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

\displaystyle \textbf{General Instructions:}
\displaystyle \text{Read the following instructions very carefully and strictly follow them:}
\displaystyle \text{(i) This question paper comprises four Sections A, B, C and D. This question}
\displaystyle \text{paper carries 36 questions. All questions are compulsory.}
\displaystyle \text{(ii) Section A: Questions no. 1 to 20 comprises of 20 questions of 1 mark each.}
\displaystyle \text{(iii) Section B: Questions no. 21 to 26 comprises of 6 questions of 2 marks each.}
\displaystyle \text{(iv) Section C: Questions no. 27 to 32 comprises of 6 questions of 4 marks each.}
\displaystyle \text{(v) Section D: Questions no. 33 to 36 comprises of 4 questions of 6 marks each.}
\displaystyle \text{(vi) There is no overall choice in the question paper. However, an internal choice}
\displaystyle \text{has been provided in 3 questions of one mark, 2 questions of two marks, 2}
\displaystyle \text{questions of four marks and 2 questions of six marks. Only one of the choices}
\displaystyle \text{in such questions have to be attempted.}
\displaystyle \text{(vii) In addition to this, separate instructions are given with each section and}
\displaystyle \text{question, wherever necessary.}
\displaystyle \text{(viii) Use of calculators is not permitted.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question numbers 1 to 20 carry 1 mark each.}
\displaystyle \text{Question numbers 1 to 10 are multiple choice type questions. Select the }  \text{correct option.}

\displaystyle \textbf{1. } \text{The area of a triangle formed by vertices O, A and B, where}
\displaystyle \overrightarrow{OA} = \widehat{i} + 2\widehat{j} + 3\widehat{k} \text{ and } \overrightarrow{OB} = -3\widehat{i} - 2\widehat{j} + \widehat{k} \text{ is}
\displaystyle \text{(a) } 3\sqrt{5} \text{ sq. units} \quad \text{(b) } 5\sqrt{5} \text{ sq. units}
\displaystyle \text{(c) } 6\sqrt{5} \text{ sq. units} \quad \text{(d) } 4 \text{ sq. units}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}\ \overrightarrow{OA}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{OB}=-3\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{OA}\times\overrightarrow{OB}= \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix}
\displaystyle =\widehat{i}\{2(1)-3(-2)\}-\widehat{j}\{1(1)-3(-3)\}+\widehat{k}\{1(-2)-2(-3)\}
\displaystyle =\widehat{i}(2-(-6))-\widehat{j}(1-(-9))+\widehat{k}(-2-(-6))
\displaystyle =8\widehat{i}-10\widehat{j}+4\widehat{k}
\displaystyle |\overrightarrow{OA}\times\overrightarrow{OB}|= \sqrt{a^{2}+b^{2}+c^{2}}
\displaystyle =\sqrt{8^{2}+(-10)^{2}+(4)^{2}}
\displaystyle =\sqrt{64+100+16}
\displaystyle =\sqrt{180}=6\sqrt{5}
\displaystyle \therefore\ \text{Area of }\triangle OAB= \frac{1}{2}\left|\overrightarrow{OA}\times\overrightarrow{OB}\right|
\displaystyle \text{Area of }\triangle OAB= \frac{1}{2}\times6\sqrt{5}=3\sqrt{5}\ \text{sq. units}

\displaystyle \textbf{2. } \text{If } \cos\left(\sin^{-1}\left(\frac{2}{\sqrt{5}}\right) + \cos^{-1} x\right) = 0, \text{ then } x \text{ is equal to}
\displaystyle \text{(a) } \frac{1}{\sqrt{5}} \quad \text{(b) } -\frac{2}{\sqrt{5}}
\displaystyle \text{(c) } \frac{2}{\sqrt{5}} \quad \text{(d) } 1
\displaystyle \text{Answer:}
\displaystyle \text{(c)}\ \cos\left(\sin^{-1}\frac{2}{\sqrt{5}}+\cos^{-1}x\right)=0
\displaystyle \Rightarrow\ \sin^{-1}\frac{2}{\sqrt{5}}+\cos^{-1}x=\cos^{-1}0
\displaystyle \Rightarrow\ \sin^{-1}\frac{2}{\sqrt{5}}+\cos^{-1}x=\frac{\pi}{2}
\displaystyle \Rightarrow\ \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}\frac{2}{\sqrt{5}}
\displaystyle \text{Taking cos on both sides}
\displaystyle \cos(\cos^{-1}x)=\cos\left(\frac{\pi}{2}-\sin^{-1}\frac{2}{\sqrt{5}}\right)
\displaystyle \Rightarrow\ x=\sin\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)
\displaystyle \Rightarrow\ x=\frac{2}{\sqrt{5}}

\displaystyle \textbf{3. } \text{The interval in which the function } f \text{ given by } f(x) = x^{2} e^{-x} \text{ is strictly increasing is}
\displaystyle \text{(a) } (-\infty, \infty) \quad \text{(b) } (-\infty, 0)
\displaystyle \text{(c) } (2, \infty) \quad \text{(d) } (0, 2)
\displaystyle \text{Answer:}
\displaystyle \text{(d)}\ f(x)=x^{2}e^{-x}
\displaystyle f'(x)=x^{2}\frac{d}{dx}(e^{-x})+e^{-x}\frac{d}{dx}(x^{2})
\displaystyle =x^{2}(-e^{-x})+e^{-x}(2x)
\displaystyle =-x^{2}e^{-x}+2xe^{-x}
\displaystyle =xe^{-x}(2-x)
\displaystyle \text{Putting }f'(x)=0
\displaystyle xe^{-x}(2-x)=0
\displaystyle e^{-x}\text{ is always positive for all }x\in R
\displaystyle \Rightarrow\ x=0\ \text{and}\ x=2\displaystyle \text{Case I: for }-\infty<x<0,\ f'(x)<0\ \Rightarrow\ f(x)\ \text{decreasing}
\displaystyle \text{Case II: for }0<x<2,\ f'(x)>0\ \Rightarrow\ f(x)\ \text{increasing}
\displaystyle \text{Case III: for }2<x<\infty,\ f'(x)<0\ \Rightarrow\ f(x)\ \text{decreasing}
\displaystyle \therefore\ f(x)\ \text{is strictly increasing in }(0,2)

\displaystyle \textbf{4. } \text{The function } f(x) = \frac{x - 1}{x(x^{2} - 1)} \text{ is discontinuous at}
\displaystyle \text{(a) exactly one point} \quad \text{(b) exactly two points}
\displaystyle \text{(c) exactly three points} \quad \text{(d) no point}
\displaystyle \text{Answer:}
\displaystyle \text{(c)}\ f(x)=\frac{x-1}{x(x^{2}-1)}
\displaystyle =\frac{x-1}{x(x-1)(x+1)}
\displaystyle \Rightarrow\ f(x)=\frac{1}{x(x+1)}
\displaystyle \text{Graph will be discontinuous at three points i.e. }x=0,-1,1

\displaystyle \textbf{5. } \text{The function } f : R \to [-1, 1] \text{ defined by } f(x) = \cos x \text{ is}
\displaystyle \text{(a) both one-one and onto} \quad \text{(b) not one-one, but onto}
\displaystyle \text{(c) one-one, but not onto} \quad \text{(d) neither one-one, nor onto}
\displaystyle \text{Answer:}
\displaystyle \text{(b)}\ f(x)=\cos x
\displaystyle f:R\rightarrow[-1,1]
\displaystyle \text{If }x_{1}=\frac{\pi}{2}
\displaystyle f(x_{1})=f\left(\frac{\pi}{2}\right)=\cos\frac{\pi}{2}=0
\displaystyle \text{If }x_{2}=-\frac{\pi}{2}
\displaystyle f(x_{2})=f\left(-\frac{\pi}{2}\right)=\cos\left(-\frac{\pi}{2}\right)=0
\displaystyle \left[\cos(-\theta)=\cos\theta\right]
\displaystyle f(x_{1})=f(x_{2})
\displaystyle \text{But }x_{1}\ne x_{2}
\displaystyle \therefore\ f(x)\ \text{is not one-one}
\displaystyle \text{Also range of }\cos x\text{ is }[-1,1]
\displaystyle \text{Range}=\text{Co-domain}
\displaystyle \therefore\ f\ \text{is onto}

