MATHEMATICS

\displaystyle \text{Series SGN/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) The question paper consists of 29 questions divided into four sections A, B, C }
\displaystyle \text{and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 }
\displaystyle \text{questions of two marks each, Section C comprises of 11 questions of four marks each }
\displaystyle \text{and Section D comprises of 6 questions of six }   \text{marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as per }
\displaystyle \text{the exact requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, internal choice has been provided in 3 questions of }
\displaystyle \text{four marks each and 3 questions }   \text{of six marks each. You have to attempt only one of the}
\displaystyle \text{alternatives in all such questions.}
\displaystyle \text{(v) Use of calculators is not permitted. You may ask for} \text{logarithmic tables, if required.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Questions number 1 to 4 carry 1 Mark each.}

\displaystyle \textbf{1. } \text{If for any } 2 \times 2 \text{ square matrix } A,\ A(\mathrm{adj}\ A)=\begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}, \text{ then write the value of } |A|.
\displaystyle \text{Answer:}
\displaystyle \text{Since } A(\mathrm{adj}\ A)=\begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}=8\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\displaystyle \Rightarrow A(\mathrm{adj}\ A)=8I_2 \quad ...(i)
\displaystyle \text{We know that } A(\mathrm{adj}\ A)=|A|I_2 \quad ...(ii)
\displaystyle \text{From (i) and (ii)}
\displaystyle |A|=8

\displaystyle \textbf{2. } \text{Determine the value of } k \text{ for which the following function is continuous at } \\  x=3:   f(x)=\begin{cases} \frac{(x+3)^2-36}{x-3}, & x \ne 3 \\ k, & x=3 \end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{Since } f(x) \text{ is continuous at } x=3
\displaystyle \therefore \lim_{x\to 3^-} f(x)=\lim_{x\to 3^+} f(x)=f(3)
\displaystyle \Rightarrow f(3)=\lim_{x\to 3} f(x)
\displaystyle \Rightarrow k=\lim_{x\to 3}\frac{(x+3)^2-36}{x-3}=\lim_{x\to 3}\frac{(x+3)^2-6^2}{x-3}
\displaystyle \Rightarrow k=\lim_{x\to 3}\frac{(x+3-6)(x+3+6)}{x-3}
\displaystyle \Rightarrow k=\lim_{x\to 3}\frac{(x-3)(x+9)}{x-3}
\displaystyle \Rightarrow k=\lim_{x\to 3}(x+9)=3+9
\displaystyle \Rightarrow k=12

\displaystyle \textbf{3. } \text{Find : } \int \frac{\sin^2 x - \cos^2 x}{\sin x \cos x}\ dx
\displaystyle \text{Answer:}
\displaystyle  \int \frac{\sin^2 x-\cos^2 x}{\sin x\cos x}\,dx
\displaystyle =\int \frac{\sin^2 x}{\sin x\cos x}\,dx-\int \frac{\cos^2 x}{\sin x\cos x}\,dx
\displaystyle =\int \tan x\,dx-\int \cot x\,dx
\displaystyle =\log |\sec x|-\log |\sin x|+C

\displaystyle \textbf{4. } \text{Find the distance between the planes } 2x - y + 2z = 5 \text{ and}
\displaystyle 5x - 2.5y + 5z = 20.
\displaystyle \text{Answer:}
\displaystyle \text{Since equations of planes are } 2x-y+2z=5 \text{ and } 5x-2.5y+5z=20
\displaystyle \text{or } 2x-y+2z=5 \text{ and } 2x-y+2z=8
\displaystyle \text{These two planes are parallel.}
\displaystyle \text{Distance between planes } ax+by+cz+d_1=0 \text{ and } ax+by+cz+d_2=0 \text{ is}
\displaystyle d=\frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}
\displaystyle \Rightarrow d=\frac{|5-8|}{\sqrt{2^2+(-1)^2+2^2}}
\displaystyle =\frac{3}{\sqrt{4+1+4}}=\frac{3}{3}=1\ \text{unit}

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Questions number 5 to 12 carry 2 Mark each.}

\displaystyle \textbf{5. } \text{If } A \text{ is a skew-symmetric matrix of order } 3,\ \text{then prove that}   \det A = 0.
\displaystyle \text{Answer:}
\displaystyle \text{Since } A \text{ is skew-symmetric of order 3}
\displaystyle A^T=-A
\displaystyle |A^T|=|-A|=|A|
\displaystyle |A|=(-1)^3|A|=-|A|
\displaystyle \Rightarrow 2|A|=0 \Rightarrow |A|=0

