\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series:    }                                                                 \displaystyle \textbf{Code No. }30/1/1
\displaystyle \text{Roll No.   }

  • \displaystyle \text{Candidates must write the Code on the title page of the answer-book.}
  • \displaystyle \text{Please check that this question paper contains 16 printed pages.}
  • \displaystyle \text{Code number given on the right hand side of the question paper should be written on the}
  • \displaystyle \text{title page of the answer-book by the candidate.}
  • \displaystyle \text{Please check that this question paper contains 34 questions.}
  • \displaystyle \text{Please write down the serial number of the question before attempting it.}
  • \displaystyle \text{15 minutes time has been allotted to read this question paper. The question paper will be}
  • \displaystyle \text{distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question}
  • \displaystyle \text{paper only and will not write any answer on the answer-script during this period.}

\displaystyle \textbf{SUMMATIVE ASSESSMENT-II}
\displaystyle \textbf{MATHEMATICS}

\displaystyle \text{Time allowed : 3 hours}                                                                 \displaystyle \text{Maximum marks : 90}

\displaystyle \textbf{General Instructions :}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) The question paper consists of 34 questions divided into four sections A,}
\displaystyle \text{B, C and D.}
\displaystyle \text{(iii) Section A contains 8 questions of one mark each, which are multiple}
\displaystyle \text{choice type questions, Section B contains 6 questions of two marks each,}
\displaystyle \text{Section C contains 10 questions of three marks each, and Section D}
\displaystyle \text{contains 10 questions of four marks each.}
\displaystyle \text{(iv) Use of calculators is not permitted.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{Question Numbers 1 to 4 carry one mark each. In each of these questions, four}
\displaystyle \text{alternative choices have been provided of which only one is correct. Select the}
\displaystyle \text{correct choice.}
\\

\displaystyle \textbf{Question 1. }\text{In the given figure if }DE\parallel BC,\ AE=8\text{ cm, }EC=2\text{ cm and }
\displaystyle BC=6\text{ cm, }\text{then find }DE.
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle AE=8\text{ cm}
\displaystyle EC=2\text{ cm}
\displaystyle BC=6\text{ cm}
\displaystyle \text{As }DE\parallel BC,
\displaystyle \triangle ADE\sim\triangle ABC
\displaystyle \text{So, the ratio of their respective sides is equal.}
\displaystyle \frac{AE}{AC}=\frac{DE}{BC}
\displaystyle \frac{8}{10}=\frac{DE}{6}
\displaystyle DE=\frac{8\times6}{10}
\displaystyle DE=4.8\text{ cm}
\displaystyle \therefore DE=4.8\text{ cm}
\\

\displaystyle \textbf{Question 2. }\text{Evaluate: }\frac{1-\cot^2 45^\circ}{1+\sin^2 90^\circ}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{1-\cot^2 45^\circ}{1+\sin^2 90^\circ}
\displaystyle =\frac{1-1}{1+1}\qquad (\because\ \cot45^\circ=1\text{ and }\sin90^\circ=1)
\displaystyle =\frac{0}{2}
\displaystyle =0
\displaystyle \therefore \frac{1-\cot^2 45^\circ}{1+\sin^2 90^\circ}=0
\\

\displaystyle \textbf{Question 3. }\text{If }\mathrm{cosec}\,\theta=\frac{5}{4},\text{ find the value of }\cot\theta.
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle \mathrm{cosec}\,\theta=\frac{5}{4}
\displaystyle \text{In a right triangle,}
\displaystyle \text{Hypotenuse}=5,\quad \text{Perpendicular}=4
\displaystyle \text{Base}=\sqrt{5^2-4^2}=\sqrt{25-16}=3
\displaystyle \cot\theta=\frac{\text{Base}}{\text{Perpendicular}}
\displaystyle \cot\theta=\frac{3}{4}
\displaystyle \therefore \cot\theta=\frac{3}{4}
\\

\displaystyle \textbf{Question 4. }\text{Following table shows sale of shoes in a store during one month:}
\displaystyle \begin{array}{|c|c|}\hline \text{Size of shoe} & \text{Number of pairs sold}\\ \hline 3&4\\ \hline 4&18\\ \hline 5&25\\ \hline 6&12\\ \hline 7&5\\ \hline 8&1\\ \hline \end{array}
\displaystyle \text{Find the modal size of the shoes sold.}
\displaystyle \text{Answer:}
\displaystyle \text{From the given table,}
\displaystyle \text{Number of pairs sold is maximum for shoe size }5.
\displaystyle \text{Therefore, the modal size of the shoes sold is }5.
\\

