Mathematics

\displaystyle \textbf{Time Allowed : 3 Hours \hspace{5cm} Maximum Marks : 100}

\displaystyle \textbf{General Instructions:}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) Please check that this Question Paper contains 26 Questions.}
\displaystyle \text{(iii) Marks for each question are indicated against it.}
\displaystyle \text{(iv) Questions 1 to 6 in Section A are Very Short Answer Type Questions carrying one mark each.}
\displaystyle \text{(v) Questions 7 to 19 in Section B are Long Answer I Type Questions carrying 4 marks each.}
\displaystyle \text{(vi) Questions 20 to 26 in Section C are Long Answer II Type Questions carrying 6 marks each.}
\displaystyle \text{(vii) Please write down the serial number of the Question before attempting it.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question numbers 1 to 6 carry 1 mark each.}

\displaystyle \text{1. If } \overrightarrow{a}=7\widehat{i}+\widehat{j}-4\widehat{k} \text{ and } \overrightarrow{b}=2\widehat{i}+6\widehat{j}+3\widehat{k},  \text{then find the projection of } \overrightarrow{a} \text{ on } \overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } \overrightarrow{a}=7\widehat{i}+\widehat{j}-4\widehat{k}
\displaystyle \overrightarrow{b}=2\widehat{i}+6\widehat{j}+3\widehat{k}
\displaystyle \therefore\ \text{Projection of } \overrightarrow{a} \text{ on } \overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}
\displaystyle =\frac{(7\widehat{i}+\widehat{j}-4\widehat{k})\cdot(2\widehat{i}+6\widehat{j}+3\widehat{k})}{\sqrt{(2)^{2}+(6)^{2}+(3)^{2}}}
\displaystyle =\frac{(7\times 2)+(1\times 6)+(-4\times 3)}{\sqrt{4+36+9}}
\displaystyle =\frac{8}{\sqrt{49}}=\frac{8}{7}

\displaystyle \text{2. Find } \lambda, \text{ if the vector } \overrightarrow{a}=\widehat{i}+3\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}-\widehat{j}-\widehat{k}  \text{ and } \overrightarrow{c}=\lambda\widehat{j}+3\widehat{k} \\ \\  \text{ are coplanar.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: } \overrightarrow{a}=\widehat{i}+3\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b}=2\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{c}=\lambda\widehat{j}+3\widehat{k}
\displaystyle \text{Since given vectors are coplanar}
\displaystyle [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]=0
\displaystyle [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]=\begin{vmatrix}1&3&1\\2&-1&-1\\0&\lambda&3\end{vmatrix}=0
\displaystyle 1(-3+\lambda)-3(6)+1(2\lambda)=0\Rightarrow 3\lambda-21=0
\displaystyle \Rightarrow \lambda=7

\displaystyle \text{3. If a line makes angles } 90^\circ,\ 60^\circ \text{ and } \theta \text{ with } x,\ y \text{ and } z\text{-axis respectively, where}
\displaystyle \theta \text{ is acute, then find } \theta.
\displaystyle \text{Answer:}
\displaystyle \text{Given angles are}
\displaystyle \alpha=90^\circ,\ \beta=60^\circ,\ \lambda=\theta
\displaystyle \text{Let } l,m,n \text{ be direction cosines of the given line}
\displaystyle l=\cos\alpha,\ m=\cos\beta,\ n=\cos\lambda
\displaystyle \text{We know that } l^{2}+m^{2}+n^{2}=1
\displaystyle \cos^{2}(90^\circ)+\cos^{2}(60^\circ)+\cos^{2}\theta=1
\displaystyle 0+\left(\frac{1}{2}\right)^{2}+\cos^{2}\theta=1
\displaystyle \cos\theta=\frac{\sqrt{3}}{2}
\displaystyle \cos\theta=\cos(30^\circ)\qquad (\because\ \theta \text{ is acute})
\displaystyle \theta=30^\circ

\displaystyle \text{4. Write the element } a_{23} \text{ of a } 3 \times 3 \text{ matrix } A=(a_{ij}) \text{ whose elements } a_{ij} \text{are given by }
\displaystyle a_{ij}=\frac{|i-j|}{2}.
\displaystyle \text{Answer:}
\displaystyle  a_{ij}=\frac{|i-j|}{2}
\displaystyle a_{23}=\frac{|2-3|}{2}\qquad (\because\ i=2,\ j=3)
\displaystyle a_{23}=\frac{|-1|}{2}=\frac{1}{2}

\displaystyle \text{5. Find the differential equation representing the family of curves }
\displaystyle y=\frac{A}{x}+B, \text{ where } A \text{ and } B \text{ are arbitrary constants.}
\displaystyle \text{Answer:}
\displaystyle y=\frac{A}{x}+B\qquad \ldots (1)
\displaystyle \text{Given equation has two arbitrary constants, so we will differentiate it two times.}
\displaystyle \text{Differentiating equation } (1) \text{ w.r.t. } x,
\displaystyle \frac{dy}{dx}=-\frac{A}{x^{2}}
\displaystyle x^{2}\frac{dy}{dx}=-A\qquad \ldots (2)
\displaystyle \text{Again differentiating equation } (2) \text{ w.r.t. } x
\displaystyle x^{2}\frac{d^{2}y}{dx^{2}}+2x\frac{dy}{dx}=0\qquad \left[\because\ \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\right]
\displaystyle x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}=0

\displaystyle \text{6. Find the integrating factor of the differential equation: } \\  \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right)\frac{dx}{dy}=1.
\displaystyle \text{Answer:}
\displaystyle \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right)\frac{dx}{dy}=1
\displaystyle \frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}
\displaystyle \frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}
\displaystyle \text{It is in the form of linear differential equation}
\displaystyle \frac{dy}{dx}+Py=Q
\displaystyle \text{where } P=\frac{1}{\sqrt{x}},\ Q=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}
\displaystyle \therefore\ \text{I.F.}=e^{\int P\,dx}
\displaystyle =e^{\int \frac{1}{\sqrt{x}}\,dx}
\displaystyle =e^{2\sqrt{x}}\qquad \left[\because\ \int x^{n}\,dx=\frac{x^{n+1}}{n+1}\right]

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question numbers 7 to 19 carry 4 marks each.}

\displaystyle \text{7. If } A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix} \text{, find } A^{2}-5A+4I \text{and hence find a matrix } X \text{ such that }
\displaystyle A^{2}-5A+4I+X=O.
\displaystyle \text{OR} \\ \text{If } A=\begin{bmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{bmatrix} \text{, find } (A^{T})^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
\displaystyle A^{2}=A\cdot A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}4+1&-1&2\\4+2+3&1-3&2+3\\2-2&-1&1-3\end{bmatrix}=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}
\displaystyle 5A=5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}=\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix}
\displaystyle 4I=4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}
\displaystyle A^{2}-5A+4I=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix}+\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}
\displaystyle A^{2}-5A+4I=\begin{bmatrix}5-10+4&-1-0+0&2-5+0\\9-10+0&-2-5+4&5-15+0\\0-5+0&-1+5+0&-2-0+4\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}
\displaystyle \text{Now } A^{2}-5A+4I+X=O
\displaystyle X=-(A^{2}-5A+4I)
\displaystyle =(-1)\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}
\displaystyle =\begin{bmatrix}1&1&3\\1&3&10\\5&-4&-2\end{bmatrix}
\displaystyle \text{OR}
\displaystyle A=\begin{bmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{bmatrix}
\displaystyle A^{T}=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}
\displaystyle |A^{T}|=1(-1-8)-0(-2-6)-2(-8+3)
\displaystyle =-9+10=1\neq 0
\displaystyle \therefore\ |A^{T}|\text{ exists or } A^{T} \text{ is invertible}
\displaystyle A^{T}_{11}=(-1)^{1+1}(-1-8)=-9
\displaystyle A^{T}_{12}=(-1)^{1+2}(-2-6)=8
\displaystyle A^{T}_{13}=(-1)^{1+3}(-8+3)=-5
\displaystyle A^{T}_{21}=(-1)^{2+1}(0+8)=-8
\displaystyle A^{T}_{22}=(-1)^{2+2}(1+6)=7
\displaystyle A^{T}_{23}=(-1)^{2+3}(4-0)=-4
\displaystyle A^{T}_{31}=(-1)^{3+1}(0-2)=-2
\displaystyle A^{T}_{32}=(-1)^{3+2}(2-4)=2
\displaystyle A^{T}_{33}=(-1)^{3+3}(-1+0)=-1
\displaystyle {Adj\ } A^{T}=\begin{bmatrix}-9&8&-5\\-8&7&-4\\-2&2&-1\end{bmatrix}^{T}=\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}
\displaystyle (A^{T})^{-1}=\frac{1}{|A^{T}|} {Adj\ } A^{T}
\displaystyle (A^{T})^{-1}=\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\qquad [\because\ |A^{T}|=1]