\displaystyle \textbf{6. } \text{The coordinates of the foot of the perpendicular drawn from the point }
\displaystyle (2, -3, 4) \text{ on the y-axis is}
\displaystyle \text{(a) } (2, 3, 4) \quad \text{(b) } (-2, -3, -4)
\displaystyle \text{(c) } (0, -3, 0) \quad \text{(d) } (2, 0, 4)
\displaystyle \text{Answer:}
\displaystyle \text{(c)}\ \text{As foot of perpendicular from }(2,-3,4)
\displaystyle \text{lies on }y\text{-axis}
\displaystyle \Rightarrow\ x=0\ \text{and}\ z=0
\displaystyle \therefore\ \text{Foot of perpendicular is }(0,-3,0)

\displaystyle \textbf{7. } \text{The relation } R \text{ in the set } \{1, 2, 3\} \text{ given by } R = \{(1, 2), (2, 1), (1, 1)\} \text{ is}
\displaystyle \text{(a) symmetric and transitive, but not reflexive}
\displaystyle \text{(b) reflexive and symmetric, but not transitive}
\displaystyle \text{(c) symmetric, but neither reflexive nor transitive}
\displaystyle \text{(d) an equivalence relation}
\displaystyle \text{Answer:}
\displaystyle \text{(a)}\ \text{Let }A=\{1,2,3\}
\displaystyle R=\{(1,2),(2,1),(1,1)\}
\displaystyle \text{(i) Relation }R\text{ is reflexive if }(a,a)\in R\ \forall a\in A
\displaystyle \text{But }(2,2)\text{ and }(3,3)\text{ are missing}
\displaystyle \therefore\ R\ \text{is not reflexive}
\displaystyle \text{(ii) Relation }R\text{ is symmetric if }(a,b)\in R\Rightarrow(b,a)\in R
\displaystyle \text{Here }(1,2)\text{ and }(2,1)\in R
\displaystyle \therefore\ R\ \text{is symmetric}
\displaystyle \text{(iii) Relation }R\text{ is transitive if }(a,b),(b,c)\in R\Rightarrow(a,c)\in R
\displaystyle (1,2)\in R,\ (2,1)\in R
\displaystyle \Rightarrow\ (1,1)\in R
\displaystyle \text{Hence, }R\ \text{is transitive}

\displaystyle \textbf{8. } \text{The angle between the vectors } \hat{i} - \hat{j} \text{ and } \hat{j} - \hat{k} \text{ is}
\displaystyle \text{(a) } -\frac{\pi}{3} \quad \text{(b) } 0
\displaystyle \text{(c) } \frac{\pi}{3} \quad \text{(d) } \frac{2\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \text{(d)}\ \overrightarrow{a}=i-j,\ \overrightarrow{b}=j-k
\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}
\displaystyle =\frac{(i-j)\cdot(j-k)}{\sqrt{1^{2}+(-1)^{2}}\ \sqrt{1^{2}+(-1)^{2}}}
\displaystyle =\frac{(i\cdot j-i\cdot k-j\cdot j+j\cdot k)}{\sqrt{2}\cdot\sqrt{2}}
\displaystyle =\frac{(0-0-1+0)}{2}=-\frac{1}{2}
\displaystyle \Rightarrow\ \theta=\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}

\displaystyle \textbf{9. } \text{If } A \text{ is a non-singular square matrix of order 3 such that } A^{2} = 3A, \text{ then } |A| \text{ is}
\displaystyle \text{(a) } -3 \quad \text{(b) } 3 \quad \text{(c) } 9 \quad \text{(d) } 27
\displaystyle \text{Answer:}
\displaystyle \text{(d)}\ A^{2}=3A
\displaystyle |A^{2}|=|3A|
\displaystyle |A|^{2}=3^{3}|A|\ \text{(Order of matrix }=3)
\displaystyle |A|^{2}=27|A|
\displaystyle \Rightarrow\ |A|=27

\displaystyle \textbf{10. } \text{If } |\vec{a}| = 4 \text{ and } -3 \leq \lambda \leq 2, \text{ then } |\lambda \vec{a}| \text{ lies in}
\displaystyle \text{(a) } [0, 12] \quad \text{(b) } [2, 3]
\displaystyle \text{(c) } [8, 12] \quad \text{(d) } [-12, 8]
\displaystyle \text{Answer:}
\displaystyle \text{(a)}\ \text{Given }|\overrightarrow{a}|=4\ \text{and }-3\le\lambda\le3
\displaystyle \text{Let }\overrightarrow{a}=x_{1}i+x_{2}j+x_{3}k
\displaystyle |\overrightarrow{a}|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}=4
\displaystyle |\lambda\overrightarrow{a}|=|\lambda|\ |\overrightarrow{a}|
\displaystyle =|\lambda|\cdot4
\displaystyle \text{Since }-3\le\lambda\le3\Rightarrow0\le|\lambda|\le3
\displaystyle \Rightarrow\ 0\le|\lambda\overrightarrow{a}|\le12
\displaystyle \text{Hence, }|\lambda\overrightarrow{a}|\in[0,12]

\displaystyle \textbf{Fill in the blanks in question numbers 11 to 15.}

\displaystyle \textbf{11. } \text{If the radius of the circle is increasing at the rate of } 0.5 \text{ cm/s, then the rate of}
\displaystyle \text{increase of its circumference is } \underline{\hspace{2cm}} \text{.}
\displaystyle \text{Answer:}
\displaystyle  \ [3.14\ \text{cm/s}]\ \frac{dr}{dt}=0.5\ \text{cm/s}
\displaystyle \text{Circumference }C=2\pi r
\displaystyle \frac{dC}{dt}=2\pi\frac{dr}{dt} =2\pi(0.5)=\pi=3.14\ \text{cm/s}

\displaystyle \textbf{12. } \text{If } \left|\begin{matrix} 2x & -9 \\ -2 & x \end{matrix}\right| =  \left|\begin{matrix} -4 & 8 \\ 1 & -2 \end{matrix}\right| \text{, then value of } x \text{ is }   \underline{\hspace{2cm}} \text{.}
\displaystyle \text{Answer:}
\displaystyle  \ \begin{vmatrix}2x & -9 \\ -2 & x\end{vmatrix}= \begin{vmatrix}-4 & 8 \\ 1 & -2\end{vmatrix}
\displaystyle \Rightarrow\ (2x)(x)-(-9)(-2)=(-4)(-2)-(8)(1)
\displaystyle \Rightarrow\ 2x^{2}-18=8-8
\displaystyle \Rightarrow\ 2x^{2}-18=0
\displaystyle \Rightarrow\ 2x^{2}=18
\displaystyle \Rightarrow\ x^{2}=9
\displaystyle \Rightarrow\ x=\pm3
\displaystyle \text{Value of }x\text{ is }-3,3

\displaystyle \textbf{13. } \text{The corner points of the feasible region of an LPP are } (0, 0), (0, 8), (2, 7), (5, 4) \text{ and } (6, 0).
\displaystyle  \text{The maximum profit}   P = 3x + 2y \text{ occurs at the point } \underline{\hspace{2cm}} \text{.}
\displaystyle \text{Answer:}
\displaystyle  \ [(5,4)]\ P=3x+2y
\displaystyle \text{At }(0,0),\ P=3(0)+2(0)=0
\displaystyle \text{At }(0,8),\ P=3(0)+2(8)=16
\displaystyle \text{At }(2,7),\ P=3(2)+2(7)=6+14=20
\displaystyle \text{At }(5,4),\ P=3(5)+2(4)=15+8=23\ (\text{Max})
\displaystyle \text{At }(6,0),\ P=3(6)+2(0)=18
\displaystyle \text{Max. profit occurs at }(5,4)

\displaystyle \textbf{14. } \text{The range of the principal value branch of the function}   y = \sec^{-1} x \text{ is } \underline{\hspace{2cm}} \text{.}
\displaystyle \text{OR}
\displaystyle \text{The principal value of } \cos^{-1}\left(-\frac{1}{2}\right) \text{ is }   \underline{\hspace{2cm}} \text{.}
\displaystyle \text{Answer:}
\displaystyle \ [0,\pi]\ \left[-\frac{\pi}{2}\right]
\displaystyle \text{OR}
\displaystyle \left[\frac{2\pi}{3}\right]
\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\pi-\cos^{-1}\left(\frac{1}{2}\right)
\displaystyle \left[\cos^{-1}(-\theta)=\pi-\cos^{-1}(\theta)\right]  =\pi-\frac{\pi}{3}=\frac{2\pi}{3}