\displaystyle \textbf{SECTION B}
\displaystyle \text{Question Numbers 5 to 10 carry two marks each.}

\displaystyle \textbf{Question 5. }\text{Find the prime factorisation of the denominator of rational number expressed as }6.\overline{21}
\displaystyle \text{in simplest form.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=6.\overline{21}
\displaystyle x=6.21212121\ldots \qquad (i)
\displaystyle 100x=621.21212121\ldots \qquad (ii)
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 99x=615
\displaystyle x=\frac{615}{99}=\frac{205}{33}
\displaystyle 33=3\times11
\displaystyle \therefore \text{Prime factorisation of the denominator}=3\times11
\\

\displaystyle \textbf{Question 6. }\text{Find a quadratic polynomial, the sum and product of whose zeroes are } \sqrt{3}
\displaystyle \text{and }\frac{1}{\sqrt{3}}\text{ respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle \text{Sum of zeroes}=\sqrt{3}
\displaystyle \text{Product of zeroes}=\frac{1}{\sqrt{3}}
\displaystyle \text{Required quadratic polynomial is}
\displaystyle x^2-(\text{Sum of zeroes})x+\text{Product of zeroes}=0
\displaystyle x^2-\sqrt{3}x+\frac{1}{\sqrt{3}}=0
\displaystyle \sqrt{3}x^2-3x+1=0
\displaystyle \therefore \text{Required quadratic polynomial is }\sqrt{3}x^2-3x+1
\\

\displaystyle \textbf{Question 7. }\text{Complete the following factor tree and find the composite number }x. \displaystyle \text{Answer:}
\displaystyle \text{From the factor tree we have,}
\displaystyle y=13\times5=65
\displaystyle x=195\times3=585
\displaystyle \therefore x=585
\\

\displaystyle \textbf{Question 8. }\text{In a rectangle }ABCD,\ E\text{ is middle point of }AD.\text{ If }AD=40\text{ m and }AB=48\text{ m,}
\displaystyle \text{then find }EB.
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle E\text{ is the midpoint of }AD
\displaystyle AD=40\text{ m and }AB=48\text{ m}
\displaystyle AE=\frac{40}{2}=20\text{ m}
\displaystyle \text{In right angled }\triangle ABE
\displaystyle (BE)^2=(AE)^2+(AB)^2
\displaystyle (BE)^2=(20)^2+(48)^2
\displaystyle (BE)^2=400+2304
\displaystyle (BE)^2=2704
\displaystyle BE=52\text{ m}
\displaystyle \therefore BE=52\text{ m}
\\

\displaystyle \textbf{Question 9. }\text{If }x=p\sec\theta+q\tan\theta\text{ and }y=p\tan\theta+q\sec\theta,
\displaystyle \text{then prove that }x^2-y^2=p^2-q^2.
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle x=p\sec\theta+q\tan\theta
\displaystyle y=p\tan\theta+q\sec\theta
\displaystyle \text{Taking LHS,}
\displaystyle x^2-y^2=(p\sec\theta+q\tan\theta)^2-(p\tan\theta+q\sec\theta)^2
\displaystyle x^2-y^2=p^2(\sec^2\theta-\tan^2\theta)+q^2(\tan^2\theta-\sec^2\theta)
\displaystyle x^2-y^2=p^2(1)+q^2(-1)\qquad [\sec^2\theta-\tan^2\theta=1]
\displaystyle x^2-y^2=p^2-q^2
\displaystyle \therefore x^2-y^2=p^2-q^2\ \text{Hence proved.}
\\

\displaystyle \textbf{Question 10. }\text{Given below is the distribution of weekly pocket money received by students of a}
\displaystyle \text{class. Calculate the pocket money that is received by most of the students.}
\displaystyle \begin{array}{|c|c|}\hline \text{Pocket Money (in Rs)} & \text{No. of Students}\\ \hline 0-20&2\\ \hline 20-40&2\\ \hline 40-60&3\\ \hline 60-80&12\\ \hline 80-100&18\\ \hline 100-120&5\\ \hline 120-140&2\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Modal class}=80-100
\displaystyle l=80,\quad f_1=18,\quad f_0=12,\quad f_2=5,\quad h=20
\displaystyle \text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h
\displaystyle =80+\left(\frac{18-12}{2(18)-12-5}\right)\times20
\displaystyle =80+\left(\frac{6}{19}\right)\times20
\displaystyle =80+6.3157
\displaystyle =86.32
\displaystyle \therefore \text{The pocket money received by most of the students is Rs }86.32
\\

