MATHEMATICS

\displaystyle \text{Series BVM/1} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 65/1/3} \hspace{1.0cm} | \hspace{1.0cm} \text{Set 3 }    

\displaystyle \text{Time Allowed : 3 hours} \hspace{5.0cm} \text{Maximum Marks : 100 }  

\displaystyle \textbf{General Instructions :}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) This question paper contains 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 8 questions of} \\ \text{two marks each,}
\displaystyle \text{Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions} \\ \text{of six marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact} \\ \text{requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, internal choice has been provided in 1 question of} \\ \text{Section A, 3 questions of Section B,}
\displaystyle \text{3 questions of Section C and 3 questions of Section D. You have to attempt only one of the} \\ \text{alternatives in all such questions.}
\displaystyle \text{(v) Use of calculators is not permitted. You may ask logarithmic tables, if required.}

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

\displaystyle \textbf{1. }\text{If } 3A-B=\begin{bmatrix}5&0\\1&1\end{bmatrix} \text{ and } B=\begin{bmatrix}4&3\\2&5\end{bmatrix},   \text{then find the matrix } A.
\displaystyle \text{Answer:}
\displaystyle  \ 3A-\begin{bmatrix}4&3\\2&5\end{bmatrix}=\begin{bmatrix}5&0\\1&1\end{bmatrix}
\displaystyle \text{Taking } B \text{ to right hand side}
\displaystyle 3A=\begin{bmatrix}5&0\\1&1\end{bmatrix}+\begin{bmatrix}4&3\\2&5\end{bmatrix}=\begin{bmatrix}9&3\\3&6\end{bmatrix}
\displaystyle A=\begin{bmatrix}3&1\\1&2\end{bmatrix}

\displaystyle \textbf{2. }\text{Write the order and the degree of the following differential equation:}
\displaystyle x^3\left(\frac{d^2y}{dx^2}\right)^2+x\left(\frac{dy}{dx}\right)^4=0
\displaystyle \text{Answer:}
\displaystyle  \text{Highest order derivative is } \frac{d^2y}{dx^2}
\displaystyle \text{So, order of the differential equation is } 2
\displaystyle \text{Power of } \frac{d^2y}{dx^2} \text{ is } 2 \text{ in the given equation}
\displaystyle \text{So, degree of the differential equation is } 2
\displaystyle \therefore \text{Order is } 2 \text{ and degree is } 2

\displaystyle \textbf{3. }\text{If } f(x)=x+1,\ \text{find } \frac{d}{dx}(f\circ f)(x).
\displaystyle \text{Answer:}
\displaystyle  \text{Given: } f(x)=x+1
\displaystyle (f\circ f)(x)=f(f(x))=f(x+1)=(x+1)+1=x+2
\displaystyle \text{Differentiating on both sides}
\displaystyle \frac{d}{dx}(f\circ f)(x)=\frac{d}{dx}(x+2)  =\frac{d}{dx}(x)+\frac{d}{dx}(2)=1

\displaystyle \textbf{4. }\text{If a line makes angles } 90^\circ,135^\circ,45^\circ \text{ with the } x,y,z  \text{axes respectively, find its } \\ \text{direction cosines.}
\displaystyle \text{OR}
\displaystyle \text{Find the vector equation of the line which passes through the point } (3,4,5)
\displaystyle \text{and is parallel to the vector } 2\widehat{i}+2\widehat{j}-3\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle  \text{Direction cosines are } \cos 90^\circ,\cos 135^\circ,\cos 45^\circ
\displaystyle \cos 90^\circ=0
\displaystyle \cos 135^\circ=\cos(90^\circ+45^\circ)=-\sin 45^\circ=-\frac{1}{\sqrt{2}}
\displaystyle \cos 45^\circ=\frac{1}{\sqrt{2}}
\displaystyle \text{Direction cosines are } 0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}
\displaystyle \text{OR}
\displaystyle \text{Vector equation of a line passing through a point having position vector } \overrightarrow{a}
\displaystyle \text{and parallel to vector } \overrightarrow{b} \text{ is } \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}
\displaystyle \overrightarrow{b}=2\widehat{i}+2\widehat{j}-3\widehat{k}
\displaystyle \text{Point } (3,4,5)\Rightarrow \overrightarrow{a}=3\widehat{i}+4\widehat{j}+5\widehat{k}
\displaystyle \therefore \overrightarrow{r}=3\widehat{i}+4\widehat{j}+5\widehat{k}+\lambda(2\widehat{i}+2\widehat{j}-3\widehat{k})

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

\displaystyle \textbf{5. }\text{Find: } \int \sin x \log(\cos x)\,dx.
\displaystyle \text{Answer:}
\displaystyle  \int \sin x \log(\cos x)\,dx
\displaystyle \text{Let } \cos x=t\Rightarrow -\sin x\,dx=dt
\displaystyle \Rightarrow \int -\log t\,dt
\displaystyle \text{Using integration by parts}
\displaystyle \int u\,dv=uv-\int v\,du
\displaystyle =-\left[t\log t-\int t\cdot \frac{1}{t}\,dt\right]
\displaystyle =-\left[t\log t-\int 1\,dt\right]=-(t\log t-t)+C
\displaystyle =-t\log t+t+C
\displaystyle =-\cos x\log(\cos x)+\cos x+C

\displaystyle \textbf{6. }\text{Evaluate: } \int_{-\pi}^{\pi}(1-x^2)\sin x \cos^2 x\,dx
\displaystyle \text{OR}
\displaystyle \text{Evaluate: } \int_{-1}^{2}\frac{|x|}{x}\,dx
\displaystyle \text{Answer:}
\displaystyle  \text{As we know that}
\displaystyle \int_{-a}^{a}f(x)\,dx=\begin{cases}2\int_{0}^{a}f(x)\,dx,& \text{if } f(x)\text{ is even}\\0,& \text{if } f(x)\text{ is odd}\end{cases}
\displaystyle \int_{-\pi}^{\pi}(1-x^2)\sin x \cos^2 x\,dx
\displaystyle \text{Let } f(x)=(1-x^2)\sin x \cos^2 x
\displaystyle f(-x)=(1-(-x)^2)\sin(-x)\cos^2(-x)
\displaystyle =(1-x^2)(-\sin x)\cos^2 x
\displaystyle =-(1-x^2)\sin x \cos^2 x=-f(x)
\displaystyle \text{Hence } f \text{ is odd function}
\displaystyle \therefore \int_{-\pi}^{\pi}(1-x^2)\sin x \cos^2 x\,dx=0
\displaystyle \text{OR}
\displaystyle |x|=x \text{ when } x\geq 0
\displaystyle =-x \text{ when } x<0
\displaystyle \therefore \frac{|x|}{x}=1 \text{ when } x\geq 0
\displaystyle =-1 \text{ when } x<0
\displaystyle \int_{-1}^{2}\frac{|x|}{x}\,dx=\int_{-1}^{0}(-1)\,dx+\int_{0}^{2}(1)\,dx
\displaystyle =-\left[x\right]_{-1}^{0}+\left[x\right]_{0}^{2}
\displaystyle =-[0-(-1)]+[2-0]
\displaystyle =-1+2=1

\displaystyle \textbf{7. }\text{Examine whether the operation } * \text{ defined on } R \text{ by } a*b=ab+1
\displaystyle \text{is (i) a binary operation or not. (ii) if a binary operation, is it associative or not?}
\displaystyle \text{Answer:}
\displaystyle  \text{(i) Let } a\in R \text{ and } b\in R
\displaystyle \text{then } ab\in R
\displaystyle \text{Also } ab+1\in R
\displaystyle \text{So } a*b\in R
\displaystyle \text{So } * \text{ satisfies the closure property}
\displaystyle \text{Since } * \text{ is defined for all } a,b\in R
\displaystyle \therefore * \text{ is a binary operation}
\displaystyle \text{(ii) For } * \text{ to be associative}
\displaystyle (a*b)*c=(ab+1)*c=(ab+1)c+1
\displaystyle =abc+c+1 \qquad \ldots (1)
\displaystyle a*(b*c)=a*(bc+1)=a(bc+1)+1
\displaystyle =abc+a+1 \qquad \ldots (2)
\displaystyle \text{From equations (1) and (2)}
\displaystyle (a*b)*c\neq a*(b*c)
\displaystyle \therefore * \text{ is not associative}