\displaystyle \textbf{15. } \text{The distance between parallel planes } 2x + y - 2z - 6 = 0
\displaystyle \text{ and } 4x + 2y - 4z = 0 \text{ is } \underline{\hspace{2cm}} \text{ units.}
\displaystyle \text{OR}
\displaystyle \text{If } P(1, 0, -3) \text{ is the foot of the perpendicular from the origin to the plane, then}
\displaystyle \text{the cartesian equation of the plane is}   \underline{\hspace{2cm}} \text{.}
\displaystyle \ [2]\ d=\frac{|d_{1}-d_{2}|}{\sqrt{a^{2}+b^{2}+c^{2}}}
\displaystyle 2x+y-2z-6=0\ \Rightarrow\ d_{1}=-6
\displaystyle 4x+2y-4z=0\ \Rightarrow\ d_{2}=0
\displaystyle \text{Also }a=2,\ b=1,\ c=-2
\displaystyle d=\frac{|-6-0|}{\sqrt{2^{2}+1^{2}+(-2)^{2}}}
\displaystyle =\frac{6}{\sqrt{4+1+4}}=\frac{6}{\sqrt{9}}=\frac{6}{3}=2
\displaystyle \therefore\ d=2
\displaystyle \text{OR}
\displaystyle [x-3z=10]
\displaystyle P(1,0,-3)
\displaystyle \text{Since }P(1,0,-3)\text{ is foot of perpendicular from origin}
\displaystyle \overrightarrow{OP}=i-3k
\displaystyle d=|\overrightarrow{OP}|=\sqrt{1^{2}+(-3)^{2}}=\sqrt{10}
\displaystyle \widehat{n}=\frac{i-3k}{\sqrt{10}}
\displaystyle \text{Equation of plane: }\overrightarrow{r}\cdot\widehat{n}=d
\displaystyle \overrightarrow{r}\cdot\frac{i-3k}{\sqrt{10}}=\sqrt{10}
\displaystyle \Rightarrow\ x-3z=10

\displaystyle \textbf{Question numbers 16 to 20 are very short answer type questions.}

\displaystyle \textbf{16. } \text{Evaluate: } \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos^{2} x \, dx
\displaystyle \text{Answer:}
\displaystyle \ I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\cos^{2}x\ dx
\displaystyle \text{Let }f(x)=x\cos^{2}x
\displaystyle f(-x)=(-x)\cos^{2}(-x)
\displaystyle =-x\cos^{2}x
\displaystyle \left[\cos(-\theta)=\cos\theta\right]
\displaystyle \therefore\ f(x)\text{ is odd function}
\displaystyle \text{For odd function }\int_{-a}^{a}f(x)\ dx=0
\displaystyle \therefore\ I=0

\displaystyle \textbf{17. } \text{Find the coordinates of the point where the line }   \frac{x - 1}{3} = \frac{y + 4}{7} = \frac{z + 4}{2} \text{ cuts} \\ \text{the xy-plane.}
\displaystyle \text{Answer:}
\displaystyle \ \text{Let }\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}=\lambda
\displaystyle \Rightarrow\ x=3\lambda+1,\ y=7\lambda-4,\ z=2\lambda-4\qquad\text{(1)}
\displaystyle \text{For }xy\text{-plane}
\displaystyle z=0
\displaystyle \Rightarrow\ 2\lambda-4=0\Rightarrow2\lambda=4\Rightarrow\lambda=2
\displaystyle \text{From (1)}
\displaystyle x=3(2)+1=7
\displaystyle y=7(2)-4=10
\displaystyle z=2(2)-4=0
\displaystyle \therefore\ \text{Line cuts the plane at }(7,10,0)

\displaystyle \textbf{18. } \text{Find the value of } k \text{, so that the function}   f(x) = \left\{\begin{matrix} kx^{2} + 5 & \text{if } x \leq 1 \\ 2 & \text{if } x > 1 \end{matrix}\right.
\displaystyle \text{is continuous at } x = 1 \text{.}
\displaystyle \text{Answer:}
\displaystyle \ f(x)=\begin{cases} kx^{2}+5 & \text{if }x\leq1 \\ 2 & \text{if }x>1 \end{cases}
\displaystyle \text{L.H.L.}=\lim_{x\to1^{-}}(kx^{2}+5)
\displaystyle \text{Let }x=1-h
\displaystyle \Rightarrow\ \lim_{h\to0}k(1-h)^{2}+5
\displaystyle =\lim_{h\to0}k(1-0)^{2}+5
\displaystyle =k+5
\displaystyle \text{R.H.L.}=\lim_{x\to1^{+}}2=2
\displaystyle \text{As function is continuous at }x=1
\displaystyle \text{L.H.L.}=\text{R.H.L.}
\displaystyle \Rightarrow\ k+5=2
\displaystyle \Rightarrow\ k=2-5=-3

\displaystyle \textbf{19. } \text{Find the integrating factor of the differential equation}   x \frac{dy}{dx} = 2x^{2} + y \text{.}
\displaystyle \text{Answer:}
\displaystyle \ x\frac{dy}{dx}=2x^{2}+y
\displaystyle \Rightarrow\ \frac{dy}{dx}=2x+\frac{y}{x}
\displaystyle \Rightarrow\ \frac{dy}{dx}-\frac{y}{x}=2x
\displaystyle \text{Here }P=-\frac{1}{x},\ Q=2x
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int-\frac{1}{x}\,dx}  =e^{-\log x}  =\frac{1}{x}

\displaystyle \textbf{20. } \text{Differentiate } \sec^{2}(x^{2}) \text{ with respect to } x^{2} \text{.}
\displaystyle \text{OR}
\displaystyle \text{If } y = f(x^{2}) \text{ and } f'(x) = e^{\sqrt{x}} \text{, then find } \frac{dy}{dx} \text{.}
\displaystyle \text{Answer:}
\displaystyle   \text{Let }f(x)=\sec(x^{2})
\displaystyle \text{Differentiate with respect to }x,\ \text{we get}
\displaystyle \frac{d}{dx}f(x)=\sec(x^{2})\tan(x^{2})\frac{d}{dx}(x^{2})
\displaystyle =\sec(x^{2})\tan(x^{2})\cdot2x
\displaystyle =2x\sec(x^{2})\tan(x^{2})
\displaystyle \text{OR}
\displaystyle f'(x)=e^{\sqrt{x}}
\displaystyle \Rightarrow\ f'(x^{2})=e^{\sqrt{x^{2}}}=e^{x}
\displaystyle \text{Let }y=f(x^{2})
\displaystyle \Rightarrow\ \frac{dy}{dx}=f'(x^{2})\frac{d}{dx}(x^{2})
\displaystyle \Rightarrow\ \frac{dy}{dx}=e^{x}\cdot2x=2xe^{x}

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question numbers 21 to 26 carry 2 marks each.}

\displaystyle \textbf{21. } \text{Find a vector } \overrightarrow{r} \text{ equally inclined to the three axes and whose} \text{magnitude is }
\displaystyle 3\sqrt{3} \text{ units.}
\displaystyle \text{OR}
\displaystyle \text{Find the angle between unit vectors } \vec{a} \text{ and } \vec{b} \text{ so that}   \sqrt{3}\vec{a} - \vec{b} \text{ is also a unit vector.}
\displaystyle \text{Answer:}
\displaystyle \ \overrightarrow{r}\ \text{is equally inclined to three axes}
\displaystyle \text{Therefore, direction cosines will be equal}
\displaystyle \cos\alpha=\cos\beta=\cos\gamma=l
\displaystyle \text{We know that}
\displaystyle l^{2}+m^{2}+n^{2}=1
\displaystyle \Rightarrow\ l^{2}+l^{2}+l^{2}=1\qquad\text{[from (1)]}
\displaystyle \Rightarrow\ 3l^{2}=1\Rightarrow l^{2}=\frac{1}{3}
\displaystyle \Rightarrow\ l=\frac{1}{\sqrt{3}}
\displaystyle |\overrightarrow{r}|=3\sqrt{3}\ \text{units}
\displaystyle \overrightarrow{r}=|\overrightarrow{r}|(l\widehat{i}+m\widehat{j}+n\widehat{k})
\displaystyle =3\sqrt{3}\left(\frac{1}{\sqrt{3}}\widehat{i}+\frac{1}{\sqrt{3}}\widehat{j}+\frac{1}{\sqrt{3}}\widehat{k}\right)
\displaystyle =3(\widehat{i}\pm\widehat{j}\pm\widehat{k})
\displaystyle \text{OR}
\displaystyle |\overrightarrow{a}|=1,\ |\overrightarrow{b}|=1\qquad \text{(}\because\ \overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are unit vectors)}
\displaystyle \text{Since }(\sqrt{3}\overrightarrow{a}-\overrightarrow{b})\text{ is unit vector}
\displaystyle |\sqrt{3}\overrightarrow{a}-\overrightarrow{b}|=1
\displaystyle \Rightarrow\ (\sqrt{3}\overrightarrow{a}-\overrightarrow{b})\cdot (\sqrt{3}\overrightarrow{a}-\overrightarrow{b})=1
\displaystyle \Rightarrow\ 3\overrightarrow{a}\cdot\overrightarrow{a}+ \overrightarrow{b}\cdot\overrightarrow{b}-2\sqrt{3}\overrightarrow{a}\cdot\overrightarrow{b}=1
\displaystyle \Rightarrow\ 3|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}- 2\sqrt{3}\overrightarrow{a}\cdot\overrightarrow{b}=1
\displaystyle \Rightarrow\ 3(1)^{2}+(1)^{2}-2\sqrt{3}\overrightarrow{a}\cdot\overrightarrow{b}=1
\displaystyle \Rightarrow\ 4-2\sqrt{3}\overrightarrow{a}\cdot\overrightarrow{b}=1
\displaystyle \Rightarrow\ 2\sqrt{3}\overrightarrow{a}\cdot\overrightarrow{b}=3
\displaystyle \Rightarrow\ \overrightarrow{a}\cdot\overrightarrow{b}=\frac{3}{2\sqrt{3}}= \frac{\sqrt{3}}{2}\qquad\text{(1)}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}= |\overrightarrow{a}||\overrightarrow{b}|\cos\theta
\displaystyle =1\cdot1\cdot\cos\theta
\displaystyle =\cos\theta\qquad\text{(2)}
\displaystyle \text{From (1) and (2)}
\displaystyle \cos\theta=\frac{\sqrt{3}}{2}\Rightarrow\theta=\cos^{-1} \left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}\ \text{or }30^{\circ}

