MATHEMATICS

\displaystyle \text{Series BVM/4} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 65/4/1} \hspace{1.0cm} | \hspace{1.0cm} \text{Set 2 }    

\displaystyle \text{Time Allowed : 3 hours} \hspace{5.0cm} \text{Maximum Marks : 80 }  

\displaystyle \textbf{Time\ Allowed : 3\ Hours \hspace{6cm} Maximum\ Marks : 100}
\displaystyle \textbf{General\ Instructions:}
\displaystyle (i)\ \text{All questions are compulsory.}
\displaystyle (ii)\ \text{The question paper consists of 29 questions divided into four sections A, B, C and D.}
\displaystyle \text{Section A comprises of 4 questions of one mark each, Section B comprises of 4 questions of}
\displaystyle \text{two marks each, Section C comprises of 11 questions of four marks each and Section D}
\displaystyle \text{comprises of 6 questions of six marks each.}
\displaystyle (iii)\ \text{All questions in Section A are to be answered in one word, one sentence or as per the}
\displaystyle \text{exact requirement of the question.}
\displaystyle (iv)\ \text{There is no overall choice. However, internal choice has been provided in 1 question of}
\displaystyle \text{Section A, 3 questions of Section B, 3 questions of Section C and 3 questions of Section D.}
\displaystyle \text{You have to attempt only one of the alternatives in all such questions.}
\displaystyle (v)\ \text{Use of calculators is not permitted. You may ask for logarithmic tables, if required.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{Question numbers 1 to 4 carry 1 mark each.}

\displaystyle \textbf{1. }\text{Form the differential equation representing the family of }   \text{curves } \\  y = \frac{A}{x} + 5\text{, by eliminating the arbitrary constant } A.
\displaystyle \text{Answer:}
\displaystyle  \text{Here, } y=\frac{A}{x}+5\qquad (1)
\displaystyle \text{Differentiate the above equation w.r.t. } x\text{, we get,}
\displaystyle \frac{dy}{dx}=-\frac{A}{x^2}\Rightarrow A=-x^2\frac{dy}{dx}
\displaystyle \text{Put the value of } A \text{ in equation (1), we get}
\displaystyle y=\frac{-x^2\frac{dy}{dx}}{x}+5\Rightarrow x\frac{dy}{dx}+y=5
\displaystyle \text{which is required differential equation}

\displaystyle \textbf{2. }\text{If } A \text{ is a square matrix of order 3, with } |A| = 9\text{, then write the }   \text{value of } |2.\mathrm{adj}A|.
\displaystyle \text{Answer:}
\displaystyle  \text{Here } |A|=9
\displaystyle \text{We know that } |\mathrm{adj}(A)|=|A|^{n-1}
\displaystyle \text{where } n=3 \text{ (order)}
\displaystyle \therefore |2.\mathrm{adj}(A)|=2^3|\mathrm{adj}(A)|=2^3|A|^{2}=8\times 9^2=648

\displaystyle \textbf{3. }\text{Find the acute angle between the planes } \overrightarrow{r}.(\widehat{i}-2\widehat{j}-2\widehat{k})=1   \text{ and } \\ \overrightarrow{r}.(3\widehat{i}-6\widehat{j}+2\widehat{k})=0.
\displaystyle \text{OR}
\displaystyle \text{Find the length of the intercept, cut off by the plane }   2x + y - z = 5\text{ on the x-axis}.
\displaystyle \text{Answer:}
\displaystyle  \text{Equation of two planes are:}
\displaystyle \overrightarrow{r}.(\widehat{i}-2\widehat{j}-2\widehat{k})=1 \text{ and } \overrightarrow{r}.(3\widehat{i}-6\widehat{j}+2\widehat{k})=0
\displaystyle \therefore \text{angle between them is given as:}
\displaystyle \cos\theta=\frac{|(\widehat{i}-2\widehat{j}-2\widehat{k}).(3\widehat{i}-6\widehat{j}+2\widehat{k})|}{\sqrt{1^2+(-2)^2+(-2)^2}\sqrt{3^2+(-6)^2+(2)^2}}
\displaystyle \Rightarrow \cos\theta=\frac{|3+12-4|}{\sqrt{9}\times\sqrt{49}}=\frac{11}{21}\Rightarrow \theta=\cos^{-1}\left(\frac{11}{21}\right)
\displaystyle \textbf{OR}
\displaystyle \text{Equation of plane is: } 2x+y-z=5\qquad (1)
\displaystyle \text{We know that on x-axis: } y=z=0
\displaystyle \text{Equation (1) becomes: } 2x=5
\displaystyle \Rightarrow x=\frac{5}{2} \text{ (length of intercept)}

\displaystyle \textbf{4. }\text{If } y = \log(\cos e^x)\text{, then find } \frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle  \text{If } y=\log(\cos e^x)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{\cos e^x}\frac{d(\cos e^x)}{dx}
\displaystyle =\frac{1}{\cos e^x}(-\sin e^x)\frac{d(e^x)}{dx}=-\tan(e^x)\cdot e^x=-e^x\tan(e^x)

\displaystyle \textbf{SECTION B}
\displaystyle \text{Question numbers 5 to 12 carry 2 marks each.}

\displaystyle \textbf{5. }\text{Find: } \int_{-\frac{\pi}{4}}^{0}\frac{1+\tan x}{1-\tan x}\,dx
\displaystyle \text{Answer:}
\displaystyle  \text{Since we know that: } \tan\left(\frac{\pi}{4}+x\right)=\frac{1+\tan x}{1-\tan x}
\displaystyle \therefore \int_{-\frac{\pi}{4}}^{0}\frac{1+\tan x}{1-\tan x}dx=\int_{-\frac{\pi}{4}}^{0}\tan\left(\frac{\pi}{4}+x\right)dx
\displaystyle =\left[\log\sec\left(\frac{\pi}{4}+x\right)\right]_{-\frac{\pi}{4}}^{0}
\displaystyle =\log\sec\left(\frac{\pi}{4}\right)-\log\sec(0)=\log\sqrt{2}-\log 1=\frac{1}{2}\log 2

\displaystyle \textbf{6. }\text{Let } * \text{ be an operation defined as } * : R \times R \to R \text{ such that} \\   a * b = 2a + b\text{, } a,b \in R\text{. Check if } * \text{ is a binary operation. If }   \text{yes, find if it is associative too}.
\displaystyle \text{Answer:}
\displaystyle  \text{Given operation: } a*b=2a+b,\ a,b\in R
\displaystyle \text{If any operation is binary operation, it must follow closure property.}
\displaystyle \text{Let } a\in R \text{ and } b\in R
\displaystyle \text{Then } 2a\in R,\ \text{also } 2a+b\in R
\displaystyle \therefore a*b\in R
\displaystyle \text{So, } * \text{ satisfies closure property.}
\displaystyle \text{Since } * \text{ is defined for all } a,b\in R,\ \therefore * \text{ is a binary operation.}
\displaystyle \text{Now, } (a*b)*c=(2a+b)*c=2(2a+b)+c=4a+2b+c
\displaystyle a*(b*c)=a*(2b+c)=2a+(2b+c)=2a+2b+c
\displaystyle \therefore (a*b)*c\neq a*(b*c)
\displaystyle \text{Hence, binary operation } * \text{ is not associative.}