\displaystyle \textbf{8. }\text{Find a matrix } A \text{ such that } 2A-3B+5C=0,\ \text{where}
\displaystyle B=\begin{bmatrix}-2&2&0\\3&1&4\end{bmatrix} \text{ and } C=\begin{bmatrix}2&0&-2\\7&1&6\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle  \text{Given: } 2A-3B+5C=0
\displaystyle \text{Both } B \text{ and } C \text{ are of order } 2\times 3\text{. Therefore } A \text{ should be of order } 2\times 3
\displaystyle 2A-3\begin{bmatrix}-2&2&0\\3&1&4\end{bmatrix}+5\begin{bmatrix}2&0&-2\\7&1&6\end{bmatrix}=0
\displaystyle 2A-\begin{bmatrix}-6&6&0\\9&3&12\end{bmatrix}+\begin{bmatrix}10&0&-10\\35&5&30\end{bmatrix}=0
\displaystyle 2A+\begin{bmatrix}16&-6&-10\\26&2&18\end{bmatrix}=0
\displaystyle 2A=-\begin{bmatrix}16&-6&-10\\26&2&18\end{bmatrix}
\displaystyle A=\begin{bmatrix}-\frac{16}{2}&\frac{6}{2}&\frac{10}{2}\\-\frac{26}{2}&-\frac{2}{2}&-\frac{18}{2}\end{bmatrix}
\displaystyle A=\begin{bmatrix}-8&3&5\\-13&-1&-9\end{bmatrix}

\displaystyle \textbf{9. }\text{A die marked 1,2,3 in red and 4,5,6 in green is tossed.}
\displaystyle \text{Let } A \text{ be the event "number is even" and } B \text{ be the event "number is marked }
\displaystyle \text{red". Find whether the events } A \text{ and } B   \text{are independent or not.}
\displaystyle \text{Answer:}
\displaystyle  \text{When a die is thrown, the sample space } (S) \text{ is}
\displaystyle S=\{1,2,3,4,5,6\}
\displaystyle \text{Let } A:\text{ the number is even }=\{2,4,6\}
\displaystyle P(A)=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}=\frac{3}{6}=\frac{1}{2}
\displaystyle B:\text{ the number is marked red}
\displaystyle B=\{1,2,3\}
\displaystyle P(B)=\frac{3}{6}=\frac{1}{2}
\displaystyle P(A).P(B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}
\displaystyle P(A\cap B)=\frac{1}{6}
\displaystyle \text{[Only 2 is a number which is even and red]}
\displaystyle \text{As we can see that}
\displaystyle P(A\cap B)\neq P(A).P(B)
\displaystyle \therefore A \text{ and } B \text{ are not independent events}

\displaystyle \textbf{10. }\text{Form the differential equation representing the family of curves}
\displaystyle y=e^{2x}(a+bx), \text{ where } a \text{ and } b \text{ are arbitrary constants.}
\displaystyle \text{Answer:}
\displaystyle  \text{Given: } y=e^{2x}(a+bx) \qquad \ldots (1)
\displaystyle \text{Differentiating on both sides with respect to } x
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[e^{2x}(a+bx)\right]
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(ae^{2x})+\frac{d}{dx}(bxe^{2x})
\displaystyle \frac{dy}{dx}=2ae^{2x}+b\left[e^{2x}\frac{dx}{dx}+x\frac{d}{dx}e^{2x}\right]
\displaystyle =2ae^{2x}+b\left[e^{2x}+2xe^{2x}\right]
\displaystyle =be^{2x}+2(a+bx)e^{2x}
\displaystyle \frac{dy}{dx}=be^{2x}+2y \qquad \left[\because y=e^{2x}(a+bx)\right] \qquad \ldots (2)
\displaystyle \text{Differentiating the above equation}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=2be^{2x}+2\frac{dy}{dx} \qquad \ldots (3)
\displaystyle \text{Substituting the value of } be^{2x} \text{ from (2) in equation (3)}
\displaystyle \frac{d^2y}{dx^2}=2\left(\frac{dy}{dx}-2y\right)+2\frac{dy}{dx}
\displaystyle \frac{d^2y}{dx^2}-4\frac{dy}{dx}+4y=0

\displaystyle \textbf{11. }\text{A die is thrown 6 times. If "getting an odd number" is a "success", what is the}
\displaystyle \text{probability of (i) 5 successes? (ii) atmost 5 successes?}
\displaystyle \text{OR}
\displaystyle \text{The random variable } X \text{ has a probability distribution } P(X) \text{of the following form, }
\displaystyle \text{where } k \text{ is some number.}
\displaystyle P(X=x)=\begin{cases}k,& x=0\\2k,& x=1\\3k,& x=2\\0,& \text{otherwise}\end{cases}
\displaystyle \text{Determine the value of } k.
\displaystyle \text{Answer:}
\displaystyle  \text{Number of trials }=6
\displaystyle \text{Random variable } (X)=0,1,2,3,4,5,6
\displaystyle p\ (\text{Probability of odd numbers})=\frac{3}{6}=\frac{1}{2}
\displaystyle q\ (\text{Probability of not odd})=\frac{3}{6}=\frac{1}{2}
\displaystyle \left[\because P(X=x)={}^{n}C_{x}\,p^xq^{\,n-x}\right]
\displaystyle \text{(i) } P(X=5)={}^{6}C_{5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{6-5}=\frac{3}{32}
\displaystyle \left[\because {}^{n}C_{p}=\frac{n!}{p!(n-p)!}\right]
\displaystyle \text{(ii) Probability of getting at most 5 successes}
\displaystyle P(X\leq 5)=1-P(X=6)
\displaystyle =1-{}^{6}C_{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^0=1-\frac{1}{64}=\frac{63}{64}
\displaystyle \text{OR}
\displaystyle \text{We know that sum of probabilities of a probability distribution of random variable is 1}
\displaystyle \therefore k+2k+3k+0=1
\displaystyle 6k=1
\displaystyle k=\frac{1}{6}

\displaystyle \textbf{12. }\text{If the sum of two unit vectors is a unit vector, prove that the magnitude of their}
\displaystyle \text{difference is } \sqrt{3}.
\displaystyle \text{OR}
\displaystyle \text{If } \overrightarrow{a}=2\widehat{i}+3\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \text{and } \overrightarrow{c}=-3\widehat{i}+\widehat{j}+2\widehat{k}, \text{ find } [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}].
\displaystyle \text{Answer:}
\displaystyle  \text{Given: } |\overrightarrow{x}|=1,\ |\overrightarrow{y}|=1 \text{ and } |\overrightarrow{x}+\overrightarrow{y}|=1
\displaystyle \text{Squaring on both sides}
\displaystyle |\overrightarrow{x}+\overrightarrow{y}|^2=1^2
\displaystyle |\overrightarrow{x}|^2+|\overrightarrow{y}|^2+2|\overrightarrow{x}||\overrightarrow{y}|\cos\theta=1
\displaystyle \left[\because |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta\right]
\displaystyle 1+1+2(1)(1)\cos\theta=1
\displaystyle 2\cos\theta=-1 \qquad \ldots (1)
\displaystyle |\overrightarrow{x}-\overrightarrow{y}|^2=|\overrightarrow{x}|^2+|\overrightarrow{y}|^2-2|\overrightarrow{x}||\overrightarrow{y}|\cos\theta
\displaystyle \left[\because |\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta\right]
\displaystyle |\overrightarrow{x}-\overrightarrow{y}|^2=1+1-2(1)(1)\cos\theta=2-2\cos\theta
\displaystyle =2-(-1)=3 \qquad \text{[from (1)]}
\displaystyle |\overrightarrow{x}-\overrightarrow{y}|=\sqrt{3}
\displaystyle \text{Hence proved.}
\displaystyle \text{OR}
\displaystyle [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]=\left|\begin{matrix}2&3&1\\1&-2&1\\-3&1&2\end{matrix}\right|
\displaystyle =2(-4-1)-3(2+3)+1(1-6)
\displaystyle =-30

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

\displaystyle \textbf{13. }\text{Using properties of determinants, prove the following:}
\displaystyle \left|\begin{matrix}a&b&c\\a-b&b-c&c-a\\b+c&c+a&a+b\end{matrix}\right|=a^3+b^3+c^3-3abc
\displaystyle \text{Answer:}
\displaystyle  \ \Delta=\left|\begin{matrix}a&b&c\\a-b&b-c&c-a\\b+c&c+a&a+b\end{matrix}\right|
\displaystyle \text{Applying } R_2\rightarrow R_2-R_1
\displaystyle \Delta=\left|\begin{matrix}a&b&c\\-b&-c&-a\\b+c&c+a&a+b\end{matrix}\right|
\displaystyle \text{Applying } R_3\rightarrow R_3+R_1
\displaystyle \Delta=\left|\begin{matrix}a&b&c\\-b&-c&-a\\a+b+c&a+b+c&a+b+c\end{matrix}\right|
\displaystyle \text{Taking } (a+b+c) \text{ common}
\displaystyle \Delta=(a+b+c)\left|\begin{matrix}a&b&c\\-b&-c&-a\\1&1&1\end{matrix}\right|
\displaystyle \text{Applying } C_2\rightarrow C_2-C_3
\displaystyle \Delta=(a+b+c)\left|\begin{matrix}a&b-c&c\\-b&-c+a&-a\\1&0&1\end{matrix}\right|
\displaystyle \text{Applying } C_1\rightarrow C_1-C_3
\displaystyle \Delta=(a+b+c)\left|\begin{matrix}a-c&b-c&c\\-b+a&-c+a&-a\\0&0&1\end{matrix}\right|
\displaystyle \Delta=(a+b+c)[(a-c)(a-c)-(b-c)(a-b)]
\displaystyle =(a+b+c)[a^2+c^2-2ac-(ab-bc-ca+cb)]
\displaystyle =(a+b+c)[a^2+b^2+c^2-ab-bc-ca]
\displaystyle \Delta=a^3+b^3+c^3-3abc
\displaystyle \left[\text{By formula } x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\right]
\displaystyle \text{Hence proved.}