\displaystyle \textbf{SECTION C}
\displaystyle \text{Question Numbers 11 to 20 carry three marks each.}

\displaystyle \textbf{Question 11. }\text{Prove that }3+2\sqrt{3}\text{ is an irrational number.}
\displaystyle \text{Answer:}
\displaystyle \text{To prove: }3+2\sqrt{3}\text{ is an irrational number.}
\displaystyle \text{If possible, let }3+2\sqrt{3}\text{ be rational.}
\displaystyle \text{Then, }(3+2\sqrt{3})-3\text{ is rational.}
\displaystyle 2\sqrt{3}\text{ is rational.}
\displaystyle \sqrt{3}\text{ is rational.}
\displaystyle \text{Let the simplest form of }\sqrt{3}\text{ be }\frac{a}{b}.
\displaystyle \text{Then, }a\text{ and }b\text{ are integers having no common factor other than }1.
\displaystyle \text{Now, }\sqrt{3}=\frac{a}{b}
\displaystyle 3b^2=a^2
\displaystyle 3\text{ divides }a^2
\displaystyle 3\text{ divides }a
\displaystyle \text{Let }a=3c\text{ for some integer }c.
\displaystyle \text{Therefore, }3b^2=9c^2
\displaystyle b^2=3c^2
\displaystyle 3\text{ divides }b^2
\displaystyle 3\text{ divides }b
\displaystyle \text{Thus, }3\text{ is a common factor of }a\text{ and }b.
\displaystyle \text{This contradicts the fact that }a\text{ and }b\text{ have no common factor other than }1.
\displaystyle \text{So, }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore 3+2\sqrt{3}\text{ is irrational.}
\\

\displaystyle \textbf{Question 12. }\text{Solve by elimination:}
\displaystyle 3x=y+5
\displaystyle 5x-y=11
\displaystyle \text{Answer:}
\displaystyle \text{To solve given equations by elimination method}
\displaystyle 3x-y=5
\displaystyle 5x-y=11
\displaystyle \text{Subtracting,}
\displaystyle -2x=-6
\displaystyle x=3
\displaystyle y=3x-5=3(3)-5=9-5=4
\displaystyle \therefore x=3\text{ and }y=4
\\

\displaystyle \textbf{Question 13. }\text{A man earns Rs }600\text{ per month more than his wife. One-tenth of the man's salary}
\displaystyle \text{and one-sixth of the wife's salary amount to Rs }1500,\text{ which is saved every month. Find their incomes.}
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle \text{Man earns Rs }600\text{ per month more than his wife.}
\displaystyle \text{Let the income of man be }x.
\displaystyle \text{Income of wife}=x-600
\displaystyle \text{According to given condition,}
\displaystyle \frac{1}{10}x+\frac{1}{6}(x-600)=1500
\displaystyle \frac{x}{10}+\frac{x}{6}-100=1500
\displaystyle \frac{8x}{30}=1600
\displaystyle x=6000
\displaystyle \text{Income of man}=\text{Rs }6000
\displaystyle \text{Income of wife}=\text{Rs }6000-\text{Rs }600=\text{Rs }5400
\\

\displaystyle \textbf{Question 14. }\text{Check whether polynomial }x-1\text{ is a factor of the polynomial }
\displaystyle x^3-8x^2+19x-12. \ \text{Verify by division algorithm.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p(x)=x^3-8x^2+19x-12.
\displaystyle \text{Dividing }p(x)\text{ by }x-1,
\displaystyle x^3-8x^2+19x-12=(x-1)(x^2-7x+12)+0
\displaystyle \text{Since the remainder is }0,\ x-1\text{ is a factor of }p(x).
\displaystyle \text{Hence, }x-1\text{ is a factor of }x^3-8x^2+19x-12.
\\