\displaystyle \text{8. If } f(x)=\begin{vmatrix}a&-1&0\\ax&a&-1\\ax^{2}&ax&a\end{vmatrix},  \text{using properties of determinants find the value of } \\ f(2x)-f(x).
\displaystyle \text{Answer:}
\displaystyle \text{Given: } f(x)=\begin{vmatrix}a&-1&0\\ax&a&-1\\ax^{2}&ax&a\end{vmatrix}
\displaystyle \text{Taking } a \text{ common from } C_{1}
\displaystyle f(x)=a\begin{vmatrix}1&-1&0\\x&a&-1\\x^{2}&ax&a\end{vmatrix}
\displaystyle \text{Applying } C_{2}\to C_{2}+C_{1}
\displaystyle f(x)=a\begin{vmatrix}1&0&0\\x&a+x&-1\\x^{2}&ax+x^{2}&a\end{vmatrix}
\displaystyle f(x)=a\left(a(a+x)+(ax+x^{2})\right)
\displaystyle f(x)=a(a^{2}+ax+ax+x^{2})
\displaystyle f(x)=a(a^{2}+2ax+x^{2})\qquad \ldots (1)
\displaystyle \text{To find } f(2x)
\displaystyle f(2x)=\begin{vmatrix}a&-1&0\\a(2x)&a&-1\\a(2x)^{2}&a(2x)&a\end{vmatrix}
\displaystyle =\begin{vmatrix}a&-1&0\\2ax&a&-1\\4ax^{2}&2ax&a\end{vmatrix}
\displaystyle \text{Taking } a \text{ common from } C_{1}
\displaystyle f(2x)=a\begin{vmatrix}1&-1&0\\2x&a&-1\\4x^{2}&2ax&a\end{vmatrix}
\displaystyle \text{Applying } C_{2}\to C_{2}+C_{1}
\displaystyle f(2x)=a\begin{vmatrix}1&0&0\\2x&2x+a&-1\\4x^{2}&2ax+4x^{2}&a\end{vmatrix}
\displaystyle f(2x)=a\left(a(2x+a)+2ax+4x^{2}\right)
\displaystyle f(2x)=a(2ax+a^{2}+2ax+4x^{2})
\displaystyle f(2x)=a(a^{2}+4ax+4x^{2})\qquad \ldots (2)
\displaystyle \therefore\ f(2x)-f(x)=a(a^{2}+4ax+4x^{2})-a(a^{2}+2ax+x^{2})
\displaystyle \text{(from (1) \& (2))}
\displaystyle =a(a^{2}+4ax+4x^{2}-a^{2}-2ax-x^{2})
\displaystyle =a(3x^{2}+2ax)

\displaystyle \text{9. Find : } \int \frac{dx}{\sin x+\sin 2x} \\ \text{ OR} \\ \text{Integrate the following w.r.t. } x : \frac{x^{2}-3x+1}{\sqrt{1-x^{2}}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{dx}{\sin x+\sin 2x}
\displaystyle I=\int \frac{dx}{\sin x+2\sin x\cos x}\qquad \left[\because\ \sin 2\theta=2\sin\theta\cos\theta\right]
\displaystyle =\int \frac{dx}{\sin x(1+2\cos x)}
\displaystyle I=\int \frac{\sin x\,dx}{\sin^{2}x(1+2\cos x)}\qquad \text{(By multiplying Num \& Den by } \sin x\text{)}
\displaystyle =\int \frac{\sin x\,dx}{(1-\cos^{2}x)(1+2\cos x)}\qquad \left[\because\ \sin^{2}\theta=1-\cos^{2}\theta\right]
\displaystyle \text{Let } \cos x=t
\displaystyle -\sin x\,dx=dt
\displaystyle I=\int \frac{-dt}{(1-t^{2})(1+2t)}
\displaystyle =\int \frac{-dt}{(1+t)(1-t)(1+2t)}
\displaystyle \text{Let } \frac{-1}{(1+t)(1-t)(1+2t)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{1+2t}
\displaystyle -1=A(1-t)(1+2t)+B(1+t)(1+2t)+C(1+t)(1-t)
\displaystyle \text{Putting } t=-1,\ -1=-2A\Rightarrow A=\frac{1}{2}
\displaystyle \text{Putting } t=1,\ -1=6B\Rightarrow B=-\frac{1}{6}
\displaystyle \text{Putting } t=-\frac{1}{2},\ C=-\frac{4}{3}
\displaystyle \frac{-1}{(1+t)(1-t)(1+2t)}=\frac{1}{2(1+t)}-\frac{1}{6(1-t)}-\frac{4}{3(1+2t)}
\displaystyle I=\int \frac{1}{2(1+t)}\,dt-\int \frac{1}{6(1-t)}\,dt-\int \frac{4}{3(1+2t)}\,dt
\displaystyle I=\frac{1}{2}\log|1+t|+\frac{1}{6}\log|1-t|-\frac{4}{3\times 2}\log|1+2t|+C
\displaystyle \qquad \left[\because\ \int \frac{1}{x}\,dx=\log|x|+C\right]
\displaystyle I=\frac{1}{2}\log|1+\cos x|+\frac{1}{6}\log|1-\cos x|-\frac{2}{3}\log|1+2\cos x|+C
\displaystyle \text{OR}
\displaystyle \text{Let } I=\int \frac{x^{2}-3x+1}{\sqrt{1-x^{2}}}\,dx
\displaystyle =\int \frac{x^{2}-3x+1+1-1}{\sqrt{1-x^{2}}}\,dx
\displaystyle =\int \frac{x^{2}-1+2-3x}{\sqrt{1-x^{2}}}\,dx
\displaystyle =\int \frac{-(1-x^{2})+(2-3x)}{\sqrt{1-x^{2}}}\,dx
\displaystyle =\int \left(-\sqrt{1-x^{2}}+\frac{2}{\sqrt{1-x^{2}}}-\frac{3x}{\sqrt{1-x^{2}}}\right)\,dx
\displaystyle I=-\int \sqrt{1-x^{2}}\,dx+\int \frac{2}{\sqrt{1-x^{2}}}\,dx-3\int \frac{x}{\sqrt{1-x^{2}}}\,dx
\displaystyle \text{Let } I_{1}=-\int \sqrt{1-x^{2}}\,dx
\displaystyle I_{1}=-\left[\frac{1}{2}x\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x\right]+C_{1}
\displaystyle I_{2}=\int \frac{2}{\sqrt{1-x^{2}}}\,dx=2\sin^{-1}x+C_{2}
\displaystyle I_{3}=-3\int \frac{x}{\sqrt{1-x^{2}}}\,dx
\displaystyle \text{Let } 1-x^{2}=t\Rightarrow -2x\,dx=dt
\displaystyle I_{3}=-3\int \frac{x}{\sqrt{t}}\,dx=\frac{3}{2}\int \frac{dt}{\sqrt{t}}=3\sqrt{t}+C_{3}
\displaystyle =3\sqrt{1-x^{2}}+C_{3}
\displaystyle \text{Putting values of } I_{1},I_{2},I_{3} \text{ in } I
\displaystyle I=-\frac{1}{2}x\sqrt{1-x^{2}}-\frac{1}{2}\sin^{-1}x+2\sin^{-1}x+3\sqrt{1-x^{2}}+C

