+4=12\implies x=\frac{8}{3}.
\displaystyle \therefore \text{the point of intersection is }\left(\frac{8}{3},1\right).
\displaystyle \text{The line }x-2y=3\text{ intersects }y=1\text{ at}
\displaystyle x-2=3\implies x=5.
\displaystyle \therefore \text{the point of intersection is }(5,1).
\displaystyle \text{Hence, the required solution is the common region satisfying all four inequalities.}
\displaystyle \text{Its lower boundary consists of the line }3x+4y=12\text{ up to }\left(\frac{8}{3},1\right),
\displaystyle \text{the line }y=1\text{ from }\left(\frac{8}{3},1\right)\text{ to }(5,1),\text{ and thereafter}
\displaystyle \text{the line }x-2y=3.\text{ The solution region is unbounded above.}
\displaystyle \text{The point }\left(\frac{8}{3},1\right)\text{ is not included, while }(5,1)\text{ is included.}
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{Solve the following system of inequalities graphically:}
\displaystyle 3x+4y\leq60,\qquad x+3y\leq30,\qquad x\geq0,\qquad y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are }3x+4y=60\text{ and }x+3y=30.
\displaystyle \text{For }3x+4y=60,\text{ the intercepts are }(20,0)\text{ and }(0,15).
\displaystyle \text{For }x+3y=30,\text{ the intercepts are }(30,0)\text{ and }(0,10).
\displaystyle \text{Both inequalities are satisfied by }(0,0),\text{ so we take the half-planes containing the origin.}
\displaystyle \text{Also, }x\geq0\text{ and }y\geq0,\text{ so the solution region lies in the first quadrant.}
\displaystyle \text{To find the intersection of the two boundary lines, solve}
\displaystyle 3x+4y=60,\qquad x+3y=30.
\displaystyle x=30-3y.
\displaystyle 3(30-3y)+4y=60.
\displaystyle 90-5y=60\implies y=6.
\displaystyle \therefore x=30-18=12.
\displaystyle \therefore \text{the two lines intersect at }(12,6).
\displaystyle \text{Hence, the vertices of the feasible region are }(0,0),\ (20,0),\ (12,6)\text{ and }(0,10).
\displaystyle \therefore \text{the required solution is the closed region bounded by these four vertices.}
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{Solve the following system of inequalities graphically:}
\displaystyle 2x+y\geq4,\qquad x+y\leq3,\qquad 2x-3y\leq6.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are}
\displaystyle 2x+y=4,\qquad x+y=3,\qquad 2x-3y=6.
\displaystyle 2x+y\geq4\implies y\geq4-2x.
\displaystyle x+y\leq3\implies y\leq3-x.
\displaystyle 2x-3y\leq6\implies y\geq\frac{2x-6}{3}.
\displaystyle \text{Now, }2x+y=4\text{ and }x+y=3\text{ intersect at}
\displaystyle x=1,\qquad y=2.
\displaystyle \therefore \text{one vertex is }(1,2).
\displaystyle \text{The lines }x+y=3\text{ and }2x-3y=6\text{ intersect at}
\displaystyle x=3,\qquad y=0.
\displaystyle \therefore \text{the second vertex is }(3,0).
\displaystyle \text{The lines }2x+y=4\text{ and }2x-3y=6\text{ intersect at}
\displaystyle y=4-2x.
\displaystyle 2x-3(4-2x)=6.
\displaystyle 8x=18\implies x=\frac{9}{4},\qquad y=-\frac{1}{2}.
\displaystyle \therefore \text{the third vertex is }\left(\frac{9}{4},-\frac{1}{2}\right).
\displaystyle \therefore \text{the common solution is the closed triangular region with vertices}
\displaystyle (1,2),\qquad(3,0),\qquad\left(\frac{9}{4},-\frac{1}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{Solve the following system of inequalities graphically:}
\displaystyle 4x+3y\leq60,\qquad y\geq2x,\qquad x\geq3,\qquad x\geq0,\qquad y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are }4x+3y=60,\quad y=2x,\quad x=3.