\displaystyle \textbf{7. }\text{X and Y are two points with position vectors } 3\overrightarrow{a}+\overrightarrow{b} \text{ and } \overrightarrow{a}-3\overrightarrow{b} \text{ respectively.}
\displaystyle \text{Write the position vector of a point } Z \text{ which divides} \text{the line segment } XY  \text{ in the ratio }
\displaystyle  2:1 \text{ externally}.
\displaystyle \textbf{OR}
\displaystyle \text{Let } \overrightarrow{a}=\widehat{i}+2\widehat{j}-3\widehat{k} \text{ and } \overrightarrow{b}=3\widehat{i}-\widehat{j}+2\widehat{k} \text{ be two vectors. Show }  \text{that the vectors }
\displaystyle (\overrightarrow{a}+\overrightarrow{b}) \text{ and } (\overrightarrow{a}-\overrightarrow{b}) \text{ are perpendicular to each}   \text{other}.
\displaystyle \text{Answer:}
\displaystyle  \text{Position vector of } X=3\overrightarrow{a}+\overrightarrow{b}
\displaystyle \text{Position vector of } Y=\overrightarrow{a}-3\overrightarrow{b}
\displaystyle \therefore Z \text{ divides } XY \text{ externally in the ratio } 2:1
\displaystyle \text{Position vector of } Z=\frac{2(\overrightarrow{a}-3\overrightarrow{b})-(3\overrightarrow{a}+\overrightarrow{b})}{2-1}
\displaystyle =\frac{2\overrightarrow{a}-6\overrightarrow{b}-3\overrightarrow{a}-\overrightarrow{b}}{1}=-\overrightarrow{a}-7\overrightarrow{b}
\displaystyle \textbf{OR}
\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}-3\widehat{k},\ \overrightarrow{b}=3\widehat{i}-\widehat{j}+2\widehat{k}
\displaystyle \therefore \overrightarrow{a}+\overrightarrow{b}=4\widehat{i}+\widehat{j}-\widehat{k},\ \overrightarrow{a}-\overrightarrow{b}=-2\widehat{i}+3\widehat{j}-5\widehat{k}
\displaystyle (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=(4)(-2)+(1)(3)+(-1)(-5)=0
\displaystyle \text{Hence the two vectors } (\overrightarrow{a}+\overrightarrow{b}) \text{ and } (\overrightarrow{a}-\overrightarrow{b}) \text{ are perpendicular.}

\displaystyle \textbf{8. }\text{If } A \text{ and } B \text{ are symmetric matrices, such that } AB \text{ and } BA \text{ are both defined, then}
\displaystyle \text{prove that } AB-BA \text{ is a skew symmetric }   \text{matrix}.
\displaystyle \text{Answer:}
\displaystyle  \text{A and B are symmetric matrices i.e. } A'=A \text{ and } B'=B
\displaystyle \text{Also } (AB)'=B'A'
\displaystyle \text{Now, } (AB-BA)'=(AB)'-(BA)'=B'A'-A'B'=BA-AB
\displaystyle =-(AB-BA)
\displaystyle \text{Hence } AB-BA \text{ is skew symmetric matrix.}

\displaystyle \textbf{9. }\text{12 cards numbered 1 to 12 (one number on one card), are placed in a box and}
\displaystyle \text{mixed up thoroughly. Then a card is}   \text{drawn at random from the box. If it is known }
\displaystyle \text{that the number on the drawn card is greater than 5, find the probability that}
\displaystyle \text{the card bears an odd number}.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A \text{ be an event "the number on the card drawn is odd"}
\displaystyle \text{and } B \text{ be an event "the number on the drawn card is greater than 5".}
\displaystyle S=\{1,2,3,4,5,6,7,8,9,10,11,12\},\ A=\{1,3,5,7,9,11\},\ B=\{6,7,8,9,10,11,12\}
\displaystyle A\cap B=\{7,9,11\}
\displaystyle \therefore P(A)=\frac{6}{12}=\frac{1}{2},\ P(B)=\frac{7}{12} \text{ and } P(A\cap B)=\frac{3}{12}=\frac{1}{4}
\displaystyle \text{Required probability }=P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{7}{12}}=\frac{1}{4}\times\frac{12}{7}=\frac{3}{7}

\displaystyle \textbf{10. }\text{Out of 8 outstanding students of a school, in which there are 3 boys and 5 girls, a}
\displaystyle \text{team of 4 students is to be selected} \text{for a quiz competition. Find the probability}
\displaystyle \text{that 2 boys and 2 girls are selected}.
\displaystyle \textbf{OR}
\displaystyle \text{In a multiple choice examination with three possible answers for each of the five }
\displaystyle \text{questions, what is the probability that a candidate would get four or more correct}
\displaystyle \text{answers just by guessing?}
\displaystyle \text{Answer:}
\displaystyle  \text{Let } A \text{ be the event of selecting 2 boys \& 2 girls and } C \text{ be the event of selecting 4 students}
\displaystyle \text{Then, } A=\{(2B,2G)\} \text{ and } C=\{(1B,3G),(2B,2G),(3B,1G),(0B,4G)\}
\displaystyle \text{where B is used for boy and G is used for girl.}
\displaystyle \text{So, required probability }=P(A)=\frac{{}^{3}C_{2}\cdot {}^{5}C_{2}}{{}^{8}C_{4}}=\frac{3}{7}
\displaystyle \text{As, } {}^{n}C_{r}=\frac{n!}{r!(n-r)!}
\displaystyle \textbf{OR}
\displaystyle \text{Let } p \text{ be the probability of answer to be correct and } q \text{ be answer to get wrong.}
\displaystyle p=\frac{1}{3},\ q=1-p=1-\frac{1}{3}=\frac{2}{3},\ n=5,\ r=4,5
\displaystyle \text{and } P(X=r)={}^{n}C_{r}p^{r}q^{(n-r)}
\displaystyle P(\text{four or more successes})=P(X=4)+P(X=5)
\displaystyle ={}^{5}C_{4}\left(\frac{1}{3}\right)^{4}\left(\frac{2}{3}\right)+{}^{5}C_{5}\left(\frac{1}{3}\right)^{5}=\frac{11}{243}

\displaystyle \textbf{11. }\text{Solve the following differential equation:}   \frac{dy}{dx}+y=\cos x-\sin x
\displaystyle \text{Answer:}
\displaystyle  \text{Given, differential equation is } \frac{dy}{dx}+y=\cos x-\sin x
\displaystyle \text{which is of the form } \frac{dy}{dx}+Py=Q
\displaystyle \text{Here, } P=1,\ Q=\cos x-\sin x
\displaystyle \Rightarrow \text{I.F.}=e^{\int Pdx}=e^{\int 1dx}=e^x
\displaystyle \text{Now, solution of diff eqn. is given by}
\displaystyle y\cdot \text{I.F.}=\int (\text{I.F.})Qdx + c
\displaystyle \Rightarrow y e^x=\int e^x(\cos x-\sin x)dx + c
\displaystyle =e^x\cos x + c
\displaystyle \therefore y=\cos x + ce^{-x}
\displaystyle \text{Hence, solution is } y=\cos x + ce^{-x}