\displaystyle \text{10. Evaluate : } \int_{-\pi}^{\pi} (\cos ax-\sin bx)^{2} dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{-\pi}^{\pi}(\cos ax-\sin bx)^{2}dx
\displaystyle I=\int_{-\pi}^{\pi}(\cos^{2}ax+\sin^{2}bx-2\cos ax\sin bx)\,dx
\displaystyle \left[\because\ (a-b)^{2}=a^{2}+b^{2}-2ab\right]
\displaystyle I=\int_{-\pi}^{\pi}\cos^{2}ax\,dx+\int_{-\pi}^{\pi}\sin^{2}bx\,dx-2\int_{-\pi}^{\pi}\cos ax\sin bx\,dx
\displaystyle \text{Here, } \cos^{2}ax \text{ and } \sin^{2}bx \text{ are even functions, } \cos ax\sin bx \text{ is odd function}
\displaystyle \therefore\ I=2\int_{0}^{\pi}\cos^{2}ax\,dx+2\int_{0}^{\pi}\sin^{2}bx\,dx
\displaystyle I=2\int_{0}^{\pi}\frac{1+\cos 2ax}{2}\,dx+2\int_{0}^{\pi}\frac{1-\cos 2bx}{2}\,dx
\displaystyle \left[\because\ \cos 2\theta=2\cos^{2}\theta-1,\ \cos 2\theta=1-2\sin^{2}\theta\right]
\displaystyle I=\int_{0}^{\pi}(1+\cos 2ax)\,dx+\int_{0}^{\pi}(1-\cos 2bx)\,dx
\displaystyle I=\int_{0}^{\pi}(2+\cos 2ax-\cos 2bx)\,dx
\displaystyle I=\left[2x+\frac{\sin 2ax}{2a}-\frac{\sin 2bx}{2b}\right]_{0}^{\pi}
\displaystyle I=2\pi+\frac{\sin 2a\pi}{2a}-\frac{\sin 2b\pi}{2b}

\displaystyle \text{11. A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 }
\displaystyle \text{red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise}
\displaystyle \text{bag B. If two  balls are drawn  at random (without replacement) from the selected}
\displaystyle \text{bag,  find the  probability of one of them being red and another black.}
\displaystyle \text{OR}
\displaystyle \text{An unbiased coin is tossed 4 times. Find the mean and variance of the number of} \\ \text{heads obtained.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_{1} \text{ be the probability of getting 1 or 2 on a die and } E_{2} \text{ be the probability}
\displaystyle \text{of getting 3, 4, 5} \text{ or } 6 \text{ on a die}
\displaystyle P(E_{1})=\frac{2}{6}=\frac{1}{3}
\displaystyle P(E_{2})=\frac{4}{6}=\frac{2}{3}
\displaystyle \text{Also let } E \text{ be the probability of getting one ball red and another one black}
\displaystyle \therefore\ P\left(\frac{E}{E_{1}}\right)=\text{Probability of getting } E \text{ when bag A is chosen}
\displaystyle =P(BR)+P(RB)
\displaystyle =\frac{4}{10}\times\frac{6}{9}+\frac{6}{10}\times\frac{4}{9}=\frac{8}{15}
\displaystyle P\left(\frac{E}{E_{2}}\right)=\text{Probability of } E \text{ when bag B is chosen}
\displaystyle =\frac{3}{10}\times\frac{7}{9}+\frac{7}{10}\times\frac{3}{9}
\displaystyle =\frac{42}{90}=\frac{7}{15}
\displaystyle P(E)=P(E_{1})P\left(\frac{E}{E_{1}}\right)+P(E_{2})P\left(\frac{E}{E_{2}}\right)
\displaystyle P(E)=\frac{1}{3}\times\frac{8}{15}+\frac{2}{3}\times\frac{7}{15}
\displaystyle P(E)=\frac{22}{45}
\displaystyle \text{OR}
\displaystyle \text{Let } X \text{ denote the number of heads in the four tosses of the coin.}
\displaystyle X \text{ can be } 0,\ 1,\ 2,\ 3,\ 4
\displaystyle P(X=0)=\text{Probability of getting no head } (TTTT)
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}
\displaystyle \text{Total number of outcomes}=2^{4}=16\qquad (\because\ \text{coin is thrown 4 times})
\displaystyle P(X=0)=\frac{1}{16}
\displaystyle P(X=1)=\text{Probability of getting 1 head}
\displaystyle [(HTTT),\ (THTT),\ (TTHT),\ (TTTH)]
\displaystyle =\frac{4}{16}=\frac{1}{4}
\displaystyle P(X=2)=\text{Probability of getting 2 heads}
\displaystyle [(HHTT),\ (HTHT),\ (HTTH),\ (THHT),\ (THTH),\ (TTHH)]
\displaystyle =\frac{6}{16}=\frac{3}{8}
\displaystyle P(X=3)=\text{Probability of getting 3 heads}
\displaystyle [(HHHT),\ (HHTH),\ (HTHH),\ (THHH)]
\displaystyle =\frac{4}{16}=\frac{1}{4}
\displaystyle P(X=4)=\text{Probability of getting 4 heads } (HHHH)
\displaystyle =\frac{1}{16}
\displaystyle \text{Probability distribution of } X \text{ is given below.}
\displaystyle \begin{array}{|c|c|c|c|}\hline X_{i}&P_{i}(X=X_{i})&P_{i}X_{i}&P_{i}X_{i}^{2}\\\hline 0&\frac{1}{16}&0&0\\\hline 1&\frac{1}{4}&\frac{1}{4}&\frac{1}{4}\\\hline 2&\frac{3}{8}&\frac{3}{4}&\frac{3}{2}\\\hline 3&\frac{1}{4}&\frac{3}{4}&\frac{9}{4}\\\hline 4&\frac{1}{16}&\frac{1}{4}&1\\\hline \sum P_{i}=1& &\sum P_{i}X_{i}=2&\sum P_{i}X_{i}^{2}=5\\\hline \end{array}
\displaystyle \therefore\ \text{Mean}=\overline{X}=\sum P_{i}X_{i}=2
\displaystyle  {Var\ }(X)=\sum P_{i}X_{i}^{2}-\left(\sum P_{i}X_{i}\right)^{2}
\displaystyle =5-(2)^{2}
\displaystyle =1
\displaystyle \text{Hence, mean }=2 \text{ and variance }=1

\displaystyle \text{12. If } \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}, \text{ find } (\overrightarrow{r}\times \widehat{i})\cdot (\overrightarrow{r}\times \widehat{j})+xy.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}
\displaystyle \text{Now, } (\overrightarrow{r}\times \widehat{i})\cdot(\overrightarrow{r}\times \widehat{j})+xy
\displaystyle =\left[(x\widehat{i}+y\widehat{j}+z\widehat{k})\times \widehat{i}\right]\cdot\left[(x\widehat{i}+y\widehat{j}+z\widehat{k})\times \widehat{j}\right]+xy
\displaystyle =\left(x\widehat{i}\times \widehat{i}+y\widehat{j}\times \widehat{i}+z\widehat{k}\times \widehat{i}\right)\cdot\left(x\widehat{i}\times \widehat{j}+y\widehat{j}\times \widehat{j}+z\widehat{k}\times \widehat{j}\right)+xy
\displaystyle \text{As we know that}
\displaystyle \widehat{i}\times \widehat{j}=\widehat{k},\ \widehat{j}\times \widehat{i}=-\widehat{k}
\displaystyle \widehat{j}\times \widehat{k}=\widehat{i},\ \widehat{k}\times \widehat{j}=-\widehat{i}
\displaystyle \widehat{k}\times \widehat{i}=\widehat{j},\ \widehat{i}\times \widehat{k}=-\widehat{j}
\displaystyle \widehat{i}\times \widehat{i}=0,\ \widehat{j}\times \widehat{j}=0,\ \widehat{k}\times \widehat{k}=0
\displaystyle \therefore\ (\overrightarrow{r}\times \widehat{i})\cdot(\overrightarrow{r}\times \widehat{j})+xy=(0-z\widehat{k}+y\widehat{j})\cdot(x\widehat{k}+0+z(-\widehat{i}))+xy
\displaystyle =-xy+xy
\displaystyle =0