\displaystyle \textbf{Question 15. }\text{If the perimeters of two similar triangles }ABC\text{ and }DEF\text{ are }50\text{ cm and }70\text{ cm}
\displaystyle \text{respectively and one side of }\triangle ABC=20\text{ cm, then find the corresponding side of }\triangle DEF.
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle \text{Perimeters of two similar triangles }ABC\text{ and }DEF\text{ are }50\text{ cm and }70\text{ cm.}
\displaystyle \text{One side of triangle }ABC=20\text{ cm}
\displaystyle \text{As both triangles are similar, the ratio of their corresponding sides is equal}
\displaystyle \text{to the ratio of their perimeters.}
\displaystyle \frac{50}{70}=\frac{20}{\text{other side}}
\displaystyle \text{Other side}=\frac{70\times20}{50}=28\text{ cm}
\displaystyle \therefore \text{Corresponding side of }\triangle DEF=28\text{ cm}
\\

\displaystyle \textbf{Question 16. }\text{In the figure if }DE\parallel OB\text{ and }EF\parallel BC,\text{ then prove that } \\ DF\parallel OC. \displaystyle \text{Answer:}
\displaystyle \text{In }\triangle AOB,\text{ we have}
\displaystyle DE\parallel OB \qquad [\text{Given}]
\displaystyle \text{Therefore, by basic proportionality theorem, we have}
\displaystyle \frac{AE}{EB}=\frac{AD}{DO}\qquad (i)
\displaystyle \text{In }\triangle ABC,\text{ we have}
\displaystyle EF\parallel BC \qquad [\text{Given}]
\displaystyle \text{Therefore, by basic proportionality theorem, we have}
\displaystyle \frac{AE}{EB}=\frac{AF}{FC}\qquad (ii)
\displaystyle \text{From (i) and (ii), we have}
\displaystyle \frac{AF}{FC}=\frac{AD}{DO}
\displaystyle \therefore DF\parallel OC
\displaystyle \text{By the converse of Basic Proportionality Theorem.}
\\

\displaystyle \textbf{Question 17. }\text{Prove the identity:}
\displaystyle (\sec A-\cos A)(\cot A+\tan A)=\tan A\sec A.
\displaystyle \text{Answer:}
\displaystyle \text{To prove: }(\sec A-\cos A)(\cot A+\tan A)=\tan A\sec A
\displaystyle \text{Taking LHS,}
\displaystyle (\sec A-\cos A)(\cot A+\tan A)
\displaystyle =\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)
\displaystyle =\left(\frac{1-\cos^2 A}{\cos A}\right)\left(\frac{\cos^2 A+\sin^2 A}{\sin A\cos A}\right)
\displaystyle =\left(\frac{\sin^2 A}{\cos A}\right)\left(\frac{1}{\sin A\cos A}\right)
\displaystyle =\left(\frac{\sin A}{\cos A}\right)\left(\frac{1}{\cos A}\right)
\displaystyle =\tan A\sec A
\displaystyle \therefore \text{LHS}=\text{RHS}\quad \text{Hence proved.}
\\

\displaystyle \textbf{Question 18. }\text{Given }2\cos 3\theta=\sqrt{3},\text{ find the value of }\theta.
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle 2\cos3\theta=\sqrt{3}
\displaystyle \cos3\theta=\frac{\sqrt{3}}{2}
\displaystyle \cos3\theta=\cos30^\circ
\displaystyle \text{On comparing,}
\displaystyle 3\theta=30^\circ
\displaystyle \theta=10^\circ
\\

\displaystyle \textbf{Question 19. }\text{For helping poor girls of their class, students saved pocket money as} \\ \text{shown in the following table:}
\displaystyle \begin{array}{|c|c|}\hline \text{Money saved (in Rs)} & \text{Number of Students}\\ \hline 5-7&6\\ \hline 7-9&3\\ \hline 9-11&9\\ \hline 11-13&5\\ \hline 13-15&7\\ \hline \end{array}
\displaystyle \text{Find mean and median for this data.}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Class Interval} & x_i & f_i & f_ix_i & \text{c.f.}\\ \hline 5-7&6&6&36&6\\ \hline 7-9&8&3&24&9\\ \hline 9-11&10&9&90&18\\ \hline 11-13&12&5&60&23\\ \hline 13-15&14&7&98&30\\ \hline \text{Total}&&\sum f_i=30&\sum f_ix_i=308&\\ \hline \end{array}
\displaystyle \text{Mean}=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle =\frac{308}{30}=10.26
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-CF}{f}\right)\times h
\displaystyle \text{Here }l=9,\quad \frac{N}{2}=15,\quad CF=9,\quad h=2,\quad f=9
\displaystyle \text{Median}=9+\left(\frac{15-9}{9}\right)\times2
\displaystyle =9+1.33=10.33
\displaystyle \therefore \text{Mean}=10.26\text{ and Median}=10.33
\\