\displaystyle 4x+3y\leq60\text{ represents the region on or below the line }4x+3y=60.
\displaystyle y\geq2x\text{ represents the region on or above the line }y=2x.
\displaystyle x\geq3\text{ represents the region on or to the right of the line }x=3.
\displaystyle \text{Also, }x\geq0,\ y\geq0\text{ restrict the solution to the first quadrant.}
\displaystyle \text{The lines }y=2x\text{ and }x=3\text{ intersect at }(3,6).
\displaystyle \text{The lines }4x+3y=60\text{ and }x=3\text{ intersect at}
\displaystyle 4(3)+3y=60\implies3y=48\implies y=16.
\displaystyle \therefore \text{the point of intersection is }(3,16).
\displaystyle \text{The lines }4x+3y=60\text{ and }y=2x\text{ intersect at}
\displaystyle 4x+3(2x)=60.
\displaystyle 10x=60\implies x=6,\qquad y=12.
\displaystyle \therefore \text{the point of intersection is }(6,12).
\displaystyle \therefore \text{the required solution is the closed triangular region with vertices}\displaystyle (3,6),\qquad(3,16),\qquad(6,12).
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{Solve the following system of inequalities graphically:}
\displaystyle 3x+2y\leq150,\qquad x+4y\leq80,\qquad x\leq15,\qquad y\geq0,\qquad x\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are }3x+2y=150,\quad x+4y=80,\quad x=15.
\displaystyle \text{Since }x\geq0\text{ and }y\geq0,\text{ the solution region lies in the first quadrant.}
\displaystyle x+4y=80\text{ has intercepts }(80,0)\text{ and }(0,20).
\displaystyle \text{The line }x=15\text{ intersects the }x\text{-axis at }(15,0).
\displaystyle \text{The lines }x=15\text{ and }x+4y=80\text{ intersect at}
\displaystyle 15+4y=80.
\displaystyle 4y=65\implies y=\frac{65}{4}.
\displaystyle \therefore \text{the point of intersection is }\left(15,\frac{65}{4}\right).
\displaystyle \text{On the }y\text{-axis, }x+4y\leq80\implies y\leq20.
\displaystyle \therefore \text{the point }(0,20)\text{ is another vertex of the feasible region.}
\displaystyle \text{For all these points, }3x+2y\leq150\text{ is also satisfied.}
\displaystyle \text{Hence, }3x+2y\leq150\text{ does not further restrict the feasible region.}
\displaystyle \therefore \text{the vertices of the feasible region are}
\displaystyle (0,0),\qquad(15,0),\qquad\left(15,\frac{65}{4}\right),\qquad(0,20).
\displaystyle \therefore \text{the required solution is the closed region bounded by these four vertices.}\displaystyle \\

\displaystyle \textbf{Question 78: }\text{Solve the following system of inequalities graphically:}
\displaystyle x+2y\leq10,\qquad x+y\geq1,\qquad x-y\leq0,\qquad x\geq0,\qquad y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are }x+2y=10,\quad x+y=1,\quad x-y=0.
\displaystyle x+2y\leq10\text{ represents the region on or below the line }x+2y=10.
\displaystyle x+y\geq1\text{ represents the region on or above the line }x+y=1.
\displaystyle x-y\leq0\implies y\geq x.
\displaystyle \therefore \text{we take the region on or above the line }y=x.
\displaystyle \text{Also, }x\geq0,\ y\geq0\text{ restrict the solution to the first quadrant.}
\displaystyle \text{The line }x+y=1\text{ intersects the }y\text{-axis at }(0,1).
\displaystyle \text{The lines }x+y=1\text{ and }y=x\text{ intersect at}
\displaystyle 2x=1\implies x=y=\frac{1}{2}.
\displaystyle \therefore \text{the point of intersection is }\left(\frac{1}{2},\frac{1}{2}\right).