\displaystyle \textbf{14. }\text{Solve: } \tan^{-1}(4x)+\tan^{-1}(6x)=\frac{\pi}{4}
\displaystyle \text{Answer:}
\displaystyle  \text{We know that}
\displaystyle \tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)
\displaystyle \tan^{-1}4x+\tan^{-1}6x=\frac{\pi}{4}
\displaystyle \tan^{-1}\left(\frac{4x+6x}{1-(4x)(6x)}\right)=\frac{\pi}{4}
\displaystyle \tan^{-1}\left(\frac{10x}{1-24x^2}\right)=\frac{\pi}{4}
\displaystyle \text{Taking tan on both sides}
\displaystyle \tan\left(\tan^{-1}\left(\frac{10x}{1-24x^2}\right)\right)=\tan\frac{\pi}{4}
\displaystyle \Rightarrow \frac{10x}{1-24x^2}=1 \qquad \left[\because \tan\frac{\pi}{4}=1\right]
\displaystyle \Rightarrow 24x^2+10x-1=0
\displaystyle \Rightarrow 24x^2+12x-2x-1=0 \qquad \text{[Middle term splitting]}
\displaystyle \Rightarrow 12x(2x+1)-1(2x+1)=0
\displaystyle \Rightarrow (2x+1)(12x-1)=0
\displaystyle \text{If } 2x+1=0 \text{ or } 12x-1=0
\displaystyle x=-\frac{1}{2} \qquad \text{or} \qquad x=\frac{1}{12}
\displaystyle \text{For } x=-\frac{1}{2},\ \text{LHS becomes negative, so it is not valid}
\displaystyle \therefore x=\frac{1}{12}

\displaystyle \textbf{15. }\text{Show that the relation } R \text{ on } R \text{ defined as } R=\{(a,b):a\leq b\}, \text{is reflexive and}
\displaystyle \text{transitive but not symmetric.}
\displaystyle \text{OR}
\displaystyle \text{Prove that the function } f:N\rightarrow N,\ f(x)=x^2+x+1 \text{ is one-one but not onto.}
\displaystyle \text{Find inverse of } f:N\rightarrow S,\ \text{where } S \text{ is range of } f.
\displaystyle \text{Answer:}
\displaystyle  \text{Given: } R=\{(a,b):a\leq b\} \text{ and } a,b\in R
\displaystyle \text{We know that } a=a
\displaystyle \therefore a\leq a \Rightarrow (a,a)\in R
\displaystyle \therefore R \text{ is reflexive}
\displaystyle \text{To check whether } R \text{ is symmetric or not}
\displaystyle \text{Let } a=3,\ b=7
\displaystyle (3,7)\in R \text{ as } 3<7
\displaystyle \text{But } (7,3)\notin R \text{ because } 7 \text{ is greater than } 3
\displaystyle \therefore R \text{ is not symmetric}
\displaystyle \text{If } a\leq b \text{ and } b\leq c\text{, then } a\leq c
\displaystyle \therefore \text{We can say that if } (a,b)\in R \text{ and } (b,c)\in R\text{, then } (a,c)\in R
\displaystyle \text{Hence } R \text{ is transitive}
\displaystyle \text{OR}
\displaystyle \text{Given function } f(x)=x^2+x+1,\quad f:N\to N
\displaystyle \text{Let } f(x_1)=f(x_2) \text{ when } x_1,x_2\in N
\displaystyle x_1^2+x_1+1=x_2^2+x_2+1
\displaystyle x_1^2-x_2^2=x_2-x_1 \qquad \left[\because (a^2-b^2)=(a-b)(a+b)\right]
\displaystyle (x_1-x_2)(x_1+x_2)+(x_1-x_2)=0
\displaystyle (x_1-x_2)(x_1+x_2+1)=0
\displaystyle x_1-x_2=0 \Rightarrow x_1=x_2
\displaystyle \therefore x_1+x_2\neq -1 \text{ (sum of natural numbers is not negative)}
\displaystyle \text{So, } f(x_2)=f(x_1) \text{ only for } x_1=x_2
\displaystyle \therefore f(x) \text{ is one-one function}
\displaystyle f(x)=x^2+x+1+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(x+\frac{1}{2}\right)^2+1-\frac{1}{4}
\displaystyle =\left(x+\frac{1}{2}\right)^2+\frac{3}{4}
\displaystyle \left(x+\frac{1}{2}\right)^2 \text{ is always positive}
\displaystyle \therefore f(x) \text{ is increasing function}
\displaystyle \text{If } x=1,\ f(1)=3
\displaystyle \text{If } x=2,\ f(2)=7
\displaystyle \text{Range of } f(x)=[3,7,\ldots]
\displaystyle \text{But } f(x) \text{ does not have } 1,2
\displaystyle \therefore f(x) \text{ is into function, not onto function}
\displaystyle \text{Let } f^{-1} \text{ denote inverse of } f
\displaystyle f\circ f^{-1}(x)=x \text{ for all } x\in \text{Range }(f)
\displaystyle f(f^{-1}(x))=x \text{ for all } x\in \text{Range }(f)
\displaystyle \Rightarrow [f^{-1}(x)]^2+f^{-1}(x)+1=x \text{ for all } x\in \text{Range }(f)
\displaystyle \Rightarrow [f^{-1}(x)]^2+f^{-1}(x)+(1-x)=0
\displaystyle f^{-1}(x)=\frac{-1\pm \sqrt{1-4(1)(1-x)}}{2} \qquad \left[\because \text{for } ax^2+bx+c=0,\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]
\displaystyle f^{-1}(x)=\frac{-1\pm \sqrt{4x-3}}{2}

\displaystyle \textbf{16. }\text{Find the equation of tangent to the curve } y=\sqrt{3x-2} \text{ which is parallel}
\displaystyle \text{to the line } 4x-2y+5=0. \text{Also, write the equation of normal to the curve}
\displaystyle \text{at the point of contact.}
\displaystyle \text{Answer:}
\displaystyle  \text{Slope of line } 4x-2y+5=0
\displaystyle m=-\frac{a}{b}=-\frac{4}{-2}=2
\displaystyle \therefore \text{Slope of given line is } 2
\displaystyle y=\sqrt{3x-2}
\displaystyle \text{Differentiate on both sides}
\displaystyle \frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}} \qquad \left[\because \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\right]
\displaystyle \text{Tangent is parallel to line } 4x-2y+5=0
\displaystyle \therefore \text{Slope of tangent }=\text{ slope of } 4x-2y+5=0
\displaystyle \frac{3}{2\sqrt{3x-2}}=2 \Rightarrow 3=4\sqrt{3x-2}
\displaystyle \text{Squaring on both sides}
\displaystyle 9=16(3x-2)\Rightarrow 3x-2=\frac{9}{16}\Rightarrow x=\frac{41}{48}
\displaystyle y=\sqrt{3\left(\frac{41}{48}\right)-2}=\sqrt{\frac{9}{16}}=\frac{3}{4}
\displaystyle \text{Equation of tangent}
\displaystyle y-y_1=m(x-x_1)
\displaystyle y-\frac{3}{4}=2\left(x-\frac{41}{48}\right)
\displaystyle \frac{4y-3}{4}=2\left(\frac{48x-41}{48}\right)=\frac{48x-41}{24}
\displaystyle 6(4y-3)=48x-41\Rightarrow 24y-18=48x-41
\displaystyle \Rightarrow 48x-24y=23
\displaystyle \text{Required equation of tangent is } 48x-24y=23
\displaystyle \text{To find equation of normal to the curve at the point of contact}
\displaystyle y-y_1=m_1(x-x_1)
\displaystyle \text{Slope of normal }=\frac{-1}{\text{Slope of tangent}}
\displaystyle m_1=-\frac{1}{2} \qquad \left[\because \text{Slope of tangent}=2\right]
\displaystyle \text{Equation of normal}
\displaystyle y-\frac{3}{4}=-\frac{1}{2}\left(x-\frac{41}{48}\right)
\displaystyle 4y-3=-\frac{48x-41}{24}
\displaystyle \Rightarrow 24(4y-3)=-48x+41\Rightarrow 48x+96y=113
\displaystyle \text{Equation of normal is } 48x+96y=113