\displaystyle \textbf{12. }\text{Find: } \int x.\tan^{-1}x\,dx
\displaystyle \text{OR}
\displaystyle \text{Find: } \int \frac{dx}{\sqrt{5-4x-2x^2}}
\displaystyle \text{Answer:}
\displaystyle  \text{Let } I=\int x\tan^{-1}x\,dx
\displaystyle =\tan^{-1}x\cdot \frac{x^2}{2}-\int \frac{1}{1+x^2}\cdot \frac{x^2}{2}\,dx+c
\displaystyle =\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int \frac{x^2}{x^2+1}\,dx+c
\displaystyle =\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int \frac{x^2+1-1}{1+x^2}\,dx+c
\displaystyle =\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int \left(1-\frac{1}{1+x^2}\right)dx+c
\displaystyle =\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}(x-\tan^{-1}x)+c
\displaystyle =\frac{x^2}{2}\tan^{-1}x-\frac{x}{2}+\frac{1}{2}\tan^{-1}x+c
\displaystyle \textbf{OR}
\displaystyle \int \frac{dx}{\sqrt{5-4x-2x^2}}=\int \frac{dx}{\sqrt{5-(2x^2+4x)}}
\displaystyle =\int \frac{dx}{\sqrt{7-2(x^2+2x+1)}}
\displaystyle =\int \frac{dx}{\sqrt{7-2(x+1)^2}}
\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\left(\frac{7}{2}\right)-(x+1)^2}}
\displaystyle =\frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{x+1}{\sqrt{\frac{7}{2}}}\right)+c
\displaystyle =\frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{7}}\right)+c

\displaystyle \textbf{SECTION C}
\displaystyle \text{Question numbers 13 to 23 carry 4 marks each.}

\displaystyle \textbf{13. }\text{Using properties of determinants, find the value of } x \text{ for which}
\displaystyle \left|\begin{matrix}4-x&4+x&4+x\\4+x&4-x&4+x\\4+x&4+x&4-x\end{matrix}\right|=0
\displaystyle \text{Answer:}
\displaystyle  \text{Here, } \left|\begin{matrix}4-x&4+x&4+x\\4+x&4-x&4+x\\4+x&4+x&4-x\end{matrix}\right|=0
\displaystyle \Rightarrow \left|\begin{matrix}12+x&4+x&4+x\\12+x&4-x&4+x\\12+x&4+x&4-x\end{matrix}\right|=0
\displaystyle \text{(Operate: } C_1\rightarrow C_1+C_2+C_3\text{)}
\displaystyle =(12+x)\left|\begin{matrix}1&4-x&4+x\\1&4-x&4+x\\1&4+x&4-x\end{matrix}\right|=0
\displaystyle \Rightarrow \left|\begin{matrix}0&0&2x\\1&4-x&4+x\\1&4+x&4-x\end{matrix}\right|=0
\displaystyle \text{(Taking } (12+x) \text{ common from } C_1\text{)}
\displaystyle \text{(Operate: } R_1\rightarrow R_1-R_3\text{)}
\displaystyle \Rightarrow (12+x)(2x)\left|\begin{matrix}1&4-x&4+x\\1&4+x&4-x\end{matrix}\right|=0
\displaystyle \Rightarrow x^2(x+12)=0\Rightarrow x=0 \text{ or } x=-12

\displaystyle \textbf{14. }\text{Solve the differential equation } \frac{dy}{dx}=1+x^2+y^2+x^2y^2\text{, given } \text{that } y=1  \\ \text{when } x=0.
\displaystyle \text{OR}
\displaystyle \text{Find the particular solution of the differential equation}  \frac{dy}{dx}=\frac{xy}{x^2+y^2}\text{, given that } \\  y=1 \text{ when } x=0.
\displaystyle \text{Answer:}
\displaystyle  \text{Given differential equation can be written as}
\displaystyle \frac{dy}{dx}=(1+x^2)+y^2(1+x^2)=(1+x^2)(1+y^2)
\displaystyle \Rightarrow \frac{1}{1+y^2}dy=(1+x^2)dx
\displaystyle \text{Integrating both sides}
\displaystyle \int \frac{1}{1+y^2}dy=\int (1+x^2)dx
\displaystyle \Rightarrow \tan^{-1}y=x+\frac{x^3}{3}+C
\displaystyle \text{Put } y=1 \text{ and } x=0
\displaystyle \Rightarrow \frac{\pi}{4}=C
\displaystyle \therefore \tan^{-1}y=x+\frac{x^3}{3}+\frac{\pi}{4}
\displaystyle \textbf{OR}
\displaystyle \frac{dy}{dx}=\frac{xy}{x^2+y^2}
\displaystyle \text{This is a homogeneous differential equation}
\displaystyle \text{Substitute } y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{x(vx)}{x^2+(vx)^2}=\frac{v}{1+v^2}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v}{1+v^2}-v=\frac{v-v-v^3}{1+v^2}=\frac{-v^3}{1+v^2}
\displaystyle \Rightarrow \frac{1+v^2}{v^3}dv=-\frac{dx}{x}
\displaystyle \Rightarrow \left(\frac{1}{v^3}+\frac{1}{v}\right)dv=-\frac{dx}{x}
\displaystyle \text{Integrating both sides}
\displaystyle -\frac{1}{2v^2}+\log|v|=-\log|x|+C
\displaystyle \Rightarrow -\frac{1}{2v^2}+\log|vx|=C
\displaystyle \Rightarrow -\frac{x^2}{2y^2}+\log|y|=C
\displaystyle \text{Given } y=1 \text{ when } x=0\Rightarrow C=0
\displaystyle \therefore \log|y|=\frac{x^2}{2y^2}

\displaystyle \textbf{15. }\text{Let } A=R-\{2\} \text{ and } B=R-\{1\}\text{. If } f:A\to B \text{ is a function } \text{defined by }
\displaystyle f(x)=\frac{x-1}{x-2}\text{, show that } f \text{ is one-one and onto. }   \text{Hence, find } f^{-1}.
\displaystyle \text{OR}
\displaystyle \text{Show that the relation } S \text{ on the set } A=\{x\in Z:0\leq x\leq 12\} \text{, }  \text{given by }
\displaystyle S=\{(a,b):a,b\in Z,\ |a-b|\text{ is divisible by 3}\}\text{ is an }   \text{equivalence relation}.
\displaystyle \text{Answer:}
\displaystyle  \text{Let } x_1,x_2 \text{ be any two elements of set } A  \text{such that } f(x_1)=f(x_2)
\displaystyle \Rightarrow \frac{x_1-1}{x_1-2}=\frac{x_2-1}{x_2-2}
\displaystyle \Rightarrow x_1x_2-2x_1-x_2+2=x_1x_2-2x_2-x_1+2
\displaystyle \Rightarrow -2x_1+x_1=-2x_2+x_2\Rightarrow x_1=x_2
\displaystyle \text{Thus, } f \text{ is one-one}
\displaystyle \text{Let } y \text{ be any element of } B\text{, then } f(x)=y
\displaystyle \Rightarrow \frac{x-1}{x-2}=y,\ x\neq 2
\displaystyle \Rightarrow x=\frac{1-2y}{1-y}
\displaystyle \text{Clearly, } x=\frac{1-2y}{1-y} \text{ is real for all } y\neq 1
\displaystyle \Rightarrow \text{For each } y\in B,\ \exists x\in A \text{ such that } f(x)=y
\displaystyle \text{Thus, } f \text{ is onto } \Rightarrow f \text{ is invertible}
\displaystyle x=\frac{1-2y}{1-y}\Rightarrow f^{-1}(y)=\frac{1-2y}{1-y}
\displaystyle \text{Hence, } f^{-1}(x)=\frac{1-2x}{1-x} \text{ for all } x\in R-\{1\}
\displaystyle \textbf{OR}
\displaystyle \text{Given set } A=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}
\displaystyle \text{and } S=\{(a,b):a,b\in Z,\ |a-b|\text{ is divisible by 3}\}

\displaystyle \text{(i) For all } a\in A,\ (a,a)\in S
\displaystyle \text{i.e., } |a-a|=0 \text{ is divisible by 3}
\displaystyle \therefore S \text{ is reflexive on } A