\displaystyle \textbf{Question 20. }\text{Monthly pocket money of students of a class is given in the following} \\ \text{frequency distribution:}
\displaystyle \begin{array}{|c|c|}\hline \text{Pocket money (in Rs)} & \text{Number of Students}\\ \hline 100-125&14\\ \hline 125-150&8\\ \hline 150-175&12\\ \hline 175-200&5\\ \hline 200-225&11\\ \hline \end{array}
\displaystyle \text{Find mean pocket money using step deviation method.}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Class Interval} & f_i & x_i & u_i=\frac{x_i-A}{h} & f_i u_i\\ \hline 100-125 & 14 & 112.5 & -2 & -28\\ \hline 125-150 & 8 & 137.5 & -1 & -8\\ \hline 150-175 & 12 & 162.5 & 0 & 0\\ \hline 175-200 & 5 & 187.5 & 1 & 5\\ \hline 200-225 & 11 & 212.5 & 2 & 22\\ \hline \end{array}
\displaystyle \text{Here, }A=162.5,\ h=25
\displaystyle \sum f_i=14+8+12+5+11=50
\displaystyle \sum f_i u_i=-28-8+0+5+22=-9
\displaystyle \text{Using the formula } \bar{x}=A+h\left(\frac{\sum f_i u_i}{\sum f_i}\right)
\displaystyle \bar{x}=162.5+25\left(\frac{-9}{50}\right)
\displaystyle =162.5-4.5
\displaystyle =158
\displaystyle \therefore \text{Mean pocket money}= \text{Rs }158
\\

\displaystyle \textbf{SECTION D}
\displaystyle \text{Question Numbers 21 to 31 carry four marks each.}

\displaystyle \textbf{Question 21. }\text{If two positive integers }x\text{ and }y\text{ are expressible in terms of primes as }
\displaystyle x=p^2q^3 \text{and }y=p^3q,\text{ what can you say about their LCM and HCF. Is LCM a} \\ \text{multiple of HCF? Explain.}
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle x=p^2q^3\text{ and }y=p^3q
\displaystyle \text{LCM}=p^3q^3
\displaystyle \text{HCF}=p^2q
\displaystyle \text{As there are common factors between HCF and LCM.}
\displaystyle \text{Hence, HCF is a multiple of the LCM.}
\\

\displaystyle \textbf{Question 22. }\text{Sita Devi wants to make a rectangular pond on the roadside for providing drinking water}
\displaystyle \text{to street animals. The area of the pond decreases by }3\text{ sq ft when its length is decreased by }2\text{ ft}
\displaystyle \text{and breadth is increased by }1\text{ ft. The area increases by }4\text{ sq ft when the length is increased}
\displaystyle \text{by }1\text{ ft and the breadth remains unchanged. Find the dimensions of the pond. What motivated}
\displaystyle \text{Sita Devi to provide a water point for street animals?}
\displaystyle \text{Answer:}
\displaystyle \text{Let length of the rectangular pond}=x\text{ ft.}
\displaystyle \text{Breadth of rectangular pond}=y\text{ ft.}
\displaystyle \text{Area of rectangular pond}=xy
\displaystyle \text{According to the question,}
\displaystyle (x-2)(y+1)=xy-3
\displaystyle xy+x-2y-2=xy-3
\displaystyle x-2y=-1 \qquad (i)
\displaystyle (x+1)y=xy+4
\displaystyle xy+y=xy+4
\displaystyle y=4 \qquad (ii)
\displaystyle \text{Putting the value of }y\text{ in equation (i), we get}
\displaystyle x-2(4)=-1
\displaystyle x-8=-1
\displaystyle x=7
\displaystyle \text{Length of rectangular pond}=7\text{ ft.}
\displaystyle \text{Breadth of rectangular pond}=4\text{ ft.}
\displaystyle \textbf{Values:}
\displaystyle \text{Water is essential for living beings.}
\displaystyle \text{Animals are also living beings and they also need basic amenities.}
\\