\displaystyle \textbf{17. }\text{If } \log(x^2+y^2)=2\tan^{-1}\left(\frac{y}{x}\right), \text{ show that }   \frac{dy}{dx}=\frac{x+y}{x-y}
\displaystyle \text{OR}
\displaystyle \text{If } x^y-y^x=a^b, \text{ find } \frac{dy}{dx}
\displaystyle \text{Answer:}
\displaystyle  \text{Given: } \log(x^2+y^2)=2\tan^{-1}\left(\frac{y}{x}\right)
\displaystyle \text{Differentiating with respect to } x
\displaystyle \frac{d}{dx}\left[\log(x^2+y^2)\right]=\frac{d}{dx}\left[2\tan^{-1}\left(\frac{y}{x}\right)\right] \qquad \ldots (1)
\displaystyle \Rightarrow \frac{1}{x^2+y^2}\frac{d}{dx}(x^2+y^2)=2\times \frac{1}{1+\left(\frac{y}{x}\right)^2}\frac{d}{dx}\left(\frac{y}{x}\right)
\displaystyle \Rightarrow \frac{1}{x^2+y^2}\left[2x+2y\frac{dy}{dx}\right]=\frac{2}{\frac{x^2+y^2}{x^2}}\left[\frac{x\frac{dy}{dx}-y(1)}{x^2}\right]
\displaystyle \Rightarrow \frac{1}{x^2+y^2}\left[2x+2y\frac{dy}{dx}\right]=\frac{2}{x^2+y^2}\left[x\frac{dy}{dx}-y\right]
\displaystyle \Rightarrow 2x+2y\frac{dy}{dx}=2\left[x\frac{dy}{dx}-y\right]
\displaystyle \Rightarrow 2x+2y\frac{dy}{dx}=2x\frac{dy}{dx}-2y
\displaystyle \Rightarrow 2x+2y=2x\frac{dy}{dx}-2y\frac{dy}{dx}
\displaystyle \Rightarrow 2(x+y)=2\frac{dy}{dx}(x-y)\Rightarrow x+y=\frac{dy}{dx}(x-y)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}
\displaystyle \text{Hence proved.}
\displaystyle \text{OR}
\displaystyle \text{Let } x^y=u,\ y^x=v
\displaystyle x^y-y^x=a^b
\displaystyle u-v=a^b
\displaystyle \text{Differentiating with respect to } x
\displaystyle \frac{du}{dx}-\frac{dv}{dx}=\frac{d}{dx}(a^b)
\displaystyle \frac{du}{dx}-\frac{dv}{dx}=0 \qquad \ldots (1) \qquad \left[\because \text{Differentiation of constant term is } 0\right]
\displaystyle \text{Now } u=x^y
\displaystyle \text{Taking log on both sides}
\displaystyle \log u=\log(x^y)
\displaystyle \log u=y\log x \qquad \left[\because \log m^n=n\log m\right]
\displaystyle \text{Differentiating on both sides}
\displaystyle \frac{d}{dx}(\log u)=\frac{d}{dx}(y\log x)
\displaystyle \frac{1}{u}\frac{du}{dx}=y\frac{d}{dx}(\log x)+\log x\frac{dy}{dx}
\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\log x\frac{dy}{dx}
\displaystyle \Rightarrow \frac{du}{dx}=u\left[\frac{y}{x}+\log x\frac{dy}{dx}\right]
\displaystyle \Rightarrow \frac{du}{dx}=x^y\left[\frac{y}{x}+\log x\frac{dy}{dx}\right] \qquad \left[\because u=x^y\right]
\displaystyle \Rightarrow \frac{du}{dx}=yx^{y-1}+x^y\log x\frac{dy}{dx} \qquad \ldots (2)
\displaystyle \text{Also } v=y^x
\displaystyle \text{Taking log on both sides}
\displaystyle \log v=\log(y^x)
\displaystyle \log v=x\log y
\displaystyle \text{Differentiating on both sides}
\displaystyle \frac{d}{dx}(\log v)=\frac{d}{dx}(x\log y)
\displaystyle \frac{1}{v}\frac{dv}{dx}=x\frac{d}{dx}(\log y)+\log y\frac{d}{dx}(x)
\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx}=x\frac{1}{y}\frac{dy}{dx}+\log y
\displaystyle \Rightarrow \frac{dv}{dx}=v\left[\frac{x}{y}\frac{dy}{dx}+\log y\right]
\displaystyle \Rightarrow \frac{dv}{dx}=y^x\left[\frac{x}{y}\frac{dy}{dx}+\log y\right] \qquad \left[\because v=y^x\right]
\displaystyle \Rightarrow \frac{dv}{dx}=xy^{x-1}\frac{dy}{dx}+y^x\log y \qquad \ldots (3)
\displaystyle \text{From (1)}
\displaystyle \frac{du}{dx}-\frac{dv}{dx}=0
\displaystyle \text{Putting the values of } \frac{du}{dx} \text{ and } \frac{dv}{dx} \text{ from (2) and (3) in (1)}
\displaystyle \left(yx^{y-1}+x^y\log x\frac{dy}{dx}\right)-\left(xy^{x-1}\frac{dy}{dx}+y^x\log y\right)=0
\displaystyle yx^{y-1}+x^y\log x\frac{dy}{dx}-xy^{x-1}\frac{dy}{dx}-y^x\log y=0
\displaystyle \frac{dy}{dx}(x^y\log x-xy^{x-1})=y^x\log y-yx^{y-1}
\displaystyle \therefore \frac{dy}{dx}=\frac{y^x\log y-yx^{y-1}}{x^y\log x-xy^{x-1}}

\displaystyle \textbf{18. }\text{If } y=(\sin^{-1}x)^2, \text{ prove that } (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0
\displaystyle \text{Answer:}
\displaystyle  \ y=(\sin^{-1}x)^2 \qquad \ldots (1)
\displaystyle \text{Differentiating both sides with respect to } x
\displaystyle \Rightarrow \frac{dy}{dx}=2\sin^{-1}x\times \frac{d}{dx}(\sin^{-1}x) \qquad \text{(Using chain rule)}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2\sin^{-1}x}{\sqrt{1-x^2}}
\displaystyle \sqrt{1-x^2}\frac{dy}{dx}=2\sin^{-1}x
\displaystyle \text{Squaring both sides}
\displaystyle (1-x^2)\left(\frac{dy}{dx}\right)^2=4(\sin^{-1}x)^2
\displaystyle (1-x^2)\left(\frac{dy}{dx}\right)^2=4y \qquad \text{(From 1)}
\displaystyle \text{Differentiating both sides, we get}
\displaystyle (-2x)\left(\frac{dy}{dx}\right)^2+(1-x^2)\cdot 2\frac{dy}{dx}\frac{d^2y}{dx^2}=4\frac{dy}{dx}
\displaystyle (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0
\displaystyle \text{Hence proved.}

\displaystyle \textbf{19. }\text{Prove that } \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx, \text{ hence evaluate:}   \int_{0}^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx
\displaystyle \text{Answer:}
\displaystyle  \text{Let } a-x=t
\displaystyle \text{Differentiate w.r.t. } x
\displaystyle -1=\frac{dt}{dx}\Rightarrow dx=-dt \qquad \left[\text{As, } x=0,\ t=a \text{ and } x=a,\ t=0\right]
\displaystyle \int_{0}^{a}f(x)\,dx=-\int_{a}^{0}f(a-t)\,dt \qquad \left[\because a-x=t,\ x=a-t\right]
\displaystyle =\int_{0}^{a}f(a-t)\,dt \qquad \left[\because \int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx\right]
\displaystyle =\int_{0}^{a}f(a-x)\,dx
\displaystyle \left[\because \text{Integration is independent of the change of variable}\right]
\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
\displaystyle \text{Hence proved.}
\displaystyle \text{Let } I=\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx \qquad \left[\because \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\right]
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx \qquad \left[\because \sin(\pi-\theta)=\sin\theta,\ \cos(\pi-\theta)=-\cos\theta\right]
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^2 x}\,dx-\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^2 x}\,dx-I
\displaystyle \Rightarrow 2I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^2 x}\,dx
\displaystyle \text{Let } \cos x=t\Rightarrow -\sin x\,dx=dt
\displaystyle \text{As, } x=0,\ t=1 \text{ and } x=\pi,\ t=-1
\displaystyle 2I=-\pi\int_{1}^{-1}\frac{dt}{1+t^2}
\displaystyle \Rightarrow I=-\frac{\pi}{2}\left[\tan^{-1}t\right]_{1}^{-1} \qquad \left[\because \int \frac{1}{1+x^2}\,dx=\tan^{-1}x\right]
\displaystyle \Rightarrow I=-\frac{\pi}{2}\left[\tan^{-1}(-1)-\tan^{-1}(1)\right]
\displaystyle \Rightarrow I=-\frac{\pi}{2}\left[-\frac{\pi}{4}-\frac{\pi}{4}\right] \qquad \left[\because \tan^{-1}(1)=\frac{\pi}{4}\right]
\displaystyle \Rightarrow I=-\frac{\pi}{2}\left(-\frac{2\pi}{4}\right)\Rightarrow I=\frac{\pi^2}{4}
\displaystyle \int_{0}^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx=\frac{\pi^2}{4}