\displaystyle \text{The lines }x+2y=10\text{ and }y=x\text{ intersect at}
\displaystyle x+2x=10\implies x=y=\frac{10}{3}.
\displaystyle \therefore \text{the point of intersection is }\left(\frac{10}{3},\frac{10}{3}\right).
\displaystyle \text{The line }x+2y=10\text{ intersects the }y\text{-axis at }(0,5).
\displaystyle \therefore \text{the vertices of the feasible region are}
\displaystyle (0,1),\qquad\left(\frac{1}{2},\frac{1}{2}\right),\qquad\left(\frac{10}{3},\frac{10}{3}\right),\qquad(0,5).
\displaystyle \therefore \text{the required solution is the closed region bounded by these four vertices.}\displaystyle \\

\displaystyle \textbf{Question 79: }\text{Solve the following system of inequalities graphically:}
\displaystyle 5x+4y\leq20,\qquad x\geq1,\qquad y\geq2.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are }5x+4y=20,\quad x=1,\quad y=2.
\displaystyle 5x+4y\leq20\text{ represents the region on or below the line }5x+4y=20.
\displaystyle x\geq1\text{ represents the region on or to the right of the line }x=1.
\displaystyle y\geq2\text{ represents the region on or above the line }y=2.
\displaystyle \text{The lines }x=1\text{ and }y=2\text{ intersect at }(1,2).
\displaystyle \text{The lines }5x+4y=20\text{ and }x=1\text{ intersect at}
\displaystyle 5+4y=20.
\displaystyle 4y=15\implies y=\frac{15}{4}.
\displaystyle \therefore \text{the point of intersection is }\left(1,\frac{15}{4}\right).
\displaystyle \text{The lines }5x+4y=20\text{ and }y=2\text{ intersect at}
\displaystyle 5x+8=20.
\displaystyle 5x=12\implies x=\frac{12}{5}.
\displaystyle \therefore \text{the point of intersection is }\left(\frac{12}{5},2\right).
\displaystyle \therefore \text{the required solution is the closed triangular region with vertices}
\displaystyle (1,2),\qquad\left(1,\frac{15}{4}\right),\qquad\left(\frac{12}{5},2\right).\displaystyle \\

\displaystyle \textbf{Question 80: }\text{Solve the following system of inequalities graphically:}
\displaystyle 2x+y\geq8,\qquad x+2y\geq10.
\displaystyle \text{Answer:}
\displaystyle \text{The corresponding boundary lines are }2x+y=8\text{ and }x+2y=10.
\displaystyle \text{For }2x+y=8,\text{ the intercepts are }(4,0)\text{ and }(0,8).
\displaystyle \text{For }x+2y=10,\text{ the intercepts are }(10,0)\text{ and }(0,5).
\displaystyle \text{For the inequality }2x+y\geq8,\text{ consider the point }(0,0).
\displaystyle 2(0)+0\geq8\implies0\geq8,\text{ which is false.}
\displaystyle \therefore \text{we take the half-plane not containing the origin.}
\displaystyle \text{Similarly, for }x+2y\geq10,\text{ the point }(0,0)\text{ does not satisfy the inequality.}
\displaystyle \therefore \text{we again take the half-plane not containing the origin.}
\displaystyle \text{To find the intersection of the two boundary lines, solve}
\displaystyle 2x+y=8,\qquad x+2y=10.
\displaystyle \text{Multiplying the second equation by }2,\text{ we get }2x+4y=20.
\displaystyle \text{Subtracting }2x+y=8,\text{ we get }3y=12.
\displaystyle \therefore y=4.
\displaystyle 2x+4=8\implies x=2.
\displaystyle \therefore \text{the two boundary lines intersect at }(2,4).
\displaystyle \text{Hence, the required solution is the common region satisfying both inequalities.}
\displaystyle \text{It lies on or above both the lines }2x+y=8\text{ and }x+2y=10.
\displaystyle \therefore \text{the solution region is unbounded and both boundary lines are included.}
\displaystyle \\

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