\displaystyle \textbf{Question 23. }\text{If a polynomial }x^4+5x^3+4x^2-10x-12\text{ has two zeroes as }
\displaystyle -2\text{ and }-3,  \ \text{then find the other zeroes.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }-2\text{ and }-3\text{ are zeroes, }(x+2)(x+3)\text{ is a factor.}
\displaystyle (x+2)(x+3)=x^2+5x+6
\displaystyle x^4+5x^3+4x^2-10x-12=(x^2+5x+6)(x^2-2)
\displaystyle \therefore x^2-2=0
\displaystyle \therefore x=\pm\sqrt2
\displaystyle \therefore \text{Other zeroes are }\sqrt2\text{ and }-\sqrt2.
\\

\displaystyle \textbf{Question 24. }\text{Find all the zeroes of the polynomial }
\displaystyle 8x^4+8x^3-18x^2-20x-5,  \ \text{if it is given that two of its zeroes are } \\ \sqrt{\frac{5}{2}}\text{ and }-\sqrt{\frac{5}{2}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\sqrt{\frac{5}{2}}\text{ and }-\sqrt{\frac{5}{2}}\text{ are zeroes,}
\displaystyle \left(x-\sqrt{\frac{5}{2}}\right)\left(x+\sqrt{\frac{5}{2}}\right)=x^2-\frac{5}{2}
\displaystyle \therefore 2x^2-5\text{ is a factor.}
\displaystyle 8x^4+8x^3-18x^2-20x-5=(2x^2-5)(4x^2+4x+1)
\displaystyle =(2x^2-5)(2x+1)^2
\displaystyle \therefore 2x^2-5=0\text{ or }2x+1=0
\displaystyle \therefore x=\pm\sqrt{\frac{5}{2}}\text{ or }x=-\frac{1}{2}
\displaystyle \therefore \text{All zeroes are }\sqrt{\frac{5}{2}},\ -\sqrt{\frac{5}{2}},\ -\frac{1}{2},\ -\frac{1}{2}.
\\

\displaystyle \textbf{Question 25. }\text{In the figure, there are two points }D\text{ and }E\text{ on side }AB\text{ of }\triangle ABC
\displaystyle \text{such that }AD=BE. \ \text{If }DP\parallel BC\text{ and }EQ\parallel AC,\text{ then prove that } \\ PQ\parallel AB. \displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,\text{ we have}
\displaystyle DP\parallel BC\text{ and }EQ\parallel AC
\displaystyle \therefore \frac{AD}{DB}=\frac{AP}{PC}
\displaystyle \frac{BE}{EA}=\frac{BQ}{QC}\qquad (i)
\displaystyle \text{Also, we have}
\displaystyle \frac{AD}{DB}=\frac{AP}{PC}
\displaystyle \frac{AD}{DB}=\frac{BQ}{QC}\qquad (ii)
\displaystyle \text{From equations (i) and (ii), we get}
\displaystyle \frac{AP}{PC}=\frac{BQ}{QC}
\displaystyle \text{So, in }\triangle ABC,\ P\text{ and }Q\text{ divide sides }CA\text{ and }CB\text{ respectively in the same ratio.}
\displaystyle \therefore PQ\parallel AB
\displaystyle \text{By the converse of Basic Proportionality Theorem.}
\\

\displaystyle \textbf{Question 26. }\text{In }\triangle ABC,\text{ altitudes }AD\text{ and }CE\text{ intersect each other at the point } \\ P.\text{ Prove that}
\displaystyle \text{(i) }\triangle APE\sim\triangle CPD
\displaystyle \text{(ii) }AP\times PD=CP\times PE
\displaystyle \text{(iii) }\triangle ADB\sim\triangle CEB
\displaystyle \text{(iv) }AB\times CE=BC\times AD
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle AEP\text{ and }\triangle CDP,
\displaystyle \angle APE=\angle CPD\qquad [\text{Vertically opposite angles}]
\displaystyle \angle AEP=\angle CDP=90^\circ
\displaystyle \therefore \triangle AEP\sim\triangle CDP\qquad [\text{By AA criterion of similarity}]
\displaystyle \text{As }\triangle AEP\sim\triangle CDP,
\displaystyle \text{the ratio of their corresponding sides is equal.}
\displaystyle \frac{AP}{CP}=\frac{PE}{PD}
\displaystyle \therefore AP\times PD=CP\times PE
\displaystyle \text{In }\triangle ABD\text{ and }\triangle CBE,
\displaystyle \angle ADB=\angle CEB=90^\circ
\displaystyle \angle B\text{ is common.}
\displaystyle \therefore \triangle ABD\sim\triangle CBE\qquad [\text{By AA criterion of similarity}]
\displaystyle \text{So, the ratio of their corresponding sides are equal.}
\displaystyle \frac{AB}{BC}=\frac{AD}{CE}
\displaystyle \therefore AB\times CE=BC\times AD
\\