\displaystyle \text{13. Find the distance between the point } (-1,-5,-10) \text{ and the point of intersection}
\displaystyle \text{of the line } \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12} \text{ and the plane } x-y+z=5.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}
\displaystyle \text{Let } \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda
\displaystyle x=3\lambda+2,\ y=4\lambda-1,\ z=12\lambda+2\qquad \ldots (1)
\displaystyle \text{Coordinates of any point on the line are } (3\lambda+2,\ 4\lambda-1,\ 12\lambda+2)
\displaystyle \text{To find point of intersection of line and plane, put these values of } \\ x,y,z \text{ in equation of plane.}
\displaystyle x-y+z=5
\displaystyle (3\lambda+2)-(4\lambda-1)+(12\lambda+2)=5\qquad \text{(from (1))}
\displaystyle 11\lambda+5=5
\displaystyle \lambda=0
\displaystyle x=3\lambda+2=2
\displaystyle y=4\lambda-1=-1
\displaystyle z=12\lambda+2=2
\displaystyle \therefore\ \text{Point of intersection of line and plane is } (2,-1,2)
\displaystyle \text{Given point is } (-1,-5,-10)
\displaystyle \text{Distance}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}
\displaystyle \text{Distance between } (2,-1,2) \text{ and } (-1,-5,-10) \text{ is}
\displaystyle \text{Distance}=\sqrt{(-1-2)^{2}+(-5+1)^{2}+(-10-2)^{2}}
\displaystyle =\sqrt{(-3)^{2}+(-4)^{2}+(-12)^{2}}
\displaystyle =\sqrt{169}=13
\displaystyle \text{Distance}=13\ \text{units}
\displaystyle \text{Hence distance between } (-1,-5,-10) \text{ and point of intersection of line and plane is } 13\ \text{units.}

\displaystyle \text{14. If } \sin[\cot^{-1}(x+1)]=\cos(\tan^{-1}x), \text{ then find } x.
\displaystyle \text{OR}
\displaystyle \text{If } (\tan^{-1}x)^{2}+(\cot^{-1}x)^{2}=\frac{5\pi^{2}}{8}, \text{ then find } x.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } \sin[\cot^{-1}(x+1)]=\cos(\tan^{-1}x)
\displaystyle \text{As, } \cot^{-1}(x+1)=\sin^{-1}\frac{1}{\sqrt{1+(x+1)^{2}}}\qquad \left[\because\ \cot^{-1}P=\sin^{-1}\frac{1}{\sqrt{1+P^{2}}}\right]
\displaystyle \text{and } \tan^{-1}x=\cos^{-1}\frac{1}{\sqrt{1+x^{2}}}\qquad \left[\because\ \tan^{-1}P=\cos^{-1}\frac{1}{\sqrt{1+P^{2}}}\right]
\displaystyle \therefore\ \sin\left(\sin^{-1}\frac{1}{\sqrt{1+(x+1)^{2}}}\right)=\cos\left(\cos^{-1}\frac{1}{\sqrt{1+x^{2}}}\right)
\displaystyle \frac{1}{\sqrt{1+(x+1)^{2}}}=\frac{1}{\sqrt{1+x^{2}}}
\displaystyle \sqrt{1+x^{2}}=\sqrt{1+(x+1)^{2}}
\displaystyle \text{Squaring on both sides, we get}
\displaystyle 1+x^{2}=1+(x^{2}+1+2x)
\displaystyle 1+x^{2}=2+x^{2}+2x
\displaystyle 2x+2+x^{2}-x^{2}-1=0
\displaystyle 2x=-1
\displaystyle x=-\frac{1}{2}
\displaystyle \text{OR}
\displaystyle (\tan^{-1}x)^{2}+(\cot^{-1}x)^{2}=\frac{5\pi^{2}}{8}
\displaystyle (\tan^{-1}x)^{2}+(\cot^{-1}x)^{2}+2\tan^{-1}x\cot^{-1}x-2\tan^{-1}x\cot^{-1}x=\frac{5\pi^{2}}{8}
\displaystyle (\tan^{-1}x+\cot^{-1}x)^{2}-2\tan^{-1}x\cot^{-1}x=\frac{5\pi^{2}}{8}\qquad \ldots (1)
\displaystyle \text{As we know that } \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}
\displaystyle \text{From (1)}
\displaystyle \left(\frac{\pi}{2}\right)^{2}-2\tan^{-1}x\left(\frac{\pi}{2}-\tan^{-1}x\right)=\frac{5\pi^{2}}{8}
\displaystyle \frac{\pi^{2}}{4}-\pi\tan^{-1}x+2(\tan^{-1}x)^{2}=\frac{5\pi^{2}}{8}
\displaystyle 2(\tan^{-1}x)^{2}-\pi\tan^{-1}x-\frac{3\pi^{2}}{8}=0
\displaystyle \text{Let } \tan^{-1}x=t
\displaystyle 2t^{2}-\pi t-\frac{3\pi^{2}}{8}=0
\displaystyle \text{Solving the quadratic equation}
\displaystyle t=\frac{\pi\pm\sqrt{\pi^{2}-4(2)\left(-\frac{3\pi^{2}}{8}\right)}}{4}
\displaystyle t=\frac{\pi\pm\sqrt{4\pi^{2}}}{4}\Rightarrow t=\frac{\pi\pm 2\pi}{4}
\displaystyle t=\frac{3\pi}{4}\ \text{or}\ -\frac{\pi}{4}
\displaystyle \text{As, } -\frac{\pi}{2}\leq \tan^{-1}x\leq \frac{\pi}{2}
\displaystyle \tan^{-1}x\neq \frac{3\pi}{4},\ \tan^{-1}x=-\frac{\pi}{4}
\displaystyle x=\tan\left(-\frac{\pi}{4}\right)
\displaystyle x=-1

\displaystyle \text{15. If } y=\tan^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right),\ x^{2}\leq 1, \text{ then find } \frac{dy}{dx}.
\displaystyle \text{Answer:}
\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
\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 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]\qquad \left[\because\ \cos 2\theta=2\cos^{2}\theta-1,\ \cos 2\theta=1-2\sin^{2}\theta\right]
\displaystyle y=\tan^{-1}\left[\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right]
\displaystyle \text{Dividing Num \& Den by } \cos\theta
\displaystyle y=\tan^{-1}\left[\frac{1+\tan\theta}{1-\tan\theta}\right]
\displaystyle =\tan^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan\theta}{1-\tan \frac{\pi}{4}\tan\theta}\right]\qquad \left[\because\ \tan \frac{\pi}{4}=1\right]
\displaystyle =\tan^{-1}\left[\tan\left(\frac{\pi}{4}+\theta\right)\right]\qquad \left[\because\ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\right]
\displaystyle =\frac{\pi}{4}+\theta
\displaystyle =\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}\qquad \left[\because\ x^{2}=\cos 2\theta\right]
\displaystyle \text{Differentiating w.r.t. } x
\displaystyle \frac{dy}{dx}=0+\frac{1}{2}\left(\frac{-1}{\sqrt{1-(x^{2})^{2}}}\right)\times 2x
\displaystyle \frac{dy}{dx}=\frac{-x}{\sqrt{1-x^{4}}}\qquad \left[\because\ \frac{d}{dx}\cos^{-1}x=\frac{-1}{\sqrt{1-x^{2}}}\right]
\displaystyle \text{Hence } \frac{dy}{dx}=\frac{-x}{\sqrt{1-x^{4}}}

\displaystyle \text{16. If } x=a\cos\theta+b\sin\theta,\ y=a\sin\theta-b\cos\theta, \text{ show that } \frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+y=0.
\displaystyle \text{Answer:}
\displaystyle x=a\cos\theta+b\sin\theta\qquad \ldots (1)
\displaystyle y=a\sin\theta-b\cos\theta\qquad \ldots (2)
\displaystyle \text{On squaring and adding (1) \& (2)}
\displaystyle x^{2}+y^{2}=(a\cos\theta+b\sin\theta)^{2}+(a\sin\theta-b\cos\theta)^{2}
\displaystyle x^{2}+y^{2}=a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta+2ab\cos\theta\sin\theta+a^{2}\sin^{2}\theta+b^{2}\cos^{2}\theta-2ab\cos\theta\sin\theta
\displaystyle x^{2}+y^{2}=a^{2}(\cos^{2}\theta+\sin^{2}\theta)+b^{2}(\sin^{2}\theta+\cos^{2}\theta)
\displaystyle x^{2}+y^{2}=a^{2}+b^{2}
\displaystyle \text{Differentiating w.r.t. } x
\displaystyle 2x+2y\frac{dy}{dx}=0\Rightarrow 2y\frac{dy}{dx}=-2x
\displaystyle \frac{dy}{dx}=-\frac{x}{y}\qquad \ldots (3)
\displaystyle \text{Again differentiating w.r.t. } x
\displaystyle \frac{d^{2}y}{dx^{2}}=-\left(\frac{y-x\frac{dy}{dx}}{y^{2}}\right)\qquad \left[\because\ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}\right]
\displaystyle \frac{d^{2}y}{dx^{2}}=-\left(\frac{y-x\left(-\frac{x}{y}\right)}{y^{2}}\right)\qquad \text{[From (3)]}
\displaystyle \frac{d^{2}y}{dx^{2}}=-\left(\frac{y^{2}+x^{2}}{y^{3}}\right)\qquad \ldots (4)
\displaystyle \text{L.H.S.}=y^{2}\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+y
\displaystyle =-y^{2}\left(\frac{y^{2}+x^{2}}{y^{3}}\right)-x\left(-\frac{x}{y}\right)+y\qquad \text{[From (3) \& (4)]}
\displaystyle =-\frac{y^{2}+x^{2}}{y}+\frac{x^{2}}{y}+y
\displaystyle =\frac{-y^{2}-x^{2}+x^{2}+y^{2}}{y}
\displaystyle =0=\text{R.H.S.}
\displaystyle \text{L.H.S.}=\text{R.H.S.}
\displaystyle \text{Hence proved.}