\displaystyle \textbf{20. }\text{Find: } \int \frac{\cos x}{(1+\sin x)(2+\sin x)}\,dx
\displaystyle \text{Answer:}
\displaystyle  \text{Let } 1+\sin x=u
\displaystyle \cos x\,dx=du
\displaystyle \int \frac{\cos x\,dx}{(1+\sin x)(2+\sin x)}=\int \frac{du}{u(1+u)} \qquad \left[\because 2+\sin x=1+1+\sin x=1+u\right]
\displaystyle =\int \left(\frac{1}{u}-\frac{1}{1+u}\right)du
\displaystyle =\int \frac{1}{u}\,du-\int \frac{1}{1+u}\,du
\displaystyle =\log u-\log(1+u)+C \qquad \left[\because \int \frac{1}{x}\,dx=\log x\right]
\displaystyle =\log(1+\sin x)-\log(2+\sin x)+C
\displaystyle =\log\left(\frac{1+\sin x}{2+\sin x}\right)+C \qquad \left[\because \log m-\log n=\log\frac{m}{n}\right]
\displaystyle \therefore \int \frac{\cos x\,dx}{(1+\sin x)(2+\sin x)}=\log\left(\frac{1+\sin x}{2+\sin x}\right)+C

\displaystyle \textbf{21. }\text{Solve the differential equation: } \frac{dy}{dx}-\frac{2x}{1+x^2}y=x^2+2
\displaystyle \text{OR}
\displaystyle \text{Solve the differential equation: } (x+1)\frac{dy}{dx}=2e^{y}-1,\ y(0)=0
\displaystyle \text{Answer:}
\displaystyle  \text{Given differential equation is}
\displaystyle \frac{dy}{dx}-\frac{2x}{1+x^2}y=x^2+2
\displaystyle \text{This is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+Py=Q,\ \text{where}
\displaystyle P=-\frac{2x}{1+x^2} \text{ and } Q=x^2+2
\displaystyle \therefore \text{I.F.}=e^{\int P\,dx}=e^{\int \frac{-2x}{1+x^2}\,dx}=e^{-\log(1+x^2)}
\displaystyle =e^{\log\left(\frac{1}{1+x^2}\right)}=\frac{1}{1+x^2}
\displaystyle \text{General solution of the given differential equation is}
\displaystyle y\cdot (\text{I.F.})=\int Q(\text{I.F.})\,dx+C
\displaystyle y\left(\frac{1}{1+x^2}\right)=\int \frac{x^2+2}{1+x^2}\,dx+C
\displaystyle =\int \left(1+\frac{1}{1+x^2}\right)\,dx+C
\displaystyle =\int 1\,dx+\int \frac{1}{1+x^2}\,dx+C
\displaystyle =x+\tan^{-1}x+C \qquad \left[\because \int \frac{1}{1+x^2}\,dx=\tan^{-1}x\right]
\displaystyle y\left(\frac{1}{1+x^2}\right)=x+\tan^{-1}x+C
\displaystyle y=(1+x^2)(x+\tan^{-1}x+C)
\displaystyle \text{OR}
\displaystyle (x+1)\frac{dy}{dx}=2e^{-y}-1 \Rightarrow (x+1)\,dy=(2e^{-y}-1)\,dx
\displaystyle \Rightarrow \frac{1}{2e^{-y}-1}\,dy=\frac{1}{x+1}\,dx
\displaystyle \text{Integrating on both sides}
\displaystyle \int \frac{1}{2e^{-y}-1}\,dy=\int \frac{1}{x+1}\,dx
\displaystyle \Rightarrow \int \frac{1}{2e^{-y}-1}\,dy=\log|x+1|+\log C \qquad \left[\because \int \frac{1}{x}\,dx=\log|x|\right]
\displaystyle \Rightarrow -\int \frac{e^y}{e^y-2}\,dy=\log|x+1|+\log C \qquad \ldots (1)
\displaystyle \text{Let } e^y-2=t \Rightarrow e^y\,dy=dt
\displaystyle \text{Substituting these values in equation (1)}
\displaystyle -\int \frac{dt}{t}=\log|x+1|+\log C
\displaystyle \Rightarrow -\log|t|=\log|x+1|+\log C
\displaystyle \Rightarrow -\log|e^y-2|=\log|x+1|+\log C
\displaystyle \Rightarrow \log\left|\frac{1}{e^y-2}\right|=\log|x+1|+\log C
\displaystyle \Rightarrow \log\left|\frac{1}{e^y-2}\right|=\log|C(x+1)|
\displaystyle \Rightarrow \left|\frac{1}{e^y-2}\right|=|C(x+1)| \qquad \ldots (2)
\displaystyle \text{Given that } x=0,\ y=0
\displaystyle \left|\frac{1}{e^0-2}\right|=C(0+1) \Rightarrow \left|\frac{1}{1-2}\right|=C \Rightarrow C=1
\displaystyle \text{Putting the value of } C \text{ in (1)}
\displaystyle \left|\frac{1}{e^y-2}\right|=|x+1| \Rightarrow |(x+1)(e^y-2)|=1
\displaystyle \Rightarrow (x+1)(e^y-2)=1
\displaystyle \Rightarrow e^y-2=\frac{1}{x+1} \Rightarrow e^y=2+\frac{1}{x+1}
\displaystyle y=\log\left(2+\frac{1}{x+1}\right)

\displaystyle \textbf{22. }\text{If } \widehat{i}+\widehat{j}+\widehat{k},\ 2\widehat{i}+5\widehat{j},\ 3\widehat{i}+2\widehat{j}-3\widehat{k}   \text{and } \widehat{i}-6\widehat{j}-\widehat{k} \text{ respectively are position vectors of points }
\displaystyle A,B,C,D,  \text{ then find the angle between the straight lines } AB \text{ and } CD. \text{Find whether}
\displaystyle AB \text{ and } CD \text{ are collinear or not.}
\displaystyle \text{Answer:}
\displaystyle  \overrightarrow{AB}=(x_2-x_1)\widehat{i}+(y_2-y_1)\widehat{j}+(z_2-z_1)\widehat{k}
\displaystyle =(2-1)\widehat{i}+(5-1)\widehat{j}+(0-1)\widehat{k}=\widehat{i}+4\widehat{j}-\widehat{k}
\displaystyle AB=|\overrightarrow{AB}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\displaystyle =\sqrt{1^2+4^2+(-1)^2}=\sqrt{18}=3\sqrt{2}
\displaystyle \overrightarrow{CD}=(x_2-x_1)\widehat{i}+(y_2-y_1)\widehat{j}+(z_2-z_1)\widehat{k}
\displaystyle =(1-3)\widehat{i}+(-6-2)\widehat{j}+(-1+3)\widehat{k}=-2\widehat{i}-8\widehat{j}+2\widehat{k}
\displaystyle CD=|\overrightarrow{CD}|=\sqrt{(-2)^2+(-8)^2+2^2}=\sqrt{72}=6\sqrt{2}
\displaystyle \cos\theta=\frac{\overrightarrow{AB}\cdot \overrightarrow{CD}}{|AB||CD|}=\frac{(\widehat{i}+4\widehat{j}-\widehat{k})\cdot(-2\widehat{i}-8\widehat{j}+2\widehat{k})}{3\sqrt{2}\times 6\sqrt{2}}
\displaystyle =\frac{1(-2)+4(-8)+(-1)(2)}{36}=\frac{-2-32-2}{36}=\frac{-36}{36}=-1
\displaystyle \cos\theta=-1 \Rightarrow \theta=180^\circ
\displaystyle \text{As the angle between } \overrightarrow{AB} \text{ and } \overrightarrow{CD} \text{ is } 180^\circ
\displaystyle \therefore \overrightarrow{AB} \text{ and } \overrightarrow{CD} \text{ are collinear}