\displaystyle \text{(ii) For all } a,b\in A
\displaystyle \text{If } (a,b)\in S,\ \text{i.e., } |a-b| \text{ is divisible by 3}
\displaystyle \Rightarrow |b-a| \text{ is also divisible by 3}
\displaystyle \therefore S \text{ is symmetric on } A

\displaystyle \text{(iii) For all } a,b,c\in A
\displaystyle \text{Let } (a,b)\in S \text{ and } (b,c)\in S
\displaystyle \text{i.e., } |a-b| \text{ is divisible by 3 and } |b-c| \text{ is divisible by 3}
\displaystyle \text{and let } (a-b)=\pm 3q,\ (b-c)=\pm 3p
\displaystyle \text{Adding we get:}
\displaystyle a-c=\pm 3(p+q)=\pm 3m \text{ (say)}
\displaystyle \Rightarrow |a-c|=3m \text{ is divisible by 3}
\displaystyle \therefore S \text{ is transitive on } A
\displaystyle \text{Hence, } S \text{ is an equivalence relation on } A
\displaystyle \text{A relation is equivalent when it is reflexive, symmetric and transitive.}

\displaystyle \textbf{16. }\text{Integrate the function } \frac{\cos(x+a)}{\sin(x+b)} \text{ w.r.t. } x.
\displaystyle \text{Answer:}
\displaystyle  \text{Let } I=\int \frac{\cos(x+a)}{\sin(x+b)}\,dx
\displaystyle =\int \frac{\cos[(x+b)+(a-b)]}{\sin(x+b)}\,dx
\displaystyle =\int \frac{\cos(x+b)\cos(a-b)-\sin(x+b)\sin(a-b)}{\sin(x+b)}\,dx
\displaystyle =\cos(a-b)\int \frac{\cos(x+b)}{\sin(x+b)}\,dx-\sin(a-b)\int dx
\displaystyle =\cos(a-b)\int \cot(x+b)\,dx-\sin(a-b)x+c
\displaystyle =\cos(a-b)\log\sin(x+b)-x\sin(a-b)+c

\displaystyle \textbf{17. }\text{If } x=\sin t\text{, } y=\sin pt\text{, prove that } (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0.
\displaystyle \text{OR}
\displaystyle \text{Differentiate } \tan^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right] \text{ with respect to }   \cos^{-1}x^2.
\displaystyle \text{Answer:}
\displaystyle  \text{Let } x=\sin t \text{ and } y=\sin pt
\displaystyle \therefore \frac{dx}{dt}=\cos t \text{ and } \frac{dy}{dt}=p\cos pt
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{p\cos pt}{\cos t}
\displaystyle \Rightarrow y'\cos t=p\cos pt \qquad \left(\frac{dy}{dx}=y'\right)
\displaystyle \text{Squaring both sides, we get:}
\displaystyle y'^2\cos^2 t=p^2\cos^2 pt
\displaystyle \Rightarrow y'^2(1-\sin^2 t)=p^2(1-\sin^2 pt)
\displaystyle \Rightarrow y'^2(1-x^2)=p^2(1-y^2)
\displaystyle \text{Differentiate w.r.t. } x \text{, we get}
\displaystyle 2y'y''(1-x^2)-2xy'^2=p^2(-2yy')
\displaystyle \Rightarrow y''(1-x^2)-xy'+p^2y=0
\displaystyle \Rightarrow (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0
\displaystyle \textbf{OR}
\displaystyle y=\tan^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]
\displaystyle \text{Putting } x^2=\cos 2\theta\text{, we have}
\displaystyle y=\tan^{-1}\left[\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right]
\displaystyle \Rightarrow y=\tan^{-1}\left[\frac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}\right]
\displaystyle \Rightarrow y=\tan^{-1}\left(\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\right)
\displaystyle \Rightarrow y=\tan^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)
\displaystyle \Rightarrow y=\tan^{-1}\left[\tan\left(\frac{\pi}{4}-\theta\right)\right]
\displaystyle \Rightarrow y=\frac{\pi}{4}-\theta
\displaystyle \therefore y=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x^2 \qquad (x^2=\cos 2\theta)
\displaystyle \text{Differentiating both sides with respect to } x\text{, we get}
\displaystyle \frac{dy}{dx}=0-\frac{1}{2}\left(\frac{-1}{\sqrt{1-(x^2)^2}}\right)\cdot 2x
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x}{\sqrt{1-x^4}}
\displaystyle \text{Now let, } z=\cos^{-1}x^2\Rightarrow \frac{dz}{dx}=\frac{-2x}{\sqrt{1-x^4}}
\displaystyle \Rightarrow \frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{x}{\sqrt{1-x^4}}\times \frac{\sqrt{1-x^4}}{-2x}=-\frac{1}{2}

\displaystyle \textbf{18. }\text{Prove that: } \cos^{-1}\left(\frac{12}{13}\right)+\sin^{-1}\left(\frac{3}{5}\right)=\sin^{-1}\left(\frac{56}{65}\right)
\displaystyle \text{Answer:}
\displaystyle  \text{Let } \cos^{-1}\left(\frac{12}{13}\right)=\alpha,\ \therefore \cos\alpha=\frac{12}{13},\ \sin\alpha=\frac{5}{13}
\displaystyle \sin^{-1}\left(\frac{3}{5}\right)=\beta,\ \therefore \sin\beta=\frac{3}{5},\ \cos\beta=\frac{4}{5}
\displaystyle \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta
\displaystyle =\frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}=\frac{20+36}{65}=\frac{56}{65}
\displaystyle \therefore \alpha+\beta=\sin^{-1}\left(\frac{56}{65}\right)
\displaystyle \Rightarrow \cos^{-1}\left(\frac{12}{13}\right)+\sin^{-1}\left(\frac{3}{5}\right)=\sin^{-1}\left(\frac{56}{65}\right)

\displaystyle \textbf{19. }\text{If } y=(x)^{\cos x}+(\cos x)^{\sin x}\text{, find } \frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle  \text{Given } y=(x)^{\cos x}+(\cos x)^{\sin x}
\displaystyle \text{Let } u=x^{\cos x}
\displaystyle \text{Taking log on both sides, we get}
\displaystyle \log u=\cos x\cdot \log x
\displaystyle \text{Differentiate w.r.t. } x
\displaystyle \frac{1}{u}\frac{du}{dx}=\cos x\cdot \frac{1}{x}+\log x(-\sin x)=\frac{\cos x}{x}-\sin x\log x
\displaystyle \Rightarrow \frac{du}{dx}=x^{\cos x}\left[\frac{\cos x}{x}-\sin x\log x\right]
\displaystyle \text{Now let } v=(\cos x)^{\sin x}
\displaystyle \Rightarrow \log v=\sin x\cdot \log(\cos x)
\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx}=\sin x\left(\frac{-\sin x}{\cos x}\right)+\log(\cos x)\cos x
\displaystyle =-\sin x\tan x+\cos x\log(\cos x)
\displaystyle \Rightarrow \frac{dv}{dx}=(\cos x)^{\sin x}\left[\cos x\log(\cos x)-\sin x\tan x\right]
\displaystyle \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
\displaystyle =x^{\cos x}\left[\frac{\cos x}{x}-\sin x\log x\right]+(\cos x)^{\sin x}\left[\cos x\log(\cos x)-\sin x\tan x\right]