\displaystyle \textbf{6. } \text{Find the value of } c \text{ in Rolle's theorem for the function}  f(x)=x^3-3x \text{ in } [-\sqrt{3}, 0].
\displaystyle \text{Answer:}
\displaystyle \text{Since } f(x)=x^3-3x
\displaystyle \text{Polynomial function is continuous and differentiable everywhere}
\displaystyle \text{Hence } f(x) \text{ is continuous on } [-\sqrt{3},0] \text{ and differentiable on } (-\sqrt{3},0)
\displaystyle f(-\sqrt{3})=(-\sqrt{3})^3-3(-\sqrt{3})=-3\sqrt{3}+3\sqrt{3}=0
\displaystyle f(0)=0^3-3\times 0=0
\displaystyle \text{Hence Rolle's theorem applies, so } \exists c\in(-\sqrt{3},0) \text{ such that } f'(c)=0
\displaystyle f'(x)=3x^2-3
\displaystyle f'(c)=0 \Rightarrow 3c^2-3=0 \Rightarrow c^2=1
\displaystyle \Rightarrow c=\pm 1
\displaystyle \text{Since } c\in(-\sqrt{3},0),\ c=-1

\displaystyle \textbf{7. } \text{The volume of a cube is increasing at the rate of } 9\ \text{cm}^3/\text{s. How fast is its surface}
\displaystyle \text{area increasing when the length of an edge is } 10\ \text{cm}?
\displaystyle \text{Answer:}
\displaystyle \text{Let } x=\text{length of side}
\displaystyle V=x^3,\ S=6x^2
\displaystyle \frac{dV}{dt}=9
\displaystyle \Rightarrow \frac{d}{dt}(x^3)=3x^2\frac{dx}{dt}=9
\displaystyle \Rightarrow \frac{dx}{dt}=\frac{3}{x^2}
\displaystyle \frac{dS}{dt}=12x\frac{dx}{dt}=12x\times \frac{3}{x^2}=\frac{36}{x}
\displaystyle \Rightarrow \frac{dS}{dt}=\frac{36}{10}=3.6\ \text{cm}^2/\text{s}

\displaystyle \textbf{8. } \text{Show that the function } f(x)=x^3-3x^2+6x-100 \text{ is increasing on } R.
\displaystyle \text{Answer:}
\displaystyle \text{Since } f(x)=x^3-3x^2+6x-100
\displaystyle \therefore f'(x)=3x^2-6x+6
\displaystyle =3(x^2-2x+2)
\displaystyle =3\left[(x-1)^2+1\right]>0 \ \forall x\in R
\displaystyle \therefore f'(x)>0 \ \forall x\in R
\displaystyle \text{Hence, the function is increasing on } R

\displaystyle \textbf{9. } \text{The } x\text{-coordinate of a point on the line joining the points } P(2,2,1) \text{ and } Q(5,1,-2)
\displaystyle \text{ is } 4.\ \text{Find its } z\text{-coordinate.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } R \text{ divide } PQ \text{ in the ratio } \lambda:1
\displaystyle \text{Coordinates of } R=\left(\frac{5\lambda+2}{\lambda+1},\frac{\lambda+2}{\lambda+1},\frac{-2\lambda+1}{\lambda+1}\right)
\displaystyle \text{Given } x\text{-coordinate of } R=4
\displaystyle \Rightarrow \frac{5\lambda+2}{\lambda+1}=4
\displaystyle \Rightarrow 5\lambda+2=4\lambda+4
\displaystyle \Rightarrow \lambda=2
\displaystyle \text{Then } z\text{-coordinate of } R=\frac{-2\lambda+1}{\lambda+1}=\frac{-4+1}{3}=-1
\displaystyle \text{Hence, } z=-1