\displaystyle \textbf{23. }\text{Find the value of } \lambda \text{ so that the lines } \frac{1-x}{3}=\frac{7y-14}{\lambda}=\frac{z-3}{2}
\displaystyle \text{and } \frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5} \text{ are at right angles. Also, find whether the lines are}
\displaystyle \text{intersecting or not.}
\displaystyle  \ \frac{1-x}{3}=\frac{7y-14}{\lambda}=\frac{z-3}{2}
\displaystyle \Rightarrow \frac{x-1}{-3}=\frac{7(y-2)}{\lambda}=\frac{z-3}{2}
\displaystyle \Rightarrow \frac{x-1}{-3}=\frac{y-2}{\lambda/7}=\frac{z-3}{2}
\displaystyle \text{Here } a_1=-3,\ b_1=\frac{\lambda}{7},\ c_1=2
\displaystyle \frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}
\displaystyle \Rightarrow \frac{-7(x-1)}{3\lambda}=\frac{y-5}{1}=\frac{-(z-6)}{5}
\displaystyle \Rightarrow \frac{x-1}{-3\lambda/7}=\frac{y-5}{1}=\frac{z-6}{-5}
\displaystyle \text{Hence } a_2=-\frac{3\lambda}{7},\ b_2=1,\ c_2=-5
\displaystyle \text{If two lines are perpendicular, then}
\displaystyle a_1a_2+b_1b_2+c_1c_2=0
\displaystyle (-3)\left(-\frac{3\lambda}{7}\right)+\left(\frac{\lambda}{7}\right)(1)+(2)(-5)=0
\displaystyle \Rightarrow \frac{9\lambda}{7}+\frac{\lambda}{7}-10=0
\displaystyle \Rightarrow \frac{10\lambda-70}{7}=0
\displaystyle \Rightarrow 10\lambda-70=0 \Rightarrow \lambda=7
\displaystyle \text{Substituting the value of } \lambda \text{ in both the equations}
\displaystyle \frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-3}{2}=A \qquad \ldots (1)
\displaystyle x-1=-3A \Rightarrow x=-3A+1
\displaystyle y-2=A \Rightarrow y=A+2
\displaystyle z-3=2A \Rightarrow z=2A+3
\displaystyle \text{Coordinates of general points on first line are } (-3A+1,\ A+2,\ 2A+3)
\displaystyle \text{Second equation: } \frac{x-1}{-3}=\frac{y-5}{1}=\frac{z-6}{-5}=B \qquad \ldots (2)
\displaystyle x-1=-3B \Rightarrow x=-3B+1
\displaystyle y-5=B \Rightarrow y=B+5
\displaystyle z-6=-5B \Rightarrow z=-5B+6
\displaystyle \text{Coordinates of general points on second line are } (-3B+1,\ B+5,\ -5B+6)
\displaystyle \text{If the lines are intersecting, then they have a common point}
\displaystyle \text{So, for some value of } A \text{ and } B,\ \text{we must have}
\displaystyle -3A+1=-3B+1 \Rightarrow A=B
\displaystyle y=A+2=B+5 \qquad \ldots (3)
\displaystyle A=B+3
\displaystyle B=B+3 \Rightarrow 0=3
\displaystyle \text{It is not possible}
\displaystyle \therefore \text{Both of these lines are not intersecting}

\displaystyle \textbf{SECTION - D}
\displaystyle \text{Question numbers 24 to 29 carry 6 marks each.}

\displaystyle \textbf{24. }\text{A tank with rectangular base and rectangular sides, open at the top is to be }
\displaystyle \text{constructed so that its depth is 2 m and volume is 8 m}^3.  \text{If building of tank costs }
\displaystyle \text{Rs } 70 \text{ per square metre for the base and } \text{Rs } 45 \text{ per square metre for the sides,}
\displaystyle \text{what is the cost of least}  \text{expensive tank?}
\displaystyle \text{Answer:}
\displaystyle  \text{Depth of tank, } h=2\text{ m}
\displaystyle \text{Volume of tank }=8\text{ m}^3
\displaystyle \text{Let length of tank be } l\text{, breadth be } b
\displaystyle \text{Volume of tank }=L\times B\times H
\displaystyle l\times b\times h=8 \Rightarrow lb(2)=8
\displaystyle l=\frac{4}{b} \qquad \ldots (i)
\displaystyle \text{Area of base }=L\times B=lb
\displaystyle \therefore \text{Cost of base }=70(lb) \qquad \left[\because \text{cost of base is } \text{Rs }70\text{ per sq mtr}\right]
\displaystyle \text{Area of 4 sides }=(H\times L)+(H\times B)+(H\times L)+(H\times B)
\displaystyle =2(H\times L+H\times B)=2h(l+b)=4(l+b) \qquad \left[\because h=2\right]
\displaystyle \text{Cost of making sides }=45[4(l+b)]=180(l+b)
\displaystyle \text{Let total cost of tank be } C(l)
\displaystyle \text{Total cost }=\text{cost of base}+\text{cost of making sides}
\displaystyle C(l)=70(lb)+180(l+b)
\displaystyle =70(4)+180\left(l+\frac{4}{l}\right) \qquad \left[\because b=\frac{4}{l}\right]
\displaystyle =280+180\left(l+\frac{4}{l}\right)
\displaystyle \text{Differentiating the equation with respect to } l
\displaystyle \frac{d}{dl}C(l)=0+180\left(1-\frac{4}{l^2}\right) \qquad \left[\because \frac{d}{dx}x^n=nx^{n-1}\right]
\displaystyle \Rightarrow \frac{d}{dl}C(l)=180\left(1-\frac{4}{l^2}\right)
\displaystyle \text{Putting } \frac{d}{dl}C(l)=0
\displaystyle 0=180\left(1-\frac{4}{l^2}\right) \Rightarrow \left(1-\frac{4}{l^2}\right)=0
\displaystyle l^2=4 \Rightarrow l=\pm 2
\displaystyle \text{Length cannot be negative}
\displaystyle \therefore \text{Length of tank is } 2 \text{ metre}
\displaystyle \text{Again differentiating } \frac{d}{dl}C(l)
\displaystyle \frac{d^2}{dl^2}C(l)=180\left[0-\frac{4(-2)}{l^3}\right]=\frac{1440}{l^3}
\displaystyle \text{Put } l=2
\displaystyle \frac{d^2}{dl^2}C(l)=\frac{1440}{2^3}>0
\displaystyle \text{So, } l=2 \text{ is a point of minima}
\displaystyle C(l) \text{ is least at } l=2
\displaystyle \text{Least cost of construction}
\displaystyle C(l)=280+180\left(2+\frac{4}{2}\right)=280+180(4)=1000
\displaystyle \text{Hence, least cost of construction is } \text{Rs }1000.