\displaystyle \textbf{20. }\text{Prove that } \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\text{, and hence evaluate}
\displaystyle \int_{0}^{1}x^2(1-x)^n\,dx.
\displaystyle \text{Answer:}
\displaystyle  \text{Put } a-x=t\Rightarrow dx=-dt
\displaystyle \text{When } x=a,\ t=0 \text{ and when } x=0,\ t=a
\displaystyle \therefore \int_{0}^{a}f(a-x)\,dx=\int_{a}^{0}f(t)(-dt)=\int_{0}^{a}f(t)\,dt=\int_{0}^{a}f(x)\,dx
\displaystyle \text{Hence proved}
\displaystyle \int_{0}^{1}x^2(1-x)^n dx=\int_{0}^{1}(1-x)^2x^n dx
\displaystyle =\int_{0}^{1}(1-2x+x^2)x^n dx=\int_{0}^{1}(x^n-2x^{n+1}+x^{n+2})dx
\displaystyle =\left[\frac{x^{n+1}}{n+1}-\frac{2x^{n+2}}{n+2}+\frac{x^{n+3}}{n+3}\right]_{0}^{1}
\displaystyle =\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}=\frac{2}{(n+1)(n+2)(n+3)}

\displaystyle \textbf{21. }\text{Find the value of } x\text{, for which the four points } A(x,-1,-1)\text{, } B(4,5,1)\text{, } C(3,9,4)
\displaystyle  \text{ and } D(-4,4,4) \text{ are coplanar}.
\displaystyle \text{Answer:}
\displaystyle  \text{Points } A(x,-1,-1),\ B(4,5,1),\ C(3,9,4),\ D(-4,4,4) \text{ are coplanar}
\displaystyle \therefore \left|\begin{matrix}x-4&-1-5&-1-1\\3-4&9-5&4-1\\-4-4&4-5&4-1\end{matrix}\right|=0
\displaystyle =\left|\begin{matrix}x-4&-6&-2\\-1&4&3\\-8&-1&3\end{matrix}\right|=0
\displaystyle \Rightarrow (x-4)(12+3)+6(-3+24)-2(1+32)=0
\displaystyle \Rightarrow 15(x-4)+6(21)-2(33)=0
\displaystyle \Rightarrow 15(x-4)+126-66=0
\displaystyle \Rightarrow 15x-60+60=0\Rightarrow x=0

\displaystyle \textbf{22. }\text{A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder}
\displaystyle \text{is dragged away from the wall along the ground at the rate of 2 cm/sec. How fast}
\displaystyle \text{is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall?}
\displaystyle \text{Answer:}
\displaystyle  \text{Let } OA=x,\ OB=y,\ x^2+y^2=169
\displaystyle \text{On differentiating: } 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0
\displaystyle \Rightarrow x\frac{dx}{dt}+y\frac{dy}{dt}=0
\displaystyle \text{When } x=5,\ x^2+y^2=169 \\ \Rightarrow 25+y^2=169\Rightarrow y=12
\displaystyle \text{Given } \frac{dx}{dt}=0.02\ \text{m/sec}
\displaystyle \text{Put values: } 5(0.02)+12\frac{dy}{dt}=0
\displaystyle \Rightarrow 0.1+12\frac{dy}{dt}=0
\displaystyle \Rightarrow \frac{dy}{dt}=-\frac{0.1}{12}=-\frac{1}{120}
\displaystyle \text{Hence, height decreases at } \frac{1}{120}\ \text{m/sec}=\frac{5}{6}\ \text{cm/sec}

\displaystyle \textbf{23. }\text{Find the vector equation of the plane determined by the } \text{points } A(3,-1,2)\text{, } B(5,2,4)
\displaystyle  \text{ and } C(-1,-1,6)\text{. Hence, find the }   \text{distance of this plane, thus obtained, from the origin}.
\displaystyle \text{Answer:}
\displaystyle  \text{Given points are } A(3,-1,2),\ B(5,2,4) \text{ and } C(-1,-1,6)
\displaystyle \therefore \overrightarrow{AB}=5\widehat{i}+2\widehat{j}+4\widehat{k}-(3\widehat{i}-\widehat{j}+2\widehat{k})=2\widehat{i}+3\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{AC}=-\widehat{i}-\widehat{j}+6\widehat{k}-(3\widehat{i}-\widehat{j}+2\widehat{k})=-4\widehat{i}+0\widehat{j}+4\widehat{k}
\displaystyle \text{Now, } \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{matrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&2\\-4&0&4\end{matrix}\right|=12\widehat{i}-16\widehat{j}+12\widehat{k}
\displaystyle \text{The required plane passes through } A(3,-1,2) \text{ and normal is along vector } \overrightarrow{n}
\displaystyle \therefore \text{The vector equation of the required plane is } \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}
\displaystyle \Rightarrow \overrightarrow{r}.(12\widehat{i}-16\widehat{j}+12\widehat{k})=(3\widehat{i}-\widehat{j}+2\widehat{k}).(12\widehat{i}-16\widehat{j}+12\widehat{k})
\displaystyle =36+16+24=76
\displaystyle \Rightarrow \overrightarrow{r}.(12\widehat{i}-16\widehat{j}+12\widehat{k})=76
\displaystyle \Rightarrow \overrightarrow{r}.(3\widehat{i}-4\widehat{j}+3\widehat{k})=19
\displaystyle \Rightarrow \overrightarrow{r}.(3\widehat{i}-4\widehat{j}+3\widehat{k})-19=0
\displaystyle \text{The distance of the origin from this plane}
\displaystyle =\frac{|3\times 0-4\times 0+3\times 0-19|}{\sqrt{3^2+(-4)^2+3^2}}=\frac{|-19|}{\sqrt{9+16+9}}=\frac{19}{\sqrt{34}} \text{ units}

\displaystyle \textbf{SECTION D}
\displaystyle \text{Question numbers 24 to 29 carry 4 marks each.}

\displaystyle \textbf{24. }\text{Using integration, find the area of the greatest rectangle that can be inscribed in}
\displaystyle \text{an ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
\displaystyle \text{Answer:}
\displaystyle  \text{Let the coordinates of the vertices of rectangle } ABCD \text{ be}
\displaystyle A(a\cos\theta,b\sin\theta),\ B(-a\cos\theta,b\sin\theta),\ C(-a\cos\theta,-b\sin\theta) \text{ and } D(a\cos\theta,-b\sin\theta)
\displaystyle \text{Length of rectangle, } AB=2a\cos\theta
\displaystyle \text{Breadth of rectangle, } AD=2b\sin\theta
\displaystyle \text{Area of rectangle } ABCD=AB\times AD=2a\cos\theta\times 2b\sin\theta\displaystyle \text{Area of rectangle } (A)=2ab\sin 2\theta
\displaystyle \text{Differentiate w.r.t. } \theta;\ \frac{dA}{d\theta}=2ab\cos 2\theta.(2)
\displaystyle \text{For maximum or minimum, put } \frac{dA}{d\theta}=0\Rightarrow 4ab\cos 2\theta=0\Rightarrow \cos 2\theta=0
\displaystyle \Rightarrow 2\theta=\frac{\pi}{2}\Rightarrow \theta=\frac{\pi}{4}
\displaystyle \text{Also, } \frac{d^2A}{d\theta^2}=-8ab\sin 2\theta<0
\displaystyle \therefore \text{Area is maximum at } \theta=\frac{\pi}{4}
\displaystyle \therefore \text{Coordinates of } A \text{ are } \left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right) \text{ and coordinates of } B \text{ are } \left(-\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)
\displaystyle \therefore \text{Equation of } AB \text{ is: } y=\frac{b}{\sqrt{2}}
\displaystyle \therefore \text{Required area}
\displaystyle =4\int_{0}^{\frac{a}{\sqrt{2}}}\frac{b}{\sqrt{2}}\,dx=\frac{4b}{\sqrt{2}}\left[x\right]_{0}^{\frac{a}{\sqrt{2}}}=\frac{4b}{\sqrt{2}}\left[\frac{a}{\sqrt{2}}\right]
\displaystyle =\frac{4ab}{(\sqrt{2})^2}=2ab \text{ sq. units}