\displaystyle \text{17. The side of an equilateral triangle is increasing at the rate of 2 cm/s. At} \\ \text{what rate is its area increasing when the side of the triangle is 20 cm?}
\displaystyle \text{Answer:}
\displaystyle\text{Let } a \text{ be the side of equilateral triangle}
\displaystyle \text{Area of equilateral triangle, } A=\frac{\sqrt{3}}{4}a^{2}
\displaystyle \text{Given that, } \frac{da}{dt}=2\ \text{cm/s}\qquad \ldots (1)
\displaystyle A=\frac{\sqrt{3}}{4}a^{2}
\displaystyle \text{Differentiating w.r.t. } t
\displaystyle \frac{dA}{dt}=\frac{d}{dt}\left(\frac{\sqrt{3}}{4}a^{2}\right)
\displaystyle =\frac{\sqrt{3}}{4}(2a)\frac{da}{dt}
\displaystyle =\frac{\sqrt{3}}{2}a(2)\qquad \text{[From (1)]}
\displaystyle =\sqrt{3}\,a\ \text{cm}^{2}\text{/s}
\displaystyle \text{When, } a=20\ \text{cm}
\displaystyle \left(\frac{dA}{dt}\right)_{a=20}=\sqrt{3}(20)\ \text{cm}^{2}\text{/s}
\displaystyle =20\sqrt{3}\ \text{cm}^{2}\text{/s}
\displaystyle \text{Hence, area is increasing at the rate of } 20\sqrt{3}\ \text{cm}^{2}\text{/s}
\displaystyle \text{when the side of triangle is } 20\ \text{cm}.

\displaystyle \text{18. Find : } \int (x+3)\sqrt{3-4x-x^{2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int (x+3)\sqrt{3-4x-x^{2}}\,dx
\displaystyle \text{Let } x+3=A\frac{d}{dx}(3-4x-x^{2})+B
\displaystyle x+3=A(-4-2x)+B
\displaystyle x+3=-2Ax-4A+B
\displaystyle \text{On comparing L.H.S. \& R.H.S.}
\displaystyle 1=-2A\Rightarrow A=-\frac{1}{2}
\displaystyle -4A+B=3\Rightarrow -4\left(-\frac{1}{2}\right)+B=3
\displaystyle B=1
\displaystyle I=\int \left[\frac{1}{2}(4+2x)+1\right]\sqrt{3-4x-x^{2}}\,dx
\displaystyle I=\frac{1}{2}\int (4+2x)\sqrt{3-4x-x^{2}}\,dx+\int \sqrt{3-4x-x^{2}}\,dx
\displaystyle I=I_{1}+I_{2}
\displaystyle I_{1}=\frac{1}{2}\int (4+2x)\sqrt{3-4x-x^{2}}\,dx
\displaystyle \text{Let } 3-4x-x^{2}=t
\displaystyle (-4-2x)\,dx=dt
\displaystyle I_{1}=-\frac{1}{2}\int \sqrt{t}\,dt
\displaystyle =-\frac{1}{2}\times \frac{2}{3}t^{3/2}+C_{1}\qquad \left[\because\ \int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\right]
\displaystyle =-\frac{1}{3}(3-4x-x^{2})^{3/2}+C
\displaystyle I_{2}=\int \sqrt{3-4x-x^{2}}\,dx
\displaystyle =\int \sqrt{3-(4x+x^{2})}\,dx
\displaystyle =\int \sqrt{3-(4x+x^{2}+2^{2}-2^{2})}\,dx
\displaystyle =\int \sqrt{7-(x^{2}+4x+2^{2})}\,dx
\displaystyle =\int \sqrt{(\sqrt{7})^{2}-(x+2)^{2}}\,dx
\displaystyle \text{As we know that } \int \sqrt{a^{2}-x^{2}}\,dx=\frac{1}{2}x\sqrt{a^{2}-x^{2}}+\frac{1}{2}a^{2}\sin^{-1}\left(\frac{x}{a}\right)+C
\displaystyle \therefore\ I_{2}=\frac{1}{2}(x+2)\sqrt{(\sqrt{7})^{2}-(x+2)^{2}}+\frac{1}{2}(\sqrt{7})^{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C_{2}
\displaystyle =\frac{1}{2}(x+2)\sqrt{3-4x-x^{2}}+\frac{7}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C_{2}
\displaystyle I=I_{1}+I_{2}
\displaystyle =-\frac{1}{3}(3-4x-x^{2})^{3/2}+\frac{1}{2}(x+2)\sqrt{3-4x-x^{2}}+\frac{7}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C
\displaystyle \text{where } C=C_{1}+C_{2}

\displaystyle \text{19. Three schools A, B and C organized a mela for collecting funds for helping} \\ \text{the rehabilitation of flood victims. They sold hand made fans, mats and} \\ \text{plates from recycled material at a cost of Rs 25, Rs 100 and Rs 50 each.}
\displaystyle \text{The number of articles sold are given below:}
\displaystyle \begin{array}{c|ccc}\text{Article/School}&A&B&C\\\hline \text{Hand-fans}&40&25&35\\ \text{Mats}&50&40&50\\ \text{Plates}&20&30&40\end{array}
\displaystyle \text{Find the funds collected by each school separately by selling the above articles.} \\ \text{Also find the total funds collected for the purpose.}
\displaystyle \text{Write one value generated by the above situation.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of articles sold can be represented as}
\displaystyle X=\begin{bmatrix}40&25&35\\50&40&50\\20&30&40\end{bmatrix}
\displaystyle \text{Cost of each article can be represented as}
\displaystyle Y=\begin{bmatrix}25&100&50\end{bmatrix}
\displaystyle \therefore\ \text{Funds collected by each school separately is given by}
\displaystyle YX=\begin{bmatrix}25&100&50\end{bmatrix}\begin{bmatrix}40&25&35\\50&40&50\\20&30&40\end{bmatrix}
\displaystyle YX=\begin{bmatrix}7000&6125&7875\end{bmatrix}
\displaystyle \text{Funds collected by school A : Rs } 7000
\displaystyle \text{Funds collected by school B : Rs } 6125
\displaystyle \text{Funds collected by school C : Rs } 7875
\displaystyle \text{Total funds collected = Rs } (7000+6125+7875)
\displaystyle =\text{Rs } 21000
\displaystyle \text{This shows helping nature of students.}

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 20 to 26 carry 6 marks each.}