\displaystyle \textbf{Question 27. }\text{Prove that:}
\displaystyle (\cot A+\sec B)^2-(\tan B-\mathrm{cosec}\,A)^2=2(\cot A\sec B+\tan B\,\mathrm{cosec}\,A).
\displaystyle \text{Answer:}
\displaystyle \text{Taking LHS,}
\displaystyle (\cot A+\sec B)^2-(\tan B-\mathrm{cosec}\,A)^2
\displaystyle =\cot^2 A+\sec^2 B+2\cot A\sec B-(\tan^2 B+\mathrm{cosec}^2 A-2\tan B\,\mathrm{cosec}\,A)
\displaystyle =\cot^2 A-\mathrm{cosec}^2 A+\sec^2 B-\tan^2 B+2\cot A\sec B+2\tan B\,\mathrm{cosec}\,A
\displaystyle =(-1)+1+2\cot A\sec B+2\tan B\,\mathrm{cosec}\,A
\displaystyle =2\cot A\sec B+2\tan B\,\mathrm{cosec}\,A
\displaystyle =2(\cot A\sec B+\tan B\,\mathrm{cosec}\,A)
\displaystyle \therefore \text{LHS}=\text{RHS}\quad \text{Hence proved.}
\\

\displaystyle \textbf{Question 28. }\text{Prove that: }(\sin\theta+\cos\theta+1)(\sin\theta-1+\cos\theta)\sec\theta\,\mathrm{cosec}\,\theta=2.
\displaystyle \text{Answer:}
\displaystyle \text{Taking LHS,}
\displaystyle (\sin\theta+\cos\theta+1)(\sin\theta-1+\cos\theta)\sec\theta\,\mathrm{cosec}\,\theta
\displaystyle =\{(\sin\theta+\cos\theta)+1\}\{(\sin\theta+\cos\theta)-1\}\sec\theta\,\mathrm{cosec}\,\theta
\displaystyle =\{(\sin\theta+\cos\theta)^2-1\}\sec\theta\,\mathrm{cosec}\,\theta
\displaystyle =(\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1)\sec\theta\,\mathrm{cosec}\,\theta
\displaystyle =(2\sin\theta\cos\theta)\left(\frac{1}{\cos\theta}\right)\left(\frac{1}{\sin\theta}\right)
\displaystyle =2
\displaystyle \therefore \text{LHS}=\text{RHS}\quad \text{Hence proved.}
\\

\displaystyle \textbf{Question 29. }\text{If }\tan(20^\circ-3\alpha)=\cot(5\alpha-20^\circ),\text{ then find the value of }\alpha\text{ and hence evaluate:}
\displaystyle \sin\alpha\sec\alpha\tan\alpha-\mathrm{cosec}\,\alpha\cos\alpha\cot\alpha.
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle \tan(20^\circ-3\alpha)=\cot(5\alpha-20^\circ)
\displaystyle \tan(20^\circ-3\alpha)=\tan(90^\circ-5\alpha+20^\circ)
\displaystyle \tan(20^\circ-3\alpha)=\tan(110^\circ-5\alpha)
\displaystyle \text{On comparing,}
\displaystyle 20^\circ-3\alpha=110^\circ-5\alpha
\displaystyle 2\alpha=90^\circ
\displaystyle \alpha=45^\circ
\displaystyle \text{Now,}
\displaystyle \sin\alpha\sec\alpha\tan\alpha-\mathrm{cosec}\,\alpha\cos\alpha\cot\alpha
\displaystyle \text{Putting }\alpha=45^\circ,
\displaystyle =\sin45^\circ\sec45^\circ\tan45^\circ-\mathrm{cosec}\,45^\circ\cos45^\circ\cot45^\circ
\displaystyle =\frac{1}{\sqrt{2}}\times\sqrt{2}\times1-\sqrt{2}\times\frac{1}{\sqrt{2}}\times1
\displaystyle =1-1
\displaystyle =0
\\