\displaystyle \textbf{22. } \text{If } A = \begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} \text{ and } I =   \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \text{, find scalar } k \text{ so that } A^{2} + I = kA \text{.}
\displaystyle \text{Answer:}
\displaystyle \ A=\begin{bmatrix}-3 & 2 \\ 1 & -1\end{bmatrix},\ I= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
\displaystyle A^{2}=AA=\begin{bmatrix}-3 & 2 \\ 1 & -1\end{bmatrix} \begin{bmatrix}-3 & 2 \\ 1 & -1\end{bmatrix}
\displaystyle \Rightarrow\ A^{2}=\begin{bmatrix}9+2 & -6-2 \\ -3-1 & 2+1\end{bmatrix}
\displaystyle \Rightarrow\ A^{2}=\begin{bmatrix}11 & -8 \\ -4 & 3\end{bmatrix}
\displaystyle A^{2}+I=kA
\displaystyle \Rightarrow\ \begin{bmatrix}11 & -8 \\ -4 & 3\end{bmatrix}+ \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=k\begin{bmatrix}-3 & 2 \\ 1 & -1\end{bmatrix}
\displaystyle \Rightarrow\ \begin{bmatrix}12 & -8 \\ -4 & 4\end{bmatrix}= \begin{bmatrix}-3k & 2k \\ k & -k\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle -3k=12
\displaystyle \Rightarrow\ k=-4

\displaystyle \textbf{23. } \text{If } f(x) = \sqrt{\frac{\sec x - 1}{\sec x + 1}} \text{, find } f'\left(\frac{\pi}{3}\right) \text{.}
\displaystyle \text{OR}
\displaystyle \text{Find } f'(x) \text{ if } f(x) = (\tan x)^{\tan x} \text{.}
\displaystyle \text{Answer:}
\displaystyle \ f(x)=\sqrt{\frac{\sec x-1}{\sec x+1}}
\displaystyle =\sqrt{\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}}
\displaystyle =\sqrt{\frac{1-\cos x}{1+\cos x}}
\displaystyle =\sqrt{\frac{2\sin^{2}\frac{x}{2}}{2\cos^{2}\frac{x}{2}}}
\displaystyle =\sqrt{\tan^{2}\frac{x}{2}}
\displaystyle =\tan\frac{x}{2}
\displaystyle f'(x)=\sec^{2}\left(\frac{x}{2}\right)\cdot\frac{1}{2}
\displaystyle \Rightarrow\ f'\left(\frac{\pi}{3}\right)=\frac{1}{2} \sec^{2}\frac{\pi}{6}
\displaystyle =\frac{1}{2}\left(\frac{2}{\sqrt{3}}\right)^{2}
\displaystyle =\frac{1}{2}\cdot\frac{4}{3}
\displaystyle =\frac{2}{3}
\displaystyle \text{OR}
\displaystyle f(x)=(\tan x)^{\tan x}
\displaystyle \text{Let }y=f(x)=(\tan x)^{\tan x}
\displaystyle \log y=\log\left((\tan x)^{\tan x}\right)
\displaystyle \Rightarrow\ \log y=\tan x\log(\tan x)
\displaystyle \text{Differentiating both sides with respect to }x
\displaystyle \Rightarrow\ \frac{1}{y}\frac{dy}{dx}= \tan x\frac{d}{dx}\log(\tan x)+\log(\tan x)\frac{d}{dx}(\tan x)
\displaystyle \Rightarrow\ \frac{1}{y}\frac{dy}{dx}= \tan x\left(\frac{1}{\tan x}\sec^{2}x\right)+\log(\tan x)\sec^{2}x
\displaystyle \Rightarrow\ \frac{1}{y}\frac{dy}{dx}= \sec^{2}x+\sec^{2}x\log(\tan x)
\displaystyle \Rightarrow\ \frac{dy}{dx}=y\sec^{2}x(1+\log(\tan x))
\displaystyle \therefore\ f'(x)=(\tan x)^{\tan x}\sec^{2}x (1+\log(\tan x))

\displaystyle \textbf{24. } \text{Find: } \int \frac{\tan^{3} x}{\cos^{3} x} \, dx
\displaystyle \text{Answer:}
\displaystyle \ \text{Let }I=\int\frac{\tan^{3}x}{\cos^{3}x}\ dx
\displaystyle =\int\tan^{3}x\cdot\sec^{3}x\ dx
\displaystyle =\int\tan^{2}x\cdot\sec^{2}x\cdot\tan x\sec x\ dx
\displaystyle =\int(\sec^{2}x-1)\sec^{2}x\cdot\tan x\sec x\ dx
\displaystyle \text{Let }\sec x=t,\ \text{then }\tan x\sec x\ dx=dt
\displaystyle \Rightarrow\ I=\int(t^{2}-1)t^{2}\ dt
\displaystyle =\int(t^{4}-t^{2})\ dt
\displaystyle =\frac{t^{5}}{5}-\frac{t^{3}}{3}+C
\displaystyle =\frac{1}{5}\sec^{5}x-\frac{1}{3}\sec^{3}x+C

\displaystyle \textbf{25. } \text{Show that the plane } x - 5y - 2z = 1 \text{ contains the line}   \frac{x - 5}{3} = y = 2 - z \text{.}
\displaystyle \text{Answer:}
\displaystyle \ \text{Equation of plane is }x-5y-2z=1\qquad\text{(1)}
\displaystyle a_{1}=1,\ b_{1}=-5,\ c_{1}=-2
\displaystyle \text{Equation of line is }\frac{x-5}{3}=y=-(z-2)
\displaystyle \frac{x-5}{3}=\frac{y-0}{1}=\frac{z-2}{-1}
\displaystyle \Rightarrow\ a_{2}=3,\ b_{2}=1,\ c_{2}=-1
\displaystyle \text{The line passes through }(5,0,2)
\displaystyle \text{Substituting }(5,0,2)\text{ in equation of plane,}
\displaystyle 5-5(0)-2(2)=1
\displaystyle \Rightarrow\ 5-4=1
\displaystyle \Rightarrow\ 1=1
\displaystyle \therefore\ (5,0,2)\text{ lies on plane}
\displaystyle \text{Now, }a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}
\displaystyle =3-5+2=0
\displaystyle \therefore\ \text{Line is perpendicular to normal of plane}
\displaystyle \text{Hence, line lies in plane}

\displaystyle \textbf{26. } \text{A fair dice is thrown two times. Find the probability distribution of the number}
\displaystyle \text{of sixes. Also determine the mean of the number of sixes.}
\displaystyle \text{Answer:}
\displaystyle \ \text{Number of trials}=2
\displaystyle \text{Number of sample space}=36
\displaystyle X\ (\text{Number of sixes})=0,1,2
\displaystyle P(X=0)=\frac{25}{36}
\displaystyle P(X=1)=\frac{10}{36}
\displaystyle P(X=2)=\frac{1}{36}
\displaystyle XP(X)\ \text{for }X=0\text{ is }0
\displaystyle XP(X)\ \text{for }X=1\text{ is }\frac{10}{36}
\displaystyle XP(X)\ \text{for }X=2\text{ is }\frac{2}{36}
\displaystyle \text{Mean}=\sum XP(X)
\displaystyle =0+\frac{10}{36}+\frac{2}{36}
\displaystyle =\frac{12}{36}=\frac{1}{3}