\displaystyle \textbf{25. }\text{If } A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix}, \text{ find } A^{-1}.
\displaystyle \text{Hence, solve the system of equations } x+y+z=6,\ x+2z=7,\ 3x+y+z=12.
\displaystyle \text{OR}
\displaystyle \text{Find the inverse of the following matrix using elementary operations.}
\displaystyle A=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle  \ A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix}
\displaystyle |A|=1(0-2)-1(1-6)+1(1-0)=-2+5+1=4
\displaystyle \therefore |A|=4\neq 0 \qquad \therefore A^{-1}\text{ exists}
\displaystyle \text{To find } \mathrm{Adj}\,A,\ \text{we have to find cofactors}
\displaystyle c_{11}=\left|\begin{matrix}0&2\\1&1\end{matrix}\right|=0-2=-2
\displaystyle c_{12}=-\left|\begin{matrix}1&2\\3&1\end{matrix}\right|=-(1-6)=5
\displaystyle c_{13}=\left|\begin{matrix}1&0\\3&1\end{matrix}\right|=1-0=1
\displaystyle c_{21}=-\left|\begin{matrix}1&1\\1&1\end{matrix}\right|=-(1-1)=0
\displaystyle c_{22}=\left|\begin{matrix}1&1\\3&1\end{matrix}\right|=1-3=-2
\displaystyle c_{23}=-\left|\begin{matrix}1&1\\3&1\end{matrix}\right|=-(1-3)=2
\displaystyle c_{31}=\left|\begin{matrix}1&1\\0&2\end{matrix}\right|=2-0=2
\displaystyle c_{32}=-\left|\begin{matrix}1&1\\1&2\end{matrix}\right|=-(2-1)=-1
\displaystyle c_{33}=\left|\begin{matrix}1&1\\1&0\end{matrix}\right|=0-1=-1
\displaystyle \mathrm{Adj}\,A=\begin{bmatrix}-2&5&1\\0&-2&2\\2&-1&-1\end{bmatrix}^{T}=\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{\mathrm{Adj}\,A}{|A|}=\frac{1}{4}\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}
\displaystyle \text{Equations are } x+y+z=6,\ x+2z=7,\ 3x+y+z=12
\displaystyle \text{It can be written as } AX=B
\displaystyle \text{where } A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix},\ B=\begin{bmatrix}6\\7\\12\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}
\displaystyle AX=B
\displaystyle X=A^{-1}B \qquad \left[\text{Multiplying } A^{-1} \text{ on both sides}\right]
\displaystyle X=\frac{1}{4}\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}\begin{bmatrix}6\\7\\12\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}-12+24\\30-14-12\\6+14-12\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}12\\4\\8\end{bmatrix}=\begin{bmatrix}3\\1\\2\end{bmatrix}
\displaystyle \therefore \text{Solutions of equations are } x=3,\ y=1,\ z=2
\displaystyle \text{OR}
\displaystyle \text{We know that } A=IA
\displaystyle \begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A
\displaystyle R_2\rightarrow R_2+R_1
\displaystyle \begin{bmatrix}1&2&-2\\0&5&-2\\0&-2&1\end{bmatrix}=\begin{bmatrix}1&0&0\\1&1&0\\0&0&1\end{bmatrix}A
\displaystyle R_2\rightarrow R_2+2R_3
\displaystyle \begin{bmatrix}1&2&-2\\0&1&0\\0&-2&1\end{bmatrix}=\begin{bmatrix}1&0&0\\1&1&2\\0&0&1\end{bmatrix}A
\displaystyle R_1\rightarrow R_1-2R_2
\displaystyle \begin{bmatrix}1&0&-2\\0&1&0\\0&-2&1\end{bmatrix}=\begin{bmatrix}-1&-2&-4\\1&1&2\\0&0&1\end{bmatrix}A
\displaystyle R_3\rightarrow R_3+2R_2
\displaystyle \begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-1&-2&-4\\1&1&2\\2&2&5\end{bmatrix}A
\displaystyle R_1\rightarrow R_1+2R_3
\displaystyle \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}A
\displaystyle \text{Hence } A^{-1}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}

\displaystyle \textbf{26. }\text{Prove that the curves } y^2=4x \text{ and } x^2=4y \text{ divide the area of the square bounded }
\displaystyle \text{by } x=0,\ x=4,\ y=4 \text{ and } y=0 \text{ into three equal parts.}
\displaystyle \text{OR}
\displaystyle \text{Using integration, find the area of the triangle whose vertices are } (2,3),(3,5),(4,4).
\displaystyle \text{Answer:}
\displaystyle  \text{Let } A_1,A_2,A_3 \text{ be the areas of regions } OABQO,\ OQBPO,\ OPBCO\displaystyle A_1=\int_{0}^{4}|y_1|\,dx \qquad \left[\because y_1>0 \therefore |y_1|=y_1\right]
\displaystyle (x,y_1) \text{ lies on } x^2=4y
\displaystyle y_1=\frac{x^2}{4}
\displaystyle A_1=\int_{0}^{4}\frac{x^2}{4}\,dx=\frac{1}{4}\left[\frac{x^3}{3}\right]_{0}^{4}=\frac{1}{4\times 3}\left[(4)^3-(0)^3\right]
\displaystyle =\frac{1}{4}\times \frac{64}{3}=\frac{16}{3}\text{ sq. units} \qquad \ldots (1)
\displaystyle \text{For } A_2,
\displaystyle A_2=\int_{0}^{4}(y_2-y_1)\,dx \qquad \left[\because y_2-y_1|=y_2-y_1 \text{ as } y_2-y_1>0\right]
\displaystyle (x,y_2) \text{ lies on } y^2=4x
\displaystyle y_2=2\sqrt{x}
\displaystyle A_2=\int_{0}^{4}\left[2\sqrt{x}-\frac{x^2}{4}\right]dx
\displaystyle =\left[\frac{4}{3}x^{3/2}-\frac{x^3}{12}\right]_{0}^{4} \qquad \left[\because \int x^n\,dx=\frac{x^{n+1}}{n+1}\right]
\displaystyle =\frac{4}{3}(4)^{3/2}-\frac{(4)^3}{12}
\displaystyle =\frac{64}{3}-\frac{16}{3}
\displaystyle A_2=\frac{16}{3}\text{ sq. units} \qquad \ldots (2)
\displaystyle \text{For } A_3,
\displaystyle \left[\because (x_1,y_1) \text{ lies on } y^2=4x \Rightarrow y^2=4x_1 \Rightarrow x_1=\frac{y^2}{4}\right]
\displaystyle A_3=\int_{0}^{4}|x_1|\,dy=\int_{0}^{4}\frac{y^2}{4}\,dy
\displaystyle =\frac{1}{4}\left[\frac{y^3}{3}\right]_{0}^{4}
\displaystyle =\frac{1}{4}\times \frac{64}{3}=\frac{16}{3} \qquad \ldots (3)
\displaystyle \text{From (1), (2) \& (3)}
\displaystyle A_1=A_2=A_3=\frac{16}{3}\text{ sq. units}
\displaystyle \text{Hence proved.}
\displaystyle \text{OR}
\displaystyle \text{Let } A(2,3),\ B(3,5) \text{ and } C(4,4) \text{ be the vertices of } \Delta ABC\displaystyle \text{Equation of line } AB \text{ is}
\displaystyle \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\Rightarrow \frac{y-3}{5-3}=\frac{x-2}{3-2}
\displaystyle \Rightarrow \frac{y-3}{2}=\frac{x-2}{1}
\displaystyle \Rightarrow y-3=2(x-2)\Rightarrow y=2x-1 \qquad \ldots (1)
\displaystyle \text{Equation of line } BC \text{ is}
\displaystyle \frac{y-5}{4-5}=\frac{x-3}{4-3}\Rightarrow \frac{y-5}{-1}=\frac{x-3}{1}
\displaystyle \Rightarrow y-5=-x+3
\displaystyle \Rightarrow y=-x+8 \qquad \ldots (2)
\displaystyle \text{Equation of line } CA \text{ is}
\displaystyle \frac{y-4}{3-4}=\frac{x-4}{2-4}\Rightarrow \frac{y-4}{-1}=\frac{x-4}{-2}
\displaystyle \Rightarrow 2y-8=x-4\Rightarrow 2y=x+4
\displaystyle \Rightarrow y=\frac{x+4}{2} \qquad \ldots (3)
\displaystyle \text{Let area of triangle } ABC \text{ be } A
\displaystyle A=\int_{2}^{3}|y_1|\,dx+\int_{3}^{4}|y_2|\,dx-\int_{2}^{4}|y_3|\,dx
\displaystyle =\int_{2}^{3}(2x-1)\,dx+\int_{3}^{4}(-x+8)\,dx-\int_{2}^{4}\left(\frac{x+4}{2}\right)dx \qquad \left[\text{From (1), (2) \& (3)}\right]
\displaystyle =\left[x^2-x\right]_{2}^{3}+\left[-\frac{x^2}{2}+8x\right]_{3}^{4}-\frac{1}{2}\left[\frac{x^2}{2}+4x\right]_{2}^{4}
\displaystyle =\left[(3)^2-3-(2)^2+2\right]+\left[-\frac{(4)^2}{2}+32+\frac{9}{2}-24\right]-\frac{1}{2}\left[\frac{16}{2}+16-\frac{4}{2}-8\right]
\displaystyle =9-3-4+2-\frac{16}{2}+32+\frac{9}{2}-24-\frac{1}{2}\times 14
\displaystyle =\frac{3}{2}\text{ sq. units}
\displaystyle \text{Area of required triangle }=\frac{3}{2}\text{ sq. units}