\displaystyle \textbf{25. }\text{An insurance company insured 3000 cyclists, 6000 scooter drivers and 9000 car}
\displaystyle \text{drivers. The probability of an accident involving a cyclist, a scooter driver and a car}
\displaystyle \text{drivers are 0.3, } 0.05 \text{ and } 0.02 \text{ respectively. One of the insured persons meets with an}
\displaystyle \text{accident. What is the probability that he is a cyclist?}
\displaystyle \text{Answer:}
\displaystyle  \text{An insurance company insured 3000 cyclists, 6000 scooter drivers and 9000 car drivers}
\displaystyle \therefore \text{Total number of drivers }=18000
\displaystyle P(E_1)=\frac{3000}{18000}=\frac{1}{6},\quad P(E_2)=\frac{6000}{18000}=\frac{1}{3},\quad P(E_3)=\frac{9000}{18000}=\frac{1}{2}
\displaystyle \text{Let } A \text{ be the event that an insured person meets with an accident}
\displaystyle P\left(\frac{A}{E_1}\right)=0.3,\quad P\left(\frac{A}{E_2}\right)=0.05,\quad P\left(\frac{A}{E_3}\right)=0.02
\displaystyle \text{Required probability }=P\left(\frac{E_1}{A}\right)=\frac{P\left(\frac{A}{E_1}\right)P(E_1)}{P\left(\frac{A}{E_1}\right)P(E_1)+P\left(\frac{A}{E_2}\right)P(E_2)+P\left(\frac{A}{E_3}\right)P(E_3)}
\displaystyle =\frac{0.3\times \frac{1}{6}}{0.3\times \frac{1}{6}+0.05\times \frac{1}{3}+0.02\times \frac{1}{2}}
\displaystyle =\frac{\frac{1}{20}}{\frac{1}{20}+\frac{1}{60}+\frac{1}{100}}
\displaystyle =\frac{\frac{1}{20}}{\frac{15+5+3}{300}}=\frac{\frac{1}{20}}{\frac{23}{300}}=\frac{1}{20}\times \frac{300}{23}=\frac{15}{23}

\displaystyle \textbf{26. }\text{Using elementary row transformations, find the inverse of}   \text{the matrix } \left[\begin{matrix}2&-3&5\\3&2&-4\\1&1&-2\end{matrix}\right].
\displaystyle \textbf{OR}
\displaystyle \text{Using matrices, solve the following system of linear }   \text{equations:}
\displaystyle x+2y-3z=-4
\displaystyle 2x+3y+2z=2
\displaystyle 3x-3y-4z=11
\displaystyle \text{Answer:}
\displaystyle  \text{Let } A=\left[\begin{matrix}2&-3&5\\3&2&-4\\1&1&-2\end{matrix}\right]
\displaystyle \text{We know that } A=IA
\displaystyle \Rightarrow \left[\begin{matrix}2&-3&5\\3&2&-4\\1&1&-2\end{matrix}\right]\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right]
\displaystyle \text{Operate: } R_1\rightarrow R_1+3R_3,\ R_2\rightarrow R_2-3R_3
\displaystyle \Rightarrow \left[\begin{matrix}5&0&-1\\0&-1&2\\1&1&-2\end{matrix}\right]\left[\begin{matrix}1&0&3\\0&1&-3\\0&0&1\end{matrix}\right]
\displaystyle \text{Operate: } R_1\rightarrow R_1-4R_3
\displaystyle \Rightarrow \left[\begin{matrix}1&-4&7\\0&-1&2\\1&1&-2\end{matrix}\right]\left[\begin{matrix}1&0&-1\\0&1&-3\\0&0&1\end{matrix}\right]
\displaystyle \text{Operate: } R_3\rightarrow R_3-4R_1
\displaystyle \Rightarrow \left[\begin{matrix}1&-4&7\\0&-1&2\\5&0&-1\end{matrix}\right]\left[\begin{matrix}1&0&-1\\0&1&-3\\0&0&3\end{matrix}\right]
\displaystyle \text{Operate: } R_1\rightarrow R_1-4R_2
\displaystyle \Rightarrow \left[\begin{matrix}1&0&-1\\0&-1&2\\5&0&-1\end{matrix}\right]\left[\begin{matrix}1&-4&11\\0&1&-3\\0&0&3\end{matrix}\right]
\displaystyle \text{Operate: } R_3\rightarrow R_3-5R_1
\displaystyle \Rightarrow \left[\begin{matrix}1&0&-1\\0&-1&2\\0&0&4\end{matrix}\right]\left[\begin{matrix}1&-4&11\\0&1&-3\\-4&20&-52\end{matrix}\right]
\displaystyle \text{Operate: } R_1\rightarrow R_1+\frac{R_3}{4}
\displaystyle \Rightarrow \left[\begin{matrix}1&0&0\\0&-1&2\\0&0&4\end{matrix}\right]\left[\begin{matrix}0&1&-2\\0&1&-3\\-4&20&-52\end{matrix}\right]
\displaystyle \text{Operate: } R_2\rightarrow R_2-\frac{R_3}{2}
\displaystyle \Rightarrow \left[\begin{matrix}1&0&0\\0&-1&0\\0&0&4\end{matrix}\right]\left[\begin{matrix}0&1&-2\\2&-9&23\\-4&20&-52\end{matrix}\right]
\displaystyle \text{Operate: } R_2\rightarrow -R_2,\ R_3\rightarrow \frac{R_3}{4}
\displaystyle \Rightarrow \left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right]\left[\begin{matrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{matrix}\right]
\displaystyle \therefore A^{-1}=\left[\begin{matrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{matrix}\right]
\displaystyle \textbf{OR}
\displaystyle \text{Given system: } x+2y-3z=-4,\ 2x+3y+2z=2,\ 3x-3y-4z=11
\displaystyle \text{Matrix form: } AX=B
\displaystyle A=\left[\begin{matrix}1&2&-3\\2&3&2\\3&-3&-4\end{matrix}\right],\ X=\left[\begin{matrix}x\\y\\z\end{matrix}\right],\ B=\left[\begin{matrix}-4\\2\\11\end{matrix}\right]
\displaystyle |A|=1(-12+6)-2(-8-6)-3(-6-9)=-6+28+45=67\neq 0
\displaystyle \therefore A^{-1} \text{ exists and system has unique solution}
\displaystyle \text{Cofactors of } A:\ A_{11}=-6,\ A_{12}=14,\ A_{13}=-15,\ A_{21}=17,\ A_{22}=5,\ A_{23}=9,\ A_{31}=13,\ A_{32}=-8,\ A_{33}=-1
\displaystyle \mathrm{adj}(A)=\left[\begin{matrix}-6&17&13\\14&5&-8\\-15&9&-1\end{matrix}\right]
\displaystyle A^{-1}=\frac{1}{67}\left[\begin{matrix}-6&17&13\\14&5&-8\\-15&9&-1\end{matrix}\right]
\displaystyle X=A^{-1}B=\frac{1}{67}\left[\begin{matrix}201\\-134\\67\end{matrix}\right]
\displaystyle \therefore x=3,\ y=-2,\ z=1