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 27 to 32 carry 4 marks each.}

\displaystyle \textbf{27. } \text{Solve the following differential equation: }   (1 + e^{\frac{y}{x}}) \, dy + e^{\frac{y}{x}} \left(1 - \frac{y}{x}\right) dx = 0 \text{ } (x \neq 0)
\displaystyle \text{Answer:}
\displaystyle \ (1+e^{\frac{y}{x}})\,dy+e^{\frac{y}{x}} \left(1-\frac{y}{x}\right)\,dx=0
\displaystyle \Rightarrow\ (1+e^{\frac{y}{x}})\,dy= -e^{\frac{y}{x}}\left(1-\frac{y}{x}\right)\,dx
\displaystyle \Rightarrow\ \frac{dy}{dx}= -\frac{e^{\frac{y}{x}}\left(1-\frac{y}{x}\right)}{1+e^{\frac{y}{x}}}\qquad\text{(1)}
\displaystyle \text{Equation is homogeneous differential equation}
\displaystyle \text{Put }y=vx
\displaystyle \Rightarrow\ \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{Substituting in (1),}
\displaystyle v+x\frac{dv}{dx}= -\frac{e^{v}(1-v)}{1+e^{v}}
\displaystyle \Rightarrow\ x\frac{dv}{dx}= -\frac{e^{v}(1-v)}{1+e^{v}}-v
\displaystyle =\frac{-e^{v}+ve^{v}-v-v e^{v}}{1+e^{v}}
\displaystyle =-\frac{e^{v}+v}{1+e^{v}}
\displaystyle \Rightarrow\ \frac{1+e^{v}}{e^{v}+v}\,dv=-\frac{dx}{x}
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{1+e^{v}}{e^{v}+v}\,dv=-\int\frac{dx}{x}
\displaystyle \text{Let }t=v+e^{v}
\displaystyle \Rightarrow\ dt=(1+e^{v})\,dv
\displaystyle \Rightarrow\ \int\frac{dt}{t}=-\log|x|+\log C
\displaystyle \Rightarrow\ \log|t|=-\log|x|+\log C
\displaystyle \Rightarrow\ \log|v+e^{v}|+\log|x|=\log C
\displaystyle \Rightarrow\ \log|(v+e^{v})x|=\log C
\displaystyle \Rightarrow\ (v+e^{v})x=C
\displaystyle \text{Putting }v=\frac{y}{x}
\displaystyle \Rightarrow\ \left(\frac{y}{x}+e^{\frac{y}{x}}\right)x=C
\displaystyle \Rightarrow\ y+xe^{\frac{y}{x}}=C

\displaystyle \textbf{28. } \text{A cottage industry manufactures pedestal lamps and wooden shades.}
\displaystyle \text{Both the products require machine time as well as craftsman time in the making.}
\displaystyle \text{The number of hours required for producing 1 unit of each and the corresponding}
\displaystyle \text{profit is given in the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Item} & \text{Machine Time} & \text{Craftsman Time} & \text{Profit (in Rs)} \\ \hline \text{Pedestal lamp} & 1.5 \text{ hours} & 3 \text{ hours} & 30 \\ \hline \text{Wooden shades} & 3 \text{ hours} & 1 \text{ hour} & 20 \\ \hline \end{array}
\displaystyle \text{In a day, the factory has availability of not more than 42 hours of machine time}
\displaystyle \text{and 24 hours of craftsman time. Assuming that all items manufactured are sold,}
\displaystyle \text{how should the manufacturer schedule his daily production in order to maximise}
\displaystyle \text{the profit? Formulate it as an LPP and solve it graphically.}
\displaystyle \text{Answer:}
\displaystyle \ \text{Let number of pedestal lamps be }x \text{ and number of wooden shades be }y
\displaystyle 1.5x+3y\leq42
\displaystyle \Rightarrow\ 0.5x+y\leq14\qquad\text{...(1)}
\displaystyle 3x+y\leq24\qquad\text{...(2)}
\displaystyle x\geq0,\ y\geq0
\displaystyle \text{Maximize profit }Z=30x+20y
\displaystyle \text{Intersection point of (1) and (2) is given by}
\displaystyle 3x+y=24
\displaystyle 0.5x+y=14
\displaystyle \text{Subtracting, }2.5x=10
\displaystyle \Rightarrow\ x=4
\displaystyle \text{Substituting in }3x+y=24,\ 3(4)+y=24
\displaystyle \Rightarrow\ y=12
\displaystyle \text{Thus, the two lines intersect at }(4,12)
\displaystyle \text{For }0.5x+y=14,\ \text{intercepts are }(0,14)\text{ and }(28,0)
\displaystyle \text{For }3x+y=24,\ \text{intercepts are }(0,24)\text{ and }(8,0)
\displaystyle \text{OABC is the feasible region}\displaystyle \text{Corner points are }O(0,0),\ A(0,14),\ B(4,12),\ C(8,0)
\displaystyle \text{At }O(0,0),\ Z=30(0)+20(0)=0
\displaystyle \text{At }A(0,14),\ Z=30(0)+20(14)=280
\displaystyle \text{At }B(4,12),\ Z=30(4)+20(12)=120+240=360
\displaystyle \text{At }C(8,0),\ Z=30(8)+20(0)=240
\displaystyle \text{Hence, profit is maximum at }(4,12)
\displaystyle \text{Number of pedestal lamps }=4
\displaystyle \text{Number of wooden shades }=12
\displaystyle \text{Maximum profit }=\text{Rs }360

\displaystyle \textbf{29. } \text{Evaluate: } \int \sin 2x \tan^{-1}(\sin x) \, dx
\displaystyle \text{Answer:}
\displaystyle \ \text{Let }\ I=\int_{0}^{\frac{\pi}{2}} \sin2x\tan^{-1}(\sin x)\ dx
\displaystyle \Rightarrow\ I=\int_{0}^{\frac{\pi}{2}} (2\sin x\cos x)\tan^{-1}(\sin x)\ dx
\displaystyle \text{Let }\sin x=t,\ \text{then }\cos x\ dx=dt
\displaystyle \text{When }x=0,\ t=\sin0=0
\displaystyle \text{When }x=\frac{\pi}{2},\ t=\sin\frac{\pi}{2}=1
\displaystyle \Rightarrow\ I=2\int_{0}^{1}t\tan^{-1}t\ dt
\displaystyle \text{Using integration by parts, }\int u\,dv=uv-\int v\,du
\displaystyle \text{Take }u=\tan^{-1}t,\ dv=t\ dt
\displaystyle \Rightarrow\ du=\frac{1}{1+t^{2}}\,dt,\ v=\frac{t^{2}}{2}
\displaystyle \Rightarrow\ I=2\left[\frac{t^{2}}{2}\tan^{-1}t\right]_{0}^{1}- 2\int_{0}^{1}\frac{t^{2}}{2(1+t^{2})}\ dt
\displaystyle \Rightarrow\ I=2\left[\frac{t^{2}}{2}\tan^{-1}t\right]_{0}^{1}- \int_{0}^{1}\frac{t^{2}}{1+t^{2}}\ dt
\displaystyle \text{Let }\ I_{1}=\int_{0}^{1}\frac{t^{2}}{1+t^{2}}\ dt
\displaystyle \Rightarrow\ I_{1}=\int_{0}^{1}\frac{t^{2}+1-1}{t^{2}+1}\ dt
\displaystyle =\int_{0}^{1}\left(1-\frac{1}{t^{2}+1}\right)\ dt
\displaystyle =\int_{0}^{1}dt-\int_{0}^{1}\frac{dt}{t^{2}+1}
\displaystyle =\left[t-\tan^{-1}t\right]_{0}^{1}
\displaystyle =\left(1-\frac{\pi}{4}\right)-0
\displaystyle =1-\frac{\pi}{4}
\displaystyle \text{Now, }\ I=2\left[\frac{t^{2}}{2}\tan^{-1}t\right]_{0}^{1}-I_{1}
\displaystyle =\left[t^{2}\tan^{-1}t\right]_{0}^{1}-\left(1-\frac{\pi}{4}\right)
\displaystyle =\left(1\cdot\frac{\pi}{4}\right)-\left(1-\frac{\pi}{4}\right)
\displaystyle =\frac{\pi}{4}-1+\frac{\pi}{4}
\displaystyle =\frac{\pi}{2}-1