\displaystyle \textbf{27. }\text{A manufacturer has employed 5 skilled men and 10 semi-skilled men}
\displaystyle \text{and makes two models A and B of an article. The making of one item of model A}
\displaystyle \text{requires 2 hours work by a skilled man and 2 hours work by a semi-skilled man.}
\displaystyle \text{One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man.}
\displaystyle \text{No man is expected to work more than 8 hours per day. The manufacturer's profit}
\displaystyle \text{on an item of model A is } \text{Rs } 15 \text{ and on an item of model B is } \text{Rs } 10. \text{How many}
\displaystyle \text{items of each model should be made per day in order to maximize daily profit? }
\displaystyle \text{Formulate the above LPP and solve it graphically and find the maximum profit.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let number of articles of model A be } x \text{ and number of articles of model B be } y\displaystyle \text{Mathematical formulation of the given linear programming problem is}
\displaystyle \text{Max } Z=15x+10y
\displaystyle \text{Subject to}
\displaystyle 2x+y\leq 40,\quad 2x+3y\leq 80
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{Let}
\displaystyle 2x+y=40
\displaystyle \begin{array}{|c|c|c|}\hline x&0&20\\ \hline y&40&0\\ \hline \end{array}
\displaystyle 2x+3y=80
\displaystyle \begin{array}{|c|c|c|}\hline x&0&40\\ \hline y&\frac{80}{3}&0\\ \hline \end{array}
\displaystyle \text{Intersection point of } 2x+y=40 \text{ and } 2x+3y=80 \text{ is } B(10,20)
\displaystyle \text{Corner points are } A\left(0,\frac{80}{3}\right),\ B(10,20),\ O(0,0) \text{ and } C(20,0)
\displaystyle \begin{array}{|c|c|}\hline \text{Corner point}&\text{Value of } Z=15x+10y\\ \hline A\left(0,\frac{80}{3}\right)&Z=\frac{800}{3}\\ \hline B(10,20)&Z=350\\ \hline O(0,0)&Z=0\\ \hline C(20,0)&Z=300\\ \hline \end{array}
\displaystyle \text{Maximum value of } Z \text{ is } \text{Rs } 350 \text{ which is attained at } B(10,20)
\displaystyle \therefore \text{Maximum profit is } \text{Rs } 350 \text{ obtained when 10 articles of model A and 20 articles of model B are produced}

\displaystyle \textbf{28. }\text{Find the vector and Cartesian equations of the plane passing through } (2,2,-1),
\displaystyle (3,4,2) \text{ and } (7,0,6). \text{Also, find the vector equation of a plane } \text{passing through}
\displaystyle (4,3,1) \text{ and parallel to the plane obtained above.}
\displaystyle \text{OR}
\displaystyle \text{Find the vector equation of the plane that contains the lines}
\displaystyle \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-\widehat{k}) \text{ and the point } (-1,3,-4). \text{Also, find the length of the}
\displaystyle \text{perpendicular drawn from the point } (2,1,4)  \text{to the plane thus obtained.}
\displaystyle \text{Answer:}
\displaystyle \textbf{28. }\text{The given points are } A(2,2,-1),\ B(3,4,2)\ \text{and}\ C(7,0,6)
\displaystyle \text{Let } \overrightarrow{a}=2\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{b}=3\widehat{i}+4\widehat{j}+2\widehat{k}\ \text{and}\ \overrightarrow{c}=7\widehat{i}+6\widehat{k}\displaystyle \text{Normal to the vector } \overrightarrow{n}\ \text{given by } \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}
\displaystyle \overrightarrow{AB}=(3\widehat{i}+4\widehat{j}+2\widehat{k})-(2\widehat{i}+2\widehat{j}-\widehat{k})=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{AC}=(7\widehat{i}+0\widehat{j}+6\widehat{k})-(2\widehat{i}+2\widehat{j}-\widehat{k})=5\widehat{i}-2\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&3\\5&-2&7\end{vmatrix}
\displaystyle =(14+6)\widehat{i}-(7-15)\widehat{j}+(-2-10)\widehat{k}
\displaystyle \overrightarrow{n}=20\widehat{i}+8\widehat{j}-12\widehat{k}
\displaystyle \text{The vector equation of the required plane is } \overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}\cdot \overrightarrow{n}
\displaystyle \overrightarrow{r}\cdot(20\widehat{i}+8\widehat{j}-12\widehat{k})=(2\widehat{i}+2\widehat{j}-\widehat{k})\cdot(20\widehat{i}+8\widehat{j}-12\widehat{k})
\displaystyle \Rightarrow \overrightarrow{r}\cdot(20\widehat{i}+8\widehat{j}-12\widehat{k})=68
\displaystyle \Rightarrow \overrightarrow{r}\cdot(5\widehat{i}+2\widehat{j}-3\widehat{k})=17
\displaystyle \text{The Cartesian equation of the plane is}
\displaystyle 5x+2y-3z=17
\displaystyle \text{Also required plane is parallel to the above plane}
\displaystyle \therefore \text{Required plane is } 5x+2y-3z+\lambda=0
\displaystyle \text{It passes through } (4,3,1)
\displaystyle 5(4)+2(3)-3(1)+\lambda=0
\displaystyle 20+6-3+\lambda=0 \Rightarrow \lambda=-23
\displaystyle \text{Equation of plane: } 5x+2y-3z-23=0
\displaystyle \Rightarrow 5x+2y-3z=23
\displaystyle \text{Vector form of equation of plane is } \overrightarrow{r}\cdot(5\widehat{i}+2\widehat{j}-3\widehat{k})=23
\displaystyle \text{OR}
\displaystyle \text{Let the vector equation of required plane is } (\overrightarrow{r}-\overrightarrow{a})\cdot \overrightarrow{n}=0
\displaystyle \text{Plane contains the line } \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-\widehat{k})
\displaystyle \therefore A(1,1,0)\ \text{and direction vector } \overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k}
\displaystyle \text{Plane passes through } A(1,1,0)\ \text{and } B(-1,3,-4)
\displaystyle \overrightarrow{AB}=(-1+1)\widehat{i}+(3-1)\widehat{j}+(-4-0)\widehat{k}=0\widehat{i}+2\widehat{j}-4\widehat{k}
\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\0&2&-4\\1&2&-1\end{vmatrix}

\displaystyle =\widehat{i}(2-(-8))-\widehat{j}(0-(-4))+\widehat{k}(0-2)
\displaystyle \overrightarrow{n}=10\widehat{i}-4\widehat{j}-2\widehat{k}
\displaystyle \text{Equation of plane is } (\overrightarrow{r}-\overrightarrow{a})\cdot \overrightarrow{n}=0
\displaystyle \overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}\cdot \overrightarrow{n}
\displaystyle \overrightarrow{r}\cdot(10\widehat{i}-4\widehat{j}-2\widehat{k})=(-\widehat{i}+3\widehat{j}-4\widehat{k})\cdot(10\widehat{i}-4\widehat{j}-2\widehat{k})
\displaystyle \Rightarrow \overrightarrow{r}\cdot(10\widehat{i}-4\widehat{j}-2\widehat{k})=-10-12+8=-14
\displaystyle \Rightarrow 10x-4y-2z=-14
\displaystyle \Rightarrow 5x-2y-z=-7
\displaystyle \text{Cartesian form of equation is } x-y-z=0
\displaystyle \text{Length of perpendicular from } P(2,1,4)\ \text{to the plane } x-y-z=0
\displaystyle d=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}
\displaystyle d=\frac{|2(1)+1(-1)+4(-1)+0|}{\sqrt{(1)^2+(-1)^2+(-1)^2}}=\frac{|2-1-4|}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}
\displaystyle \therefore d=\sqrt{3}\text{ units}

\displaystyle \textbf{29. }\text{Two cards are drawn simultaneously (or successively without replacement)}
\displaystyle \text{from a well shuffled pack of 52 cards. Find the mean and variance of the number of kings.}
\displaystyle \text{Answer:}
\displaystyle  \text{No. of trials }=2
\displaystyle X\ (\text{No. of king cards})=0,1,2
\displaystyle \text{No. of king cards }=4 \text{ and no. of other cards }=48
\displaystyle P(X=0)=P(\text{Two cards other than king})
\displaystyle =\frac{{}^{48}C_{2}}{{}^{52}C_{2}}=\frac{1128}{1326}
\displaystyle P(X=1)=P(\text{One king and one other card})
\displaystyle =\frac{{}^{4}C_{1}\times {}^{48}C_{1}}{{}^{52}C_{2}}
\displaystyle =\frac{192}{1326}
\displaystyle P(X=2)=P(\text{Two king cards})
\displaystyle =\frac{{}^{4}C_{2}}{{}^{52}C_{2}}=\frac{6}{1326}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline X&0&1&2&\text{Total}\\ \hline P(X)&\frac{1128}{1326}&\frac{192}{1326}&\frac{6}{1326}&1\\ \hline XP(X)&0&\frac{192}{1326}&\frac{12}{1326}&\frac{204}{1326}\\ \hline X^2P(X)&0&\frac{192}{1326}&\frac{24}{1326}&\frac{216}{1326}=\frac{36}{221}\\ \hline \end{array}
\displaystyle \text{Mean }=\sum XP(X)=\frac{204}{1326}=\frac{2}{13}
\displaystyle \text{Variance }=\sum X^2P(X)-\left(\sum XP(X)\right)^2
\displaystyle =\frac{36}{221}-\left(\frac{34}{221}\right)^2=\frac{1}{221}\left[36-\frac{34^2}{221}\right]
\displaystyle =\frac{1}{221}\left[\frac{36\times 221-1156}{221}\right]
\displaystyle =\frac{6800}{(221)^2}
\displaystyle \therefore \text{Variance }=\frac{6800}{(221)^2}=0.139

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