\displaystyle \textbf{10. } \text{A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in } \text{green, is tossed. Let }
\displaystyle A \text{ be the event ``number obtained is even''} \text{and } B \text{ be the event ``number obtained is red''.}
\displaystyle \text{Find if } A \text{ and } B \text{ are independent events.}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes }=6
\displaystyle \text{Favourable outcomes for } A=\{2,4,6\} \Rightarrow n(A)=3
\displaystyle P(A)=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{Favourable outcomes for } B=\{1,2,3\} \Rightarrow n(B)=3
\displaystyle P(B)=\frac{3}{6}=\frac{1}{2}
\displaystyle P(A)P(B)=\frac{1}{4}
\displaystyle \text{Now } A\cap B=\{2\} \Rightarrow n(A\cap B)=1
\displaystyle P(A\cap B)=\frac{1}{6} \ne \frac{1}{4}
\displaystyle \text{Hence, } A \text{ and } B \text{ are not independent}

\displaystyle \textbf{11. } \text{Two tailors, A and B earn Rs 300 and Rs 400 per day respectively. A can stitch }
\displaystyle \text{6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers}
\displaystyle \text{ per day. To find how many days each of them should work and if it is desired to produce }
\displaystyle \text{at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this }
\displaystyle \text{as an LPP.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,B \text{ work for } x,y \text{ days respectively}
\displaystyle \text{Shirts: } 6x+10y \ge 60
\displaystyle \text{Trousers: } 4x+4y \ge 32
\displaystyle \text{Cost } Z=300x+400y
\displaystyle x,y \ge 0
\displaystyle \text{Minimise } Z=300x+400y
\displaystyle \text{Subject to } 6x+10y \ge 60,\ 4x+4y \ge 32,\ x,y \ge 0

\displaystyle \textbf{12. } \text{Find : } \int \frac{dx}{5-8x-x^2}
\displaystyle \text{Answer:}
\displaystyle  \int \frac{dx}{5-8x-x^2}
\displaystyle =\int \frac{dx}{5-(x^2+8x)}
\displaystyle =\int \frac{dx}{5-\left(x^2+8x+16-16\right)}
\displaystyle =\int \frac{dx}{21-(x+4)^2}
\displaystyle =\int \frac{dx}{(\sqrt{21})^2-(x+4)^2}
\displaystyle =\frac{1}{2\sqrt{21}}\log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Questions number 13 to 23 carry 4 Mark each.}

\displaystyle \textbf{13. } \text{If } \tan^{-1}\left(\frac{x-3}{x-4}\right)+\tan^{-1}\left(\frac{x+3}{x+4}\right)=\frac{\pi}{4},  \text{then find the value of } x.
\displaystyle \text{Answer:}
\displaystyle \text{Since } \tan^{-1}\left(\frac{x-3}{x-4}\right)+\tan^{-1}\left(\frac{x+3}{x+4}\right)=\frac{\pi}{4}
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x-3}{x-4}\right)=\frac{\pi}{4}-\tan^{-1}\left(\frac{x+3}{x+4}\right)
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x-3}{x-4}\right)=\tan^{-1}(1)-\tan^{-1}\left(\frac{x+3}{x+4}\right)
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x-3}{x-4}\right)=\tan^{-1}\left(\frac{1-\frac{x+3}{x+4}}{1+\frac{x+3}{x+4}}\right)
\displaystyle \text{Using } \tan^{-1}A-\tan^{-1}B=\tan^{-1}\left(\frac{A-B}{1+AB}\right)
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x-3}{x-4}\right)=\tan^{-1}\left(\frac{\frac{x+4-x-3}{x+4}}{\frac{x+4+x+3}{x+4}}\right)
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x-3}{x-4}\right)=\tan^{-1}\left(\frac{1}{2x+7}\right)
\displaystyle \Rightarrow \frac{x-3}{x-4}=\frac{1}{2x+7}
\displaystyle \Rightarrow (x-3)(2x+7)=x-4
\displaystyle \Rightarrow 2x^2+x-21=x-4
\displaystyle \Rightarrow 2x^2=17
\displaystyle \Rightarrow x=\pm \sqrt{\frac{17}{2}}

\displaystyle \textbf{14. } \text{Using properties of determinants, prove that}
\displaystyle \left|\begin{matrix} a^2+2a & 2a+1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{matrix}\right|=(a-1)^3
\displaystyle \text{OR}
\displaystyle \text{Find matrix } A \text{ such that}   \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} A = \begin{bmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{Let } \Delta=\left|\begin{matrix} a^2+2a & 2a+1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{matrix}\right|
\displaystyle \text{Applying } R_1\rightarrow R_1-R_3 \text{ and } R_2\rightarrow R_2-R_3
\displaystyle \Rightarrow \Delta=