\displaystyle \textbf{30. } \text{Check whether the relation } R \text{ in the set } N \text{ of natural numbers} \text{given by } 
\displaystyle R = \{(a, b) : a \text{ is a divisor of } b\} \text{ is reflexive, symmetric or transitive. Also }
\displaystyle \text{determine whether } R \text{ is an equivalence relation.}
\displaystyle \text{OR}
\displaystyle \text{Prove that } \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) =   \frac{1}{2} \sin^{-1}\left(\frac{4}{5}\right) \text{.}
\displaystyle \text{Answer:}
\displaystyle \ R=\{(a,b):a\text{ is divisor of }b\}
\displaystyle \text{(i) Reflexive:}
\displaystyle \text{Since every natural number is a divisor of itself}
\displaystyle \therefore\ \forall a\in N,\ (a,a)\in R
\displaystyle \therefore\ R\text{ is reflexive}
\displaystyle \text{(ii) Symmetric:}
\displaystyle \text{Let }(2,4)\in R\qquad\text{(2 is divisor of 4)}
\displaystyle \text{But }(4,2)\notin R\qquad\text{(4 is not divisor of 2)}
\displaystyle \therefore\ R\text{ is not symmetric}
\displaystyle \text{(iii) Transitive:}
\displaystyle \text{Let }(a,b)\in R,\ (b,c)\in R\ \forall a,b,c\in N
\displaystyle \Rightarrow\ b=ma,\ c=nb\ \text{for some }m,n\in N
\displaystyle \Rightarrow\ c=n(ma)=mna
\displaystyle \Rightarrow\ a\text{ is divisor of }c
\displaystyle \Rightarrow\ (a,c)\in R
\displaystyle \therefore\ R\text{ is transitive}
\displaystyle \text{Hence, }R\text{ is reflexive and transitive but not symmetric}
\displaystyle \therefore\ R\text{ is not an equivalence relation}
\displaystyle \text{OR}
\displaystyle \tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{2}{9}\right)= \frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right)
\displaystyle \text{L.H.S.}=\tan^{-1}\left(\frac{1}{4}\right)+ \tan^{-1}\left(\frac{2}{9}\right)
\displaystyle =\tan^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}} {1-\frac{1}{4}\cdot\frac{2}{9}}\right)
\displaystyle \left[\tan^{-1}x+\tan^{-1}y= \tan^{-1}\left(\frac{x+y}{1-xy}\right)\right]
\displaystyle =\tan^{-1}\left(\frac{9+8}{36-2}\right)
\displaystyle =\tan^{-1}\left(\frac{17}{34}\right)
\displaystyle =\tan^{-1}\left(\frac{1}{2}\right)
\displaystyle =\frac{1}{2}\times2\tan^{-1}\left(\frac{1}{2}\right)
\displaystyle =\frac{1}{2}\sin^{-1}\left(\frac{2\cdot\frac{1}{2}} {1+\left(\frac{1}{2}\right)^{2}}\right)
\displaystyle \left[\because 2\tan^{-1}x=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\right]
\displaystyle =\frac{1}{2}\sin^{-1}\left(\frac{1}{1+\frac{1}{4}}\right)
\displaystyle =\frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right)
\displaystyle \text{L.H.S.}=\text{R.H.S.}
\displaystyle \text{Hence proved}

\displaystyle \textbf{31. } \text{Find the equation of the plane passing through the points} (1, 0, -2), (3, -1, 0) 
\displaystyle \text{ and perpendicular to the plane } 2x - y + z = 8 \text{. Also find the distance of the plane thus}
\displaystyle \text{obtained from the origin.}
\displaystyle \text{Answer:}
\displaystyle \ \text{Let plane pass through }P(1,0,-2),\ Q(3,-1,0)
\displaystyle \text{Position vectors are }\overrightarrow{a}=i-2k,\ \overrightarrow{b}=3i-j
\displaystyle \overrightarrow{PQ}=\overrightarrow{b}-\overrightarrow{a}= (3i-j)-(i-2k)
\displaystyle =2i-j+2k
\displaystyle \text{Plane is perpendicular to given plane }2x-y+z=8
\displaystyle \text{Normal vector of given plane, }\overrightarrow{n_{1}}=2i-j+k
\displaystyle \text{Normal vector of required plane will be}
\displaystyle \overrightarrow{n}=\overrightarrow{n_{1}}\times\overrightarrow{PQ}= \begin{vmatrix} i & j & k \\ 2 & -1 & 1 \\ 2 & -1 & 2 \end{vmatrix}
\displaystyle =i((-1)(2)-1(-1))-j(2\cdot2-1\cdot2)+k(2(-1)-(-1)(2))
\displaystyle =i(-2+1)-j(4-2)+k(-2+2)
\displaystyle =-i-2j
\displaystyle \text{Equation of required plane }(\overrightarrow{r}-\overrightarrow{a}) \cdot\overrightarrow{n}=0,\ \text{where }\overrightarrow{r}=xi+yj+zk
\displaystyle \Rightarrow\ (xi+yj+zk-i+2k)\cdot(-i-2j)=0
\displaystyle \Rightarrow\ ((x-1)i+yj+(z+2)k)\cdot(-i-2j)=0
\displaystyle \Rightarrow\ -(x-1)-2y=0
\displaystyle \Rightarrow\ -x+1-2y=0
\displaystyle \Rightarrow\ x+2y=1
\displaystyle \text{Distance from origin }= \frac{|0+0-1|}{\sqrt{1^{2}+2^{2}}}=\frac{1}{\sqrt{5}}

\displaystyle \textbf{32. } \text{If } \tan^{-1}\left(\frac{y}{x}\right) = \log \sqrt{x^{2} + y^{2}} \text{, prove that}   \frac{dy}{dx} = \frac{x + y}{x - y} \text{.}
\displaystyle \text{OR}
\displaystyle \text{If } y = e^{a \cos^{-1} x} \text{, } -1 < x < 1 \text{, then show that}   (1 - x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} - a^{2} y = 0 \text{.}
\displaystyle \text{Answer:}
\displaystyle \ \tan^{-1}\left(\frac{y}{x}\right)= \log\sqrt{x^{2}+y^{2}}
\displaystyle \Rightarrow\ \tan^{-1}\left(\frac{y}{x}\right)= \frac{1}{2}\log(x^{2}+y^{2})
\displaystyle \text{Differentiating both sides with respect to }x
\displaystyle \frac{1}{1+\left(\frac{y}{x}\right)^{2}} \frac{d}{dx}\left(\frac{y}{x}\right)= \frac{1}{2}\cdot\frac{1}{x^{2}+y^{2}}\frac{d}{dx}(x^{2}+y^{2})
\displaystyle \Rightarrow\ \frac{x^{2}}{x^{2}+y^{2}} \cdot\frac{x\frac{dy}{dx}-y}{x^{2}}= \frac{1}{2(x^{2}+y^{2})}\left(2x+2y\frac{dy}{dx}\right)
\displaystyle \Rightarrow\ \frac{x\frac{dy}{dx}-y}{x^{2}+y^{2}}= \frac{x+y\frac{dy}{dx}}{x^{2}+y^{2}}
\displaystyle \Rightarrow\ x\frac{dy}{dx}-y=x+y\frac{dy}{dx}
\displaystyle \Rightarrow\ x\frac{dy}{dx}-y\frac{dy}{dx}=x+y
\displaystyle \Rightarrow\ \frac{dy}{dx}(x-y)=x+y
\displaystyle \therefore\ \frac{dy}{dx}=\frac{x+y}{x-y}
\displaystyle \text{Hence, proved}
\displaystyle \text{OR}
\displaystyle y=e^{a\cos^{-1}x}
\displaystyle \Rightarrow\ \frac{dy}{dx}= e^{a\cos^{-1}x}\frac{d}{dx}(a\cos^{-1}x)
\displaystyle \Rightarrow\ \frac{dy}{dx}= ae^{a\cos^{-1}x}\left(\frac{-1}{\sqrt{1-x^{2}}}\right)
\displaystyle \left[\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^{2}}}\right]
\displaystyle \Rightarrow\ \frac{dy}{dx}= -\frac{ay}{\sqrt{1-x^{2}}}
\displaystyle \text{Squaring on both sides}
\displaystyle \left(\frac{dy}{dx}\right)^{2}= \frac{a^{2}y^{2}}{1-x^{2}}
\displaystyle \Rightarrow\ (1-x^{2})\left(\frac{dy}{dx}\right)^{2}=a^{2}y^{2}
\displaystyle \text{Differentiating with respect to }x
\displaystyle \frac{d}{dx}\left[(1-x^{2})\left(\frac{dy}{dx}\right)^{2}\right]= \frac{d}{dx}(a^{2}y^{2})
\displaystyle \Rightarrow\ (-2x)\left(\frac{dy}{dx}\right)^{2}+ (1-x^{2})\cdot2\frac{dy}{dx}\frac{d^{2}y}{dx^{2}}= a^{2}\cdot2y\frac{dy}{dx}
\displaystyle \Rightarrow\ -x\left(\frac{dy}{dx}\right)+ (1-x^{2})\frac{d^{2}y}{dx^{2}}=a^{2}y
\displaystyle \therefore\ (1-x^{2})\frac{d^{2}y}{dx^{2}}- x\frac{dy}{dx}-a^{2}y=0