\displaystyle \text{20. Let N denote the set of all natural numbers and R be the relation on } \\ N\times N \text{ defined by } (a,b)R(c,d)  \text{if } ad(b+c)=bc(a+d). \text{ Show that R is an } \\ \text{equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle \text{If relation } R \text{ is reflexive, symmetric and transitive relation, then it will} \\ \text{be equivalence relation.}
\displaystyle \text{(i) Let } (a,b) \text{ be an arbitrary element of } N\times N
\displaystyle \text{Now } (a,b)\in N\times N
\displaystyle a,b\in N
\displaystyle \Rightarrow ab(a+b)=ba(a+b)
\displaystyle \Rightarrow (a,b)\ R\ (a,b)
\displaystyle \therefore\ (a,b)\ R\ (a,b)\ \forall\ (a,b)\in N\times N
\displaystyle \therefore\ R \text{ is reflexive on } N\times N
\displaystyle \text{(ii) Let } (a,b),\ (c,d) \text{ be arbitrary element of } N\times N
\displaystyle \text{Such that } (a,b)\ R\ (c,d)
\displaystyle \text{Now } (a,b)\ R\ (c,d)
\displaystyle \Rightarrow ad(b+c)=bc(a+d)
\displaystyle \Rightarrow cb(d+a)=da(c+b)
\displaystyle \Rightarrow (c,d)\ R\ (a,b)
\displaystyle \Rightarrow (a,b)\ R\ (c,d)\Rightarrow (c,d)\ R\ (a,b)\ \forall\ (a,b),\ (c,d)\in N\times N
\displaystyle \therefore\ R \text{ is symmetric on } N\times N
\displaystyle \text{(iii) Let } (a,b),\ (c,d),\ (e,f)\in N\times N \text{ such that}
\displaystyle (a,b)\ R\ (c,d) \text{ and } (c,d)\ R\ (e,f)
\displaystyle (a,b)\ R\ (c,d)
\displaystyle \Rightarrow ad(b+c)=bc(a+d)
\displaystyle \frac{b+c}{bc}=\frac{a+d}{ad}
\displaystyle \frac{1}{c}+\frac{1}{b}=\frac{1}{d}+\frac{1}{a}\qquad \ldots (1)
\displaystyle \text{Also } (c,d)\ R\ (e,f)
\displaystyle cf(d+e)=de(c+f)
\displaystyle \frac{d+e}{de}=\frac{c+f}{cf}
\displaystyle \frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\qquad \ldots (2)
\displaystyle \text{Adding (1) and (2)}
\displaystyle \frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=\frac{1}{a}+\frac{1}{d}+\frac{1}{c}+\frac{1}{f}
\displaystyle \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}
\displaystyle \frac{b+e}{be}=\frac{a+f}{af}
\displaystyle af(b+e)=be(a+f)
\displaystyle (a,b)\ R\ (e,f)
\displaystyle \text{Thus } (a,b)\ R\ (c,d) \text{ and } (c,d)\ R\ (e,f)\Rightarrow (a,b)\ R\ (e,f)\ \forall
\displaystyle (a,b),\ (c,d),\ (e,f)\in N\times N
\displaystyle \therefore\ R \text{ is transitive on } N\times N
\displaystyle \text{Hence } R \text{ being reflexive, symmetric and transitive is an equivalence} \\ \text{relation on } N\times N.

\displaystyle \text{21. Using integration find the area of the triangle formed by positive x-axis and} \\ \text{tangent and normal to the circle } x^{2}+y^{2}=4 \text{ at } (1,\sqrt{3}).
\displaystyle \text{OR}
\displaystyle \text{Evaluate } \int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx \text{ as a limit of a sum.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: } x^{2}+y^{2}=4
\displaystyle \text{Normal at } (1,\sqrt{3}) \text{ will also pass through } (0,0). \text{ So, equation of normal:}
\displaystyle y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})
\displaystyle y-0=\frac{\sqrt{3}-0}{1-0}(x-0)
\displaystyle y=\sqrt{3}x\qquad \ldots (1)
\displaystyle \text{Slope of normal is } \sqrt{3}
\displaystyle \text{Slope of tangent }=-\frac{1}{\text{Slope of normal}}=-\frac{1}{\sqrt{3}}
\displaystyle \text{Equation of tangent at } (1,\sqrt{3}) \text{ is}
\displaystyle y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x-1)\qquad \left[\because\ y-y_{1}=m(x-x_{1})\right]
\displaystyle \sqrt{3}y-3=-x+1
\displaystyle y=\frac{-x+4}{\sqrt{3}}\qquad \ldots (2)\displaystyle \text{If } y=0 \text{ then } x=4
\displaystyle \text{Thus, } \triangle AOB \text{ is formed by the tangent, normal and the positive } x\text{-axis}
\displaystyle \text{Area of } \triangle AOB=\text{Area of } \triangle AOC+\text{Area of } \triangle ACB
\displaystyle =\int_{0}^{1}y\,dx+\int_{1}^{4}y\,dx
\displaystyle \text{Area of } \triangle AOB=\int_{0}^{1}\sqrt{3}x\,dx+\int_{1}^{4}\left(\frac{-x+4}{\sqrt{3}}\right)dx
\displaystyle =\sqrt{3}\left(\frac{x^{2}}{2}\right)_{0}^{1}+\frac{1}{\sqrt{3}}\left(-\frac{x^{2}}{2}+4x\right)_{1}^{4}
\displaystyle =\frac{\sqrt{3}}{2}(1-0)+\frac{1}{\sqrt{3}}\left[\left(-\frac{16}{2}+16\right)-\left(-\frac{1}{2}+4\right)\right]
\displaystyle =\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\left[\frac{16}{2}-\frac{7}{2}\right]=\frac{\sqrt{3}}{2}+\frac{9}{2\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}
\displaystyle =\frac{\sqrt{3}}{2}+\frac{9\sqrt{3}}{6}=2\sqrt{3}\ \text{sq. units}
\displaystyle \text{Hence area of triangle is } 2\sqrt{3}\ \text{sq. units.}
\displaystyle \text{OR}
\displaystyle \int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0 \atop n\to \infty}h[f(a)+f(a+h)+f(a+2h)+\ldots+f(a+(n-1)h)]\qquad \ldots (1)
\displaystyle \text{where, } h=\frac{b-a}{n}
\displaystyle \text{Here, } b=3,\ a=1
\displaystyle h=\frac{3-1}{n}=\frac{2}{n}\Rightarrow nh=2
\displaystyle f(x)=e^{2-3x}+x^{2}+1
\displaystyle f(a)=f(1)=e^{2-3(1)}+1^{2}+1
\displaystyle f(a+h)=f(1+h)=e^{2-3(1+h)}+(1+h)^{2}+1
\displaystyle f(a+2h)=f(1+2h)=e^{2-3(1+2h)}+(1+2h)^{2}+1
\displaystyle f(a+(n-1)h)=f(1+(n-1)h)=e^{2-3[1+(n-1)h]}+[1+(n-1)h]^{2}+1
\displaystyle \int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx=\lim_{h\to 0 \atop n\to \infty}h[e^{2-3\times 1}+1^{2}+1]
\displaystyle +[e^{2-3(1+h)}+(1+h)^{2}+1]+[e^{2-3(1+2h)}+(1+2h)^{2}+1]+\ldots
\displaystyle +[e^{2-3(1+(n-1)h)}+(1+(n-1)h)^{2}+1]
\displaystyle \text{[By using (1)]}
\displaystyle =\lim_{h\to 0 \atop n\to \infty}h[e^{2}(e^{-3}+e^{-3(1+h)}+e^{-3(1+2h)}+\ldots+e^{-3(1+(n-1)h)})]
\displaystyle +\lim_{h\to 0 \atop n\to \infty}h[1^{2}+(1+h)^{2}+(1+2h)^{2}+\ldots+(1+(n-1)h)^{2}]
\displaystyle +\lim_{h\to 0 \atop n\to \infty}h[1+1+1+\ldots n \text{ times}]
\displaystyle =\lim_{h\to 0 \atop n\to \infty}he^{-1}[1+e^{-3h}+e^{-3(2h)}+\ldots+e^{-3(n-1)h}]
\displaystyle +\lim_{h\to 0 \atop n\to \infty}h[n+2h\{1+2+3+\ldots+(n-1)\}+h^{2}\{1^{2}+2^{2}+3^{2}+\ldots+(n-1)^{2}\}]
\displaystyle +\lim_{h\to 0 \atop n\to \infty}nh
\displaystyle =\lim_{h\to 0 \atop n\to \infty}h\left[e^{-1}\left(\frac{(e^{-3h})^{n}-1}{e^{-3h}-1}\right)\right]
\displaystyle +\lim_{h\to 0 \atop n\to \infty}h\left[n+2h\frac{n(n-1)}{2}+h^{2}\frac{n(n-1)(2n-1)}{6}\right]+\lim_{h\to 0 \atop n\to \infty}nh
\displaystyle \left[\because\ a+ar+ar^{2}+\ldots+ar^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right),\ r\neq 1\right]
\displaystyle \left[1+2+3+\ldots+(n-1)=\frac{n(n-1)}{2}\right]
\displaystyle \left[1^{2}+2^{2}+3^{2}+\ldots+(n-1)^{2}=\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle =\lim_{h\to 0}e^{-1}\left[\frac{e^{-3nh}-1}{\left(\frac{e^{-3h}-1}{h}\right)}\right]
\displaystyle +\lim_{h\to 0}\left[nh+\frac{2hnh(nh-h)}{2}+\frac{nh(nh-h)(2nh-h)}{6}\right]+\lim_{h\to 0}nh
\displaystyle =\lim_{h\to 0}e^{-1}\left[\frac{e^{-6}-1}{-3\left(\frac{e^{-3h}-1}{-3h}\right)}\right]
\displaystyle +\lim_{h\to 0}\left[2+2(2-h)+\frac{2}{6}(2-h)(4-h)\right]+\lim_{h\to 0}2\qquad \left[\because\ nh=2\right]
\displaystyle =-\frac{e^{-1}}{3}(e^{-6}-1)+2+2(2-0)+\frac{2}{6}(2)(4)+2
\displaystyle =-\frac{e^{-1}}{3}(e^{-6}-1)+8+\frac{8}{3}
\displaystyle =-\frac{1}{3}(e^{-7}-e^{-1})+\frac{32}{3}
\displaystyle \therefore\ \int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx=-\frac{1}{3}(e^{-7}-e^{-1})+\frac{32}{3}