\displaystyle \textbf{27. }\text{Using integration, find the area of the region bounded by } \text{the parabola }
\displaystyle y^2=4x \text{ and the circle } 4x^2+4y^2=9.
\displaystyle \text{OR}
\displaystyle \text{Using the method of integration, find the area of the region } \text{bounded by the lines }
\displaystyle  3x-2y+1=0\text{, } 2x+3y-21=0 \text{ and}  x-5y+9=0.
\displaystyle \text{Answer:}
\displaystyle  \text{Given equation of circle is } 4x^2+4y^2=9
\displaystyle \Rightarrow x^2+y^2=\frac{9}{4}\Rightarrow (x-0)^2+(y-0)^2=\left(\frac{3}{2}\right)^2
\displaystyle \therefore \text{Centre of the circle is } (0,0) \text{ and radius is } \frac{3}{2}
\displaystyle \text{Now, equation of parabola is given by } y^2=4x  \text{ which is symmetric about x-axis}\displaystyle \text{Now, } x^2+y^2=\frac{9}{4}\Rightarrow y=\sqrt{\left(\frac{3}{2}\right)^2-x^2}
\displaystyle \text{Also, } 4x^2+4y^2=9
\displaystyle \Rightarrow 4x^2+4(4x)=9 \qquad (\because y^2=4x)
\displaystyle \Rightarrow 4x^2+16x-9=0
\displaystyle \Rightarrow (2x+9)(2x-1)=0\Rightarrow x=-\frac{9}{2},\ x=\frac{1}{2}
\displaystyle \text{Now, when } x=\frac{1}{2},\ y=\sqrt{2}
\displaystyle \therefore \text{Area of shaded region}
\displaystyle =2\left[\int_{0}^{\frac{1}{2}}(\text{y of parabola})\,dx+\int_{\frac{1}{2}}^{\frac{3}{2}}(\text{y of circle})\,dx\right]
\displaystyle =2\left[\int_{0}^{\frac{1}{2}}2\sqrt{x}\,dx+\int_{\frac{1}{2}}^{\frac{3}{2}}\sqrt{\left(\frac{3}{2}\right)^2-x^2}\,dx\right]
\displaystyle =2\times \frac{4}{3}\left[x^{\frac{3}{2}}\right]_{0}^{\frac{1}{2}}+2\left[\frac{x}{2}\sqrt{\frac{9}{4}-x^2}+\frac{9}{8}\sin^{-1}\left(\frac{x}{\frac{3}{2}}\right)\right]_{\frac{1}{2}}^{\frac{3}{2}}
\displaystyle \left[\because \int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{1}{2}a^2\sin^{-1}\left(\frac{x}{a}\right)+c\right]
\displaystyle =\frac{8}{3}\left(\frac{1}{2}\right)^{\frac{3}{2}}+2\left[\frac{9}{8}\sin^{-1}1-\frac{1}{4}\sqrt{2}-\frac{9}{8}\sin^{-1}\left(\frac{1}{3}\right)\right]
\displaystyle \left[\because \sin\frac{\pi}{2}=1\right]
\displaystyle =\frac{8}{3}\times \frac{1}{2\sqrt{2}}+\frac{9\pi}{8}-\frac{1}{\sqrt{2}}-\frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)
\displaystyle =\frac{4}{3\sqrt{2}}-\frac{1}{\sqrt{2}}+\frac{9\pi}{8}-\frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)
\displaystyle =\frac{1}{3\sqrt{2}}+\frac{9}{4}\left(\frac{\pi}{2}-\sin^{-1}\left(\frac{1}{3}\right)\right)
\displaystyle =\frac{\sqrt{2}}{6}+\frac{9}{4}\cos^{-1}\left(\frac{1}{3}\right)\text{ sq. units}
\displaystyle \textbf{OR}
\displaystyle \text{Given equation of lines are}
\displaystyle 3x-2y=-1 \qquad (i)
\displaystyle 2x+3y=21 \qquad (ii)
\displaystyle x-5y=-9 \qquad (iii)
\displaystyle \text{Solving eqs. } (i) \text{ and } (ii) \text{ as follows}
\displaystyle 2(3x-2y=-1)
\displaystyle 3(2x+3y=21)
\displaystyle \Rightarrow 6x-4y=-2
\displaystyle \underline{6x+9y=63}
\displaystyle \Rightarrow -13y=-65
\displaystyle \Rightarrow y=5
\displaystyle \text{Putting } y=5 \text{ in eq. } (i)
\displaystyle 3x-2(5)=-1\Rightarrow 3x-10=-1\Rightarrow x=\frac{9}{3}=3
\displaystyle \therefore \text{Lines } (i) \text{ and } (ii) \text{ intersect each other at point } (3,5)
\displaystyle \text{Now, solving eqs. } (ii) \text{ and } (iii)\text{, we get}
\displaystyle y=\frac{39}{13}=3
\displaystyle \text{Putting } y=3 \text{ in eq. } (iii)\text{, we get } x-5(3)=-9\Rightarrow x=6
\displaystyle \therefore \text{Lines } (ii) \text{ and } (iii) \text{ intersect each other at point } (6,3)
\displaystyle \text{Now, solving eqs. } (i) \text{ and } (iii)\text{, we get}
\displaystyle y=\frac{26}{13}=2
\displaystyle \text{Put } y=2 \text{ in eq. } (iii)\text{, we get } x-5(2)=-9\Rightarrow x=1
\displaystyle \therefore \text{Lines } (i) \text{ and } (iii) \text{ intersect each other at point } (1,2)\displaystyle \text{Now required area of triangle}
\displaystyle =\text{Area of }(ABED)+\text{Area of }(BEFC)-\text{Area of }(ADFC)
\displaystyle =\int_{1}^{3}\frac{3x+1}{2}\,dx+\int_{3}^{6}\frac{21-2x}{3}\,dx-\int_{1}^{6}\frac{x+9}{5}\,dx
\displaystyle =\frac{1}{2}\left[\frac{3x^2}{2}+x\right]_{1}^{3}+\frac{1}{3}\left[21x-x^2\right]_{3}^{6}-\frac{1}{5}\left[\frac{x^2}{2}+9x\right]_{1}^{6}
\displaystyle =\frac{1}{2}\left[\left(\frac{27}{2}+3\right)-\left(\frac{3}{2}+1\right)\right]+\frac{1}{3}\left[(126-36)-(63-9)\right]
\displaystyle \qquad -\frac{1}{5}\left[(18+54)-\left(\frac{1}{2}+9\right)\right]
\displaystyle =\frac{1}{2}\left[\frac{33}{2}-\frac{5}{2}\right]+\frac{1}{3}[90-54]-\frac{1}{5}\left[72-\frac{19}{2}\right]
\displaystyle =\frac{1}{2}\left[\frac{28}{2}\right]+\frac{1}{3}(36)-\frac{1}{5}\left(\frac{125}{2}\right)
\displaystyle =7+12-\frac{25}{2}=\frac{14+24-25}{2}=\frac{13}{2}\text{ sq. units}