\displaystyle \textbf{SECTION - D}
\displaystyle \text{Question numbers 33 to 36 carry 6 marks each.}

\displaystyle \textbf{33. } \text{Amongst all open (from the top) right circular cylindrical boxes of}
\displaystyle \text{volume } 125\pi \text{ cm}^{3} \text{, find the dimensions of the box which has the}   \text{least surface area.}
\displaystyle \text{Answer:}
\displaystyle \ \text{Volume of cylinder }=\pi r^{2}h
\displaystyle \text{where }r\text{ is radius and }h\text{ is height of cylinder}
\displaystyle \Rightarrow\ V=\pi r^{2}h
\displaystyle \Rightarrow\ V=125\pi\ \text{cm}^{3}\ \text{(given)}
\displaystyle \Rightarrow\ \pi r^{2}h=125\pi\Rightarrow h=\frac{125\pi}{\pi r^{2}}
\displaystyle \Rightarrow\ h=\frac{125}{r^{2}}\qquad\text{...(1)}
\displaystyle \text{Surface area }S=2\pi rh+\pi r^{2}
\displaystyle \Rightarrow\ S=2\pi r\left(\frac{125}{r^{2}}\right)+\pi r^{2}\qquad\text{[from (1)]}
\displaystyle \Rightarrow\ S=\frac{250\pi}{r}+\pi r^{2}
\displaystyle \text{Differentiating both sides with respect to }r
\displaystyle \Rightarrow\ \frac{dS}{dr}=250\pi\left(-\frac{1}{r^{2}}\right)+2\pi r
\displaystyle \Rightarrow\ \frac{dS}{dr}=-\frac{250\pi}{r^{2}}+2\pi r
\displaystyle \text{For least surface area, }\frac{dS}{dr}=0
\displaystyle \Rightarrow\ -\frac{250\pi}{r^{2}}+2\pi r=0
\displaystyle \Rightarrow\ 2\pi r=\frac{250\pi}{r^{2}}
\displaystyle \Rightarrow\ 2r^{3}=250
\displaystyle \Rightarrow\ r^{3}=125\Rightarrow r=5\ \text{cm}
\displaystyle \frac{d^{2}S}{dr^{2}}=\frac{500\pi}{r^{3}}+2\pi
\displaystyle \text{At }r=5,\ \frac{d^{2}S}{dr^{2}}>0
\displaystyle \therefore\ \text{Surface area is minimum at }r=5\ \text{cm}
\displaystyle \text{From (1), }\ h=\frac{125}{r^{2}}=\frac{125}{(5)^{2}}=\frac{125}{25}=5
\displaystyle \text{Radius }=5\ \text{cm, Height }=5\ \text{cm}

\displaystyle \textbf{34. } \text{Using integration, find the area lying above the x-axis and} \text{included between the circle }
\displaystyle  x^{2} + y^{2} = 8x \text{ and inside the parabola }   y^{2} = 4x \text{.}
\displaystyle \text{OR}
\displaystyle \text{Using the method of integration, find the area of the triangle ABC, coordinates of whose}
\displaystyle \text{vertices are } A(2, 0), B(4, 5) \text{ and } C(6, 3) \text{.}
\displaystyle \text{Answer:}
\displaystyle \ \text{Equation of circle }x^{2}+y^{2}-8x=0
\displaystyle \Rightarrow\ x^{2}-8x+y^{2}=0
\displaystyle \text{Using completing square method}
\displaystyle \Rightarrow\ x^{2}-8x+y^{2}+4^{2}-4^{2}=0
\displaystyle \Rightarrow\ (x-4)^{2}+y^{2}=4^{2}\qquad\text{...(1)}
\displaystyle \text{Equation of parabola }y^{2}=4x\qquad\text{...(2)}
\displaystyle \text{To find point of intersection, solve }x^{2}+y^{2}-8x=0 \text{ and }y^{2}=4x
\displaystyle \Rightarrow\ x^{2}+4x-8x=0
\displaystyle \Rightarrow\ x^{2}-4x=0
\displaystyle \Rightarrow\ x(x-4)=0
\displaystyle \Rightarrow\ x=0\ \text{or}\ x=4
\displaystyle \text{When }x=0,\ y^{2}=0\Rightarrow y=0
\displaystyle \text{When }x=4,\ y^{2}=4(4)=16\Rightarrow y=\pm4
\displaystyle \text{Required area }=\int_{0}^{4}y_{\text{parabola}}\ dx+\int_{4}^{8}y_{\text{circle}}\ dx
\displaystyle \text{From }y^{2}=4x,\ y=\sqrt{4x}=2\sqrt{x}
\displaystyle \text{From }(x-4)^{2}+y^{2}=4^{2},\ y=\sqrt{4^{2}-(x-4)^{2}}
\displaystyle \Rightarrow\ \text{Required area }=\int_{0}^{4}2\sqrt{x}\ dx+ \int_{4}^{8}\sqrt{16-(x-4)^{2}}\ dx\displaystyle \ \text{Required area}=\int_{0}^{4}2\sqrt{x}\ dx+ \int_{4}^{8}\sqrt{4^{2}-(x-4)^{2}}\ dx
\displaystyle =2\int_{0}^{4}x^{\frac{1}{2}}\ dx+ \left[\frac{x-4}{2}\sqrt{4^{2}-(x-4)^{2}}+ \frac{16}{2}\sin^{-1}\left(\frac{x-4}{4}\right)\right]_{4}^{8}
\displaystyle \left[\because\ \int\sqrt{a^{2}-u^{2}}\ du= \frac{u}{2}\sqrt{a^{2}-u^{2}}+\frac{a^{2}}{2}\sin^{-1}\left(\frac{u}{a}\right)\right]
\displaystyle =2\cdot\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{4}+ \left[\frac{x-4}{2}\sqrt{16-(x-4)^{2}}+ 8\sin^{-1}\left(\frac{x-4}{4}\right)\right]_{4}^{8}
\displaystyle =\frac{4}{3}\left[4^{\frac{3}{2}}-0\right]+ \left[0+8\sin^{-1}(1)\right]-\left[0+8\sin^{-1}(0)\right]
\displaystyle =\frac{4}{3}(8)+8\left(\frac{\pi}{2}\right)
\displaystyle =\frac{32}{3}+4\pi
\displaystyle =\left(\frac{32}{3}+4\pi\right)\ \text{sq. units}
\displaystyle \text{OR}\displaystyle \text{Equation of a line is } y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})
\displaystyle \text{Equation of }AB
\displaystyle y-0=\frac{5-0}{4-2}(x-2)
\displaystyle \Rightarrow\ y=\frac{5}{2}(x-2)=\frac{5x-10}{2}
\displaystyle \text{Equation of }AC
\displaystyle y-0=\frac{3-0}{6-2}(x-2)
\displaystyle \Rightarrow\ y=\frac{3}{4}(x-2)=\frac{3x-6}{4}
\displaystyle \text{Equation of }BC
\displaystyle y-3=\frac{5-3}{4-6}(x-6)
\displaystyle \Rightarrow\ y-3=\frac{2}{-2}(x-6)
\displaystyle \Rightarrow\ y=-x+6+3=-x+9
\displaystyle \text{Area of }\triangle ABC=\int_{2}^{4}y_{AB}\ dx+ \int_{4}^{6}y_{BC}\ dx-\int_{2}^{6}y_{AC}\ dx
\displaystyle \text{Area of }\triangle ABC= \int_{2}^{4}\frac{5x-10}{2}\ dx+\int_{4}^{6}(9-x)\ dx- \int_{2}^{6}\frac{3x-6}{4}\ dx
\displaystyle =\frac{1}{2}\left[\frac{5x^{2}}{2}-10x\right]_{2}^{4}+ \left[9x-\frac{x^{2}}{2}\right]_{4}^{6}- \frac{1}{4}\left[\frac{3x^{2}}{2}-6x\right]_{2}^{6}
\displaystyle =\frac{1}{2}\left[\left(\frac{5\cdot16}{2}-40\right)- \left(\frac{5\cdot4}{2}-20\right)\right]+\left[\left(9\cdot6-\frac{36}{2}\right)- \left(9\cdot4-\frac{16}{2}\right)\right]
\displaystyle \quad-\frac{1}{4}\left[\left(\frac{3\cdot36}{2}-36\right)- \left(\frac{3\cdot4}{2}-12\right)\right]
\displaystyle =\frac{1}{2}[0+10]+[36-28]-\frac{1}{4}[18+6]
\displaystyle =5+8-6=7\ \text{sq. units}