\displaystyle \text{22. Solve the differential equation: } (\tan^{-1}y-x)\,dy=(1+y^{2})\,dx.
\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 (\tan^{-1}y-x)\,dy=(1+y^{2})\,dx
\displaystyle \frac{dy}{dx}=\frac{1+y^{2}}{\tan^{-1}y-x}
\displaystyle \frac{dx}{dy}=\frac{\tan^{-1}y-x}{1+y^{2}}
\displaystyle \frac{dx}{dy}=\frac{\tan^{-1}y}{1+y^{2}}-\frac{x}{1+y^{2}}
\displaystyle \frac{dx}{dy}+\frac{1}{1+y^{2}}x=\frac{\tan^{-1}y}{1+y^{2}}
\displaystyle \text{This is linear differential equation of the form}
\displaystyle \frac{dx}{dy}+Px=Q
\displaystyle P=\frac{1}{1+y^{2}},\ Q=\frac{\tan^{-1}y}{1+y^{2}}
\displaystyle \text{I.F.}=e^{\int P\,dy}
\displaystyle =e^{\int \frac{1}{1+y^{2}}\,dy}=e^{\tan^{-1}y}
\displaystyle \text{Required solution is given by}
\displaystyle x(\text{I.F.})=\int Q(\text{I.F.})\,dy+C
\displaystyle xe^{\tan^{-1}y}=\int \left(\frac{\tan^{-1}y}{1+y^{2}}e^{\tan^{-1}y}\right)\,dy+C\qquad \ldots (1)
\displaystyle \text{Let } I=\int \frac{\tan^{-1}y}{1+y^{2}}e^{\tan^{-1}y}\,dy
\displaystyle \text{Let } \tan^{-1}y=t
\displaystyle \frac{1}{1+y^{2}}\,dy=dt
\displaystyle I=\int te^{t}\,dt
\displaystyle I=t\int e^{t}\,dt-\int \left[\frac{d}{dt}(t)\times \int e^{t}\,dt\right]dt
\displaystyle \qquad \left[\because\ \int u\,dv=u\int v\,dx-\int \left(\frac{d}{dx}u\int v\,dx\right)dx\right]
\displaystyle =te^{t}-\int e^{t}\,dt
\displaystyle =te^{t}-e^{t}
\displaystyle =\tan^{-1}y\,e^{\tan^{-1}y}-e^{\tan^{-1}y}
\displaystyle =e^{\tan^{-1}y}(\tan^{-1}y-1)
\displaystyle \text{Putting value of } I \text{ in (1)}
\displaystyle xe^{\tan^{-1}y}=e^{\tan^{-1}y}(\tan^{-1}y-1)+C
\displaystyle \text{OR}
\displaystyle \frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}}\qquad \ldots (1)
\displaystyle \text{This is a homogeneous equation.}
\displaystyle \text{Putting } y=vx
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{Putting the value of } y \text{ and } \frac{dy}{dx} \text{ in eq. (1)}
\displaystyle \frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}}
\displaystyle v+x\frac{dv}{dx}=\frac{x(vx)}{x^{2}+(vx)^{2}}
\displaystyle v+x\frac{dv}{dx}=\frac{v}{1+v^{2}}\Rightarrow x\frac{dv}{dx}=\frac{v}{1+v^{2}}-v
\displaystyle x\frac{dv}{dx}=-\frac{v^{3}}{1+v^{2}}\Rightarrow \frac{1+v^{2}}{v^{3}}\,dv=-\frac{1}{x}\,dx
\displaystyle \left(\frac{1}{v^{3}}+\frac{1}{v}\right)dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides}
\displaystyle \int \left(\frac{1}{v^{3}}+\frac{1}{v}\right)dv=-\int \frac{1}{x}\,dx
\displaystyle -\frac{v^{-2}}{2}+\log|v|=-\log|x|+C
\displaystyle \qquad \left[\because\ \int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C,\ \int \frac{1}{x}\,dx=\log|x|+C\right]
\displaystyle -\frac{1}{2v^{2}}+\log|v|+\log|x|=C
\displaystyle -\frac{1}{2v^{2}}+\log|vx|=C
\displaystyle -\frac{1}{2\left(\frac{y}{x}\right)^{2}}+\log\left|\frac{y}{x}\cdot x\right|=C\qquad \left[\because\ v=\frac{y}{x}\right]
\displaystyle -\frac{x^{2}}{2y^{2}}+\log|y|=C\qquad \ldots (2)
\displaystyle y=1 \text{ when } x=0\qquad \text{(given)}
\displaystyle C=0\qquad \text{(from (2))}
\displaystyle -\frac{x^{2}}{2y^{2}}+\log|y|=0\Rightarrow \log|y|=\frac{x^{2}}{2y^{2}}
\displaystyle \text{Hence } x^{2}=2y^{2}\log|y| \text{ is the solution of given equation.}

\displaystyle \text{23. If lines } \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} \text{ and } \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} \text{ intersect,}
\displaystyle \text{then find the value of } k \text{ and hence find the equation of the plane} \\ \text{containing these lines.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda
\displaystyle x=2\lambda+1,\ y=3\lambda-1,\ z=4\lambda+1
\displaystyle \text{Coordinates of any point on first line are } (2\lambda+1,\ 3\lambda-1,\ 4\lambda+1)
\displaystyle \text{Let } \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu
\displaystyle x=\mu+3,\ y=2\mu+k,\ z=\mu
\displaystyle \text{Coordinates of any point on second line are } (\mu+3,\ 2\mu+k,\ \mu)
\displaystyle \text{Comparing coordinates of both the lines}
\displaystyle 2\lambda+1=\mu+3\Rightarrow 2\lambda-\mu=2\qquad \ldots (1)
\displaystyle 3\lambda-1=2\mu+k\Rightarrow 3\lambda-2\mu=k+1\qquad \ldots (2)
\displaystyle 4\lambda+1=\mu\Rightarrow 4\lambda-\mu=-1\qquad \ldots (3)
\displaystyle \text{Solving (1) \& (3)}
\displaystyle \lambda=-\frac{3}{2}\ \text{and}\ \mu=-5
\displaystyle \text{Putting values of } \lambda \text{ \& } \mu \text{ in (2)}
\displaystyle 3\left(-\frac{3}{2}\right)-2(-5)=k+1
\displaystyle -\frac{9}{2}+10=k+1
\displaystyle k=\frac{9}{2}
\displaystyle \text{Point of intersection of both the lines}
\displaystyle P(2\lambda+1,\ 3\lambda-1,\ 4\lambda+1)
\displaystyle P\left(2\left(-\frac{3}{2}\right)+1,\ 3\left(-\frac{3}{2}\right)-1,\ 4\left(-\frac{3}{2}\right)+1\right)
\displaystyle P\left(-2,\ -\frac{11}{2},\ -5\right)
\displaystyle \text{Direction vectors of both the lines are}
\displaystyle \overrightarrow{n}_{1}=2\widehat{i}+3\widehat{j}+4\widehat{k}
\displaystyle \overrightarrow{n}_{2}=\widehat{i}+2\widehat{j}+\widehat{k}
\displaystyle \text{Direction vector of normal of plane is given as}
\displaystyle \overrightarrow{n}=\overrightarrow{n}_{1}\times \overrightarrow{n}_{2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&4\\1&2&1\end{vmatrix}
\displaystyle =\widehat{i}(3-8)-\widehat{j}(2-4)+\widehat{k}(4-3)
\displaystyle =-5\widehat{i}+2\widehat{j}+\widehat{k}
\displaystyle \text{Plane will also pass through } (1,-1,1), \text{ then } \overrightarrow{a}=\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \text{Equation of plane is given as}
\displaystyle (\overrightarrow{r}-\overrightarrow{a})\cdot \overrightarrow{n}=0
\displaystyle [\overrightarrow{r}-(\widehat{i}-\widehat{j}+\widehat{k})]\cdot(-5\widehat{i}+2\widehat{j}+\widehat{k})=0
\displaystyle \overrightarrow{r}\cdot(-5\widehat{i}+2\widehat{j}+\widehat{k})=(\widehat{i}-\widehat{j}+\widehat{k})\cdot(-5\widehat{i}+2\widehat{j}+\widehat{k})
\displaystyle \overrightarrow{r}\cdot(-5\widehat{i}+2\widehat{j}+\widehat{k})=-5-2+1
\displaystyle \overrightarrow{r}\cdot(-5\widehat{i}+2\widehat{j}+\widehat{k})=-6
\displaystyle \text{Cartesian form}
\displaystyle 5x-2y-z-6=0