\displaystyle \textbf{Question 30. }\text{The frequency distribution of weekly pocket money received by a} \\ \text{group of students is given below:}
\displaystyle \begin{array}{|c|c|}\hline \text{Pocket money (in Rs)} & \text{Number of Students}\\ \hline \text{More than or equal to }20&90\\ \hline \text{More than or equal to }40&76\\ \hline \text{More than or equal to }60&60\\ \hline \text{More than or equal to }80&55\\ \hline \text{More than or equal to }100&51\\ \hline \text{More than or equal to }120&49\\ \hline \text{More than or equal to }140&33\\ \hline \text{More than or equal to }160&12\\ \hline \text{More than or equal to }180&8\\ \hline \text{More than or equal to }200&4\\ \hline \end{array}
\displaystyle \text{Draw a 'more than type' ogive and from it, find median. Verify median by actual} \\ \text{calculations.}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Pocket money} & \text{More than cumulative frequency} & \text{Frequency}\\ \hline 20-40 & 90-76=14 & 14\\ \hline 40-60 & 76-60=16 & 16\\ \hline 60-80 & 60-55=5 & 5\\ \hline 80-100 & 55-51=4 & 4\\ \hline 100-120 & 51-49=2 & 2\\ \hline 120-140 & 49-33=16 & 16\\ \hline 140-160 & 33-12=21 & 21\\ \hline 160-180 & 12-8=4 & 4\\ \hline 180-200 & 8-4=4 & 4\\ \hline \end{array}
\displaystyle \text{For more than type ogive, plot the points }(20,90),(40,76),(60,60),(80,55),
\displaystyle (100,51),(120,49),(140,33),(160,12),(180,8),(200,4)\text{ and join them smoothly.}
\displaystyle \text{Total number of students }N=90
\displaystyle \therefore \frac{N}{2}=45
\displaystyle \text{From the ogive, the value corresponding to cumulative frequency }45\text{ gives median.}
\displaystyle \text{Median}\approx125
\displaystyle \text{Verification by calculation:}
\displaystyle \text{The median class is }120-140.
\displaystyle l=120,\quad cf=41,\quad f=16,\quad h=20
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)h
\displaystyle =120+\left(\frac{45-41}{16}\right)20
\displaystyle =120+5=125
\displaystyle \therefore \text{Median weekly pocket money}=\text{Rs }125
\\

\displaystyle \textbf{Question 31. }\text{Cost of living Index for some period is given in the following} \\ \text{frequency distribution:}
\displaystyle \begin{array}{|c|c|}\hline \text{Index} & \text{Number of weeks}\\ \hline 1500-1600&3\\ \hline 1600-1700&11\\ \hline 1700-1800&12\\ \hline 1800-1900&7\\ \hline 1900-2000&9\\ \hline 2000-2100&8\\ \hline 2100-2200&2\\ \hline \end{array}
\displaystyle \text{Find the mode and median for above data.}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Index} & \text{Number of weeks }(f) & \text{Cumulative frequency }(Cf)\\ \hline 1500-1600&3&3\\ \hline 1600-1700&11&14\\ \hline 1700-1800&12&26\\ \hline 1800-1900&7&33\\ \hline 1900-2000&9&42\\ \hline 2000-2100&8&50\\ \hline 2100-2200&2&52\\ \hline \text{Total}&N=52&\\ \hline \end{array}
\displaystyle \text{Modal class}=1700-1800
\displaystyle \text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h
\displaystyle l=1700,\quad f_1=12,\quad f_0=11,\quad f_2=7,\quad h=100
\displaystyle \text{Mode}=1700+\left(\frac{12-11}{2(12)-11-7}\right)\times100
\displaystyle =1700+\frac{1}{6}\times100
\displaystyle =1700+16.66
\displaystyle =1716.66
\displaystyle \frac{N}{2}=26
\displaystyle \text{Median class}=1700-1800
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-CF}{f}\right)\times h
\displaystyle l=1700,\quad \frac{N}{2}=26,\quad CF=14,\quad h=100,\quad f=12
\displaystyle \text{Median}=1700+\left(\frac{26-14}{12}\right)\times100
\displaystyle =1700+\left(\frac{12}{12}\right)\times100
\displaystyle =1700+100
\displaystyle =1800
\displaystyle \therefore \text{Mode}=1716.66\text{ and Median}=1800
\\

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