\displaystyle \textbf{35. } \text{If } A = \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix} \text{, find } A^{-1}   \text{and use it to solve the following system of equations:}
\displaystyle 5x - y + 4z = 5
\displaystyle 2x + 3y + 5z = 2
\displaystyle 5x - 2y + 6z = -1
\displaystyle \text{OR}
\displaystyle \text{If } x, y, z \text{ are different and } \left|\begin{matrix} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{matrix}\right| = 0 \text{,}
\displaystyle \text{then using properties of determinants show that } 1 + xyz = 0 \text{.}
\displaystyle \text{Answer:}
\displaystyle \ A=\begin{bmatrix}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{bmatrix}
\displaystyle |A|=5(18-(-10))+1(12-25)+4(-4-15)
\displaystyle =140-13-76=140-89=51
\displaystyle A_{11}=(-1)^{2}\begin{vmatrix}3 & 5 \\ -2 & 6\end{vmatrix}=18+10=28
\displaystyle A_{12}=(-1)^{3}\begin{vmatrix}2 & 5 \\ 5 & 6\end{vmatrix} =-1(12-25)=13
\displaystyle A_{13}=(-1)^{4}\begin{vmatrix}2 & 3 \\ 5 & -2\end{vmatrix} =-4-15=-19
\displaystyle A_{21}=(-1)^{3}\begin{vmatrix}-1 & 4 \\ -2 & 6\end{vmatrix} =-1(-6+8)=-2
\displaystyle A_{22}=(-1)^{4}\begin{vmatrix}5 & 4 \\ 5 & 6\end{vmatrix} =30-20=10
\displaystyle A_{23}=(-1)^{5}\begin{vmatrix}5 & -1 \\ 5 & -2\end{vmatrix} =-1(-10+5)=5
\displaystyle A_{31}=(-1)^{4}\begin{vmatrix}-1 & 4 \\ 3 & 5\end{vmatrix} =-5-12=-17
\displaystyle A_{32}=(-1)^{5}\begin{vmatrix}5 & 4 \\ 2 & 5\end{vmatrix} =-1(25-8)=-17
\displaystyle A_{33}=(-1)^{6}\begin{vmatrix}5 & -1 \\ 2 & 3\end{vmatrix} =15+2=17
\displaystyle \text{adj }A=\begin{bmatrix}28 & 13 & -19 \\ -2 & 10 & 5 \\ -17 & -17 & 17\end{bmatrix}^{T} =\begin{bmatrix}28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17\end{bmatrix}
\displaystyle A^{-1}=\frac{\text{adj }A}{|A|}= \frac{1}{51}\begin{bmatrix}28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17\end{bmatrix}
\displaystyle \text{Given }5x-y+4z=5,\ 2x+3y+5z=2,\ 5x-2y+6z=-1
\displaystyle A=\begin{bmatrix}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{bmatrix}, \ B=\begin{bmatrix}5 \\ 2 \\ -1\end{bmatrix}, \ X=\begin{bmatrix}x \\ y \\ z\end{bmatrix}
\displaystyle AX=B
\displaystyle \Rightarrow\ X=A^{-1}B
\displaystyle \Rightarrow\ \begin{bmatrix}x \\ y \\ z\end{bmatrix}= \frac{1}{51}\begin{bmatrix}28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17\end{bmatrix} \begin{bmatrix}5 \\ 2 \\ -1\end{bmatrix}
\displaystyle =\frac{1}{51}\begin{bmatrix}140-4+17 \\ 65+20+17 \\ -95+10-17\end{bmatrix}
\displaystyle =\frac{1}{51}\begin{bmatrix}153 \\ 102 \\ -102\end{bmatrix} =\begin{bmatrix}3 \\ 2 \\ -2\end{bmatrix}
\displaystyle \therefore\ x=3,\ y=2,\ z=-2
\displaystyle \text{OR}
\displaystyle \Delta= \begin{vmatrix}1 & x & x^{2}+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{vmatrix} =\begin{vmatrix}1 & x & x^{2} & x^{3} \\ y & y^{2} & 1 & y^{3} \\ z & z^{2} & 1 & z^{3}\end{vmatrix}
\displaystyle \text{Swapping }C_{2}\leftrightarrow C_{3},\ \text{then } C_{2}\leftrightarrow C_{1}\ \text{in 1st determinant}
\displaystyle =(-1)^{2}\begin{vmatrix}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix}+xyz\begin{vmatrix}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix}
\displaystyle =(1+xyz)\begin{vmatrix}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix}
\displaystyle R_{1}\rightarrow R_{1}-R_{2}
\displaystyle \Delta=(1+xyz)\begin{vmatrix}0 & x-y & x^{2}-y^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix}
\displaystyle R_{2}\rightarrow R_{2}-R_{3}
\displaystyle \Delta=(1+xyz)\begin{vmatrix}0 & x-y & x^{2}-y^{2} \\ 0 & y-z & y^{2}-z^{2} \\ 1 & z & z^{2}\end{vmatrix}
\displaystyle =(1+xyz)\begin{vmatrix}0 & x-y & (x-y)(x+y) \\ 0 & y-z & (y-z)(y+z) \\ 1 & z & z^{2}\end{vmatrix}
\displaystyle =(1+xyz)(x-y)(y-z) \begin{vmatrix}0 & 1 & x+y \\ 0 & 1 & y+z \\ 1 & z & z^{2}\end{vmatrix}
\displaystyle =-(1+xyz)(x-y)(y-z)(z-x)
\displaystyle \text{If }\Delta=0
\displaystyle (1+xyz)(x-y)(y-z)(z-x)=0
\displaystyle \text{Since }x,\ y,\ z\text{ are different}
\displaystyle x-y\ne0,\ y-z\ne0,\ z-x\ne0
\displaystyle \therefore\ 1+xyz=0
\displaystyle \text{Hence proved}

\displaystyle \textbf{36. } \text{A card from a pack of 52 cards is lost. From the remaining cards of the pack, two}
\displaystyle \text{cards are drawn randomly one-by-one without replacement and are found to be both}
\displaystyle \text{kings. Find the probability of the lost card}   \text{being a king.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }E_{1}\text{ be the event that the lost card is a king}
\displaystyle \text{Let }E_{2}\text{ be the event that the lost card is not a king}
\displaystyle \text{Let }A\text{ be the event that the two cards drawn are kings}
\displaystyle P(E_{1})=\frac{4}{52}=\frac{1}{13}
\displaystyle P(E_{2})=1-P(E_{1})=1-\frac{1}{13}=\frac{12}{13}
\displaystyle P\left(\frac{A}{E_{1}}\right)= \text{Probability of getting 2 kings if lost card is king}
\displaystyle \text{Remaining kings}=4-1=3
\displaystyle P\left(\frac{A}{E_{1}}\right)=\frac{{}^{3}C_{2}}{{}^{51}C_{2}}  =\frac{\frac{3!}{2!1!}}{\frac{51!}{2!49!}}  =\frac{3}{\frac{51\cdot50}{2}}  =\frac{6}{51\cdot50}
\displaystyle P\left(\frac{A}{E_{2}}\right)= \text{Probability of getting 2 kings if lost card is not a king}
\displaystyle \text{Remaining kings}=4
\displaystyle P\left(\frac{A}{E_{2}}\right)=\frac{{}^{4}C_{2}}{{}^{51}C_{2}} =\frac{\frac{4!}{2!2!}}{\frac{51!}{2!49!}}  =\frac{6}{\frac{51\cdot50}{2}}  =\frac{12}{51\cdot50}
\displaystyle \text{By Bayes' theorem}
\displaystyle P\left(\frac{E_{1}}{A}\right)= \frac{P(E_{1})P\left(\frac{A}{E_{1}}\right)} {P(E_{1})P\left(\frac{A}{E_{1}}\right)+P(E_{2})P\left(\frac{A}{E_{2}}\right)}
\displaystyle =\frac{\frac{1}{13}\cdot\frac{6}{51\cdot50}} {\frac{1}{13}\cdot\frac{6}{51\cdot50}+\frac{12}{13}\cdot\frac{12}{51\cdot50}}
\displaystyle =\frac{6}{6+144}=\frac{6}{150}=\frac{2}{50}=\frac{1}{25}
\displaystyle \therefore\ P\left(\frac{E_{1}}{A}\right)=\frac{1}{25}

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