\displaystyle \textbf{28. }\text{A dietician wishes to mix two types of food in such a way that the vitamin contents}
\displaystyle \text{of the mixture contains at least}   8 \text{ units of vitamin A and 10 units of vitamin C.}
\displaystyle \text{Food I contains} 2 \text{ units/kg of vitamin A and 1 unit/kg of vitamin C. It costs}
\displaystyle \text{Rs } 50 \text{ per kg to produce food I. Food II contains 1 unit/kg of vitamin A and 2 units/kg}
\displaystyle \text{of vitamin C and it costs } \text{Rs } 70 \text{ per kg to produce food II. Formulate this problem }
\displaystyle \text{as a LPP to minimise the cost of a mixture that will produce the diet. Also find the}
\displaystyle \text{minimum cost}.
\displaystyle \text{Answer:}
\displaystyle  \text{The given data can be put in the tabular form as follows:}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Food} & \text{Vitamin A} & \text{Vitamin C} & \text{Cost/Unit} \\ \hline I & 2 & 1 & \text{Rs } 50 \\ \hline II & 1 & 2 & \text{Rs } 70 \\ \hline \text{Min. requirement} & 8 & 10 & \\ \hline \end{array}
\displaystyle \text{Suppose the diet contains } x \text{ units of food I and } y \text{ units of food II.}
\displaystyle \text{Then, the required LPP is}
\displaystyle \text{Minimize } Z=50x+70y
\displaystyle \text{Subject to the constraints,}
\displaystyle 2x+y\geq 8
\displaystyle x+2y\geq 10
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{Let us draw the lines,}
\displaystyle 2x+y=8 \qquad (i)
\displaystyle x+2y=10 \qquad (ii)
\displaystyle 2x+y=8 \text{ passes through points } (0,8) \text{ and } (4,0)
\displaystyle \text{For } 2x+y=8
\displaystyle \begin{array}{|c|c|c|}\hline x & 0 & 4 \\ \hline y & 8 & 0 \\ \hline \end{array}
\displaystyle \text{and the line } x+2y=10 \text{ passes through points } (10,0) \text{ and } (0,5)
\displaystyle \text{For } x+2y=10
\displaystyle \begin{array}{|c|c|c|}\hline x & 10 & 0 \\ \hline y & 0 & 5 \\ \hline \end{array}
\displaystyle \text{Graph of above LPP is given as follows:}\displaystyle \text{Multiplying eq. (i) by 2 and subtracting eq. (ii) from eq. (i), we get}
\displaystyle 4x+2y=16
\displaystyle \underline{x+2y=10}
\displaystyle 3x=6
\displaystyle \Rightarrow x=2
\displaystyle \text{Putting } x=2 \text{ in eq. (i), we get}
\displaystyle 2(2)+y=8 \Rightarrow y=8-4=4
\displaystyle \text{These lines intersect at } P(2,4)
\displaystyle \therefore \text{The solution set is shaded region.}
\displaystyle \text{We have the table with corner points and values of } Z
\displaystyle \begin{array}{|c|c|}\hline \text{Corner point} & \text{Value of objective function } Z=50x+70y \\ \hline C(10,0) & 50(10)+70(0)=500 \\ \hline P(2,4) & 50(2)+70(4)=100+280=380\ (\text{minimum}) \\ \hline B(0,8) & 50(0)+70(8)=560 \\ \hline \end{array}
\displaystyle \text{Hence, the minimum cost is } \text{Rs } 380 \text{ when } x=2 \text{ and } y=4.

\displaystyle \textbf{29. }\text{Find the vector equation of a line passing through the point}   (2,3,2) \text{ and } \\ \text{parallel to the line } \overrightarrow{r}=(-2\widehat{i}+3\widehat{j})+\lambda(2\widehat{i}-3\widehat{j}+6\widehat{k})\text{.}   \text{Also, find the distance between} \\ \text{these two lines}.
\displaystyle \textbf{OR} \displaystyle \text{Find the coordinates of the foot of the perpendicular } Q \text{ drawn } \text{from } P(3,2,1)
\displaystyle \text{ to the plane } 2x-y+z+1=0\text{. Also, find the } \text{distance } PQ \text{ and the image of the point } 
\displaystyle P \text{ treating this plane} \text{as a mirror}.
\displaystyle \text{Answer:}
\displaystyle \text{Given line is } \overrightarrow{r}=(-2\widehat{i}+3\widehat{j})+\lambda(2\widehat{i}-3\widehat{j}+6\widehat{k})
\displaystyle \text{It is parallel to } 2\widehat{i}-3\widehat{j}+6\widehat{k}
\displaystyle \text{Required line is also parallel to } 2\widehat{i}-3\widehat{j}+6\widehat{k}
\displaystyle \text{Required line passes through the point } (2,3,2) \text{ or } (2\widehat{i}+3\widehat{j}+2\widehat{k})
\displaystyle \therefore \text{Equation of required line is } \overrightarrow{r}=(2\widehat{i}+3\widehat{j}+2\widehat{k})+\mu(2\widehat{i}-3\widehat{j}+6\widehat{k})
\displaystyle \text{Distance between parallel lines } \overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b} \text{ and } \overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b}
\displaystyle d=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{|\overrightarrow{b}|}
\displaystyle \text{Here, } \overrightarrow{a_1}=-2\widehat{i}+3\widehat{j},\ \overrightarrow{a_2}=2\widehat{i}+3\widehat{j}+2\widehat{k},\ \overrightarrow{b}=2\widehat{i}-3\widehat{j}+6\widehat{k}
\displaystyle |\overrightarrow{b}|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7,\quad (\overrightarrow{a_2}-\overrightarrow{a_1})=4\widehat{i}+2\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}=\left|\begin{matrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&0&2\\2&-3&6\end{matrix}\right|
\displaystyle =\widehat{i}(0\times 6-2\times(-3))-\widehat{j}(4\times 6-2\times 2)+\widehat{k}(4\times(-3)-0\times 2)
\displaystyle =6\widehat{i}-20\widehat{j}-12\widehat{k}
\displaystyle |(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|=\sqrt{6^2+(-20)^2+(-12)^2}=2\sqrt{145}
\displaystyle \therefore d=\frac{2\sqrt{145}}{7}\text{ units}
\displaystyle \textbf{OR}
\displaystyle \text{Given plane: } 2x-y+z+1=0
\displaystyle \text{Let } Q \text{ be foot of perpendicular and } R(x_1,y_1,z_1) \text{ be image of } P(3,2,1)
\displaystyle \text{Direction ratios of normal are } (2,-1,1)\displaystyle \text{Equation of line PQ is } \frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}
\displaystyle \text{Let } \frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}=\lambda
\displaystyle \Rightarrow x=2\lambda+3,\ y=2-\lambda,\ z=\lambda+1
\displaystyle \text{Coordinates of } Q \text{ are } (2\lambda+3,2-\lambda,\lambda+1)
\displaystyle \text{Since } Q \text{ lies on plane, substitute in } 2x-y+z+1=0
\displaystyle 2(2\lambda+3)-(2-\lambda)+(\lambda+1)+1=0
\displaystyle \Rightarrow 4\lambda+6-2+\lambda+\lambda+1+1=0
\displaystyle \Rightarrow 6\lambda+6=0\Rightarrow \lambda=-1
\displaystyle \text{Thus, } Q=(1,3,0)
\displaystyle \text{Distance } PQ=\sqrt{(1-3)^2+(3-2)^2+(0-1)^2}=\sqrt{4+1+1}=\sqrt{6}
\displaystyle \text{Since } Q \text{ is mid-point of } PR
\displaystyle \frac{x_1+3}{2}=1,\ \frac{y_1+2}{2}=3,\ \frac{z_1+1}{2}=0
\displaystyle \Rightarrow x_1=2-3=-1,\ y_1=6-2=4,\ z_1=0-1=-1
\displaystyle \therefore R(-1,4,-1)
\displaystyle \text{Hence, image of point } P(3,2,1) \text{ is } (-1,4,-1)

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