\displaystyle \text{24. If A and B are two independent events such that } P(\overline{A}\cap B)=\frac{2}{15} \\ \text{ and } P(A\cap \overline{B})=\frac{1}{6},  \text{then find } P(A) \text{ and } P(B).
\displaystyle \text{Answer:}
\displaystyle \text{Let } P(A)=x \text{ and } P(B)=y
\displaystyle \therefore\ P(\overline{A})=1-x\qquad \left[\because\ P(\overline{A})=1-P(A)\right]
\displaystyle P(\overline{B})=1-y\qquad \left[\because\ P(\overline{B})=1-P(B)\right]
\displaystyle P(\overline{A}\cap B)=\frac{2}{15}
\displaystyle P(\overline{A})\cdot P(B)=\frac{2}{15}\qquad \left[\because\ A \text{ and } B \text{ are independent}\right]
\displaystyle (1-x)y=\frac{2}{15}\qquad \ldots (1)
\displaystyle P(A\cap \overline{B})=\frac{1}{6}
\displaystyle P(A)\cdot P(\overline{B})=\frac{1}{6}
\displaystyle x(1-y)=\frac{1}{6}
\displaystyle x=\frac{1}{6-6y}\qquad \ldots (2)
\displaystyle \text{Putting the value of } x \text{ in (1)}
\displaystyle \left(1-\frac{1}{6-6y}\right)y=\frac{2}{15}
\displaystyle \left(\frac{5-6y}{6-6y}\right)y=\frac{2}{15}
\displaystyle -90y^{2}+87y=12
\displaystyle 30y^{2}-29y+4=0
\displaystyle y=\frac{29\pm \sqrt{(29)^{2}-4(30)(4)}}{60}
\displaystyle y=\frac{4}{5}\ \text{or}\ y=\frac{1}{6}
\displaystyle \text{From (2)}
\displaystyle x=\frac{1}{6-6y}
\displaystyle \text{if } y=\frac{4}{5},\ x=\frac{1}{6-6\left(\frac{4}{5}\right)}=\frac{5}{6}
\displaystyle \text{if } y=\frac{1}{6},\ x=\frac{1}{6-6\left(\frac{1}{6}\right)}=\frac{1}{5}
\displaystyle \therefore\ P(A)=\frac{5}{6},\ P(B)=\frac{4}{5}
\displaystyle \text{or, } P(A)=\frac{1}{5},\ P(B)=\frac{1}{6}

\displaystyle \text{25. Find the local maxima and local minima of the function } f(x)=\sin x-\cos x,\ 0<x<2\pi. \text{ Also find the local maximum and local minimum values.}
\displaystyle \text{Answer:}
\displaystyle  f(x)=\sin x-\cos x,\ 0<x<2\pi
\displaystyle \text{On differentiating } f(x)
\displaystyle \frac{d}{dx}f(x)=\cos x+\sin x\qquad \ldots (1)
\displaystyle \text{To find local maxima or local minima}
\displaystyle \frac{d}{dx}f(x)=0
\displaystyle \cos x=-\sin x
\displaystyle \tan x=-1
\displaystyle \Rightarrow x=\frac{3\pi}{4}\ \text{or}\ \frac{7\pi}{4}\qquad \left[\because\ 0<x<2\pi\right]
\displaystyle \text{Again differentiating (1)}
\displaystyle \frac{d^{2}}{dx^{2}}f(x)=-\sin x+\cos x
\displaystyle \text{when, } x=\frac{3\pi}{4}
\displaystyle \frac{d^{2}}{dx^{2}}f\left(\frac{3\pi}{4}\right)=-\sin\left(\frac{3\pi}{4}\right)+\cos\left(\frac{3\pi}{4}\right)
\displaystyle =-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0
\displaystyle \therefore\ \frac{d^{2}}{dx^{2}}f\left(\frac{3\pi}{4}\right)<0
\displaystyle \therefore\ x=\frac{3\pi}{4} \text{ is the point of local maximum}
\displaystyle \text{When } x=\frac{7\pi}{4}
\displaystyle \frac{d^{2}}{dx^{2}}f\left(\frac{7\pi}{4}\right)=-\sin\left(\frac{7\pi}{4}\right)+\cos\left(\frac{7\pi}{4}\right)
\displaystyle =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0
\displaystyle \therefore\ \frac{d^{2}}{dx^{2}}f\left(\frac{7\pi}{4}\right)>0
\displaystyle \therefore\ x=\frac{7\pi}{4} \text{ is the point of local minimum}
\displaystyle \text{Local maximum value }=f\left(\frac{3\pi}{4}\right)
\displaystyle =\sin\left(\frac{3\pi}{4}\right)-\cos\left(\frac{3\pi}{4}\right)
\displaystyle =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}
\displaystyle \text{Local minimum value }=f\left(\frac{7\pi}{4}\right)
\displaystyle =\sin\left(\frac{7\pi}{4}\right)-\cos\left(\frac{7\pi}{4}\right)
\displaystyle =-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}

\displaystyle \text{26. Find graphically, the maximum value of } z=2x+5y, \text{ subject to constraints given below:}
\displaystyle 2x+4y\leq 8;\ 3x+y\leq 6;\ x+y\leq 4;\ x\geq 0,\ y\geq 0.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } 2x+4y\leq 8
\displaystyle 3x+y\leq 6
\displaystyle x+y\leq 4
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{We first convert inequalities into equations to obtain lines}
\displaystyle 2x+4y=8
\displaystyle x+2y=4
\displaystyle \text{For } 2x+4y=8
\displaystyle \begin{array}{|c|c|c|c|}\hline x&0&4&2\\\hline y&2&0&1\\\hline \end{array}
\displaystyle x+y=4
\displaystyle \text{For } x+y=4
\displaystyle \begin{array}{|c|c|c|c|}\hline x&0&4&1\\\hline y&4&0&3\\\hline \end{array}
\displaystyle 3x+y=6
\displaystyle \text{For } 3x+y=6
\displaystyle \begin{array}{|c|c|c|c|}\hline x&0&2&1\\\hline y&6&0&3\\\hline \end{array}
\displaystyle \text{Point of intersection of } 3x+y=6 \text{ and } 2x+4y=8 \text{ is } (1.6,\ 1.2)
\displaystyle \text{Corner points are } O(0,0),\ A(0,2),\ B(1.6,1.2),\ C(2,0)\displaystyle \text{We have to maximize } Z=2x+5y
\displaystyle \begin{array}{|c|c|}\hline \text{Corner Points}&Z=2x+5y\\\hline O(0,0)&Z=0\\\hline A(0,2)&Z=10\ \text{(maximum)}\\\hline B(1.6,1.2)&Z=1.6(2)+5(1.2)=9.2\\\hline C(2,0)&Z=4\\\hline \end{array}
\displaystyle Z \text{ will be maximum at } A(0,2)
\displaystyle \therefore\ \text{Maximized value of } Z \text{ is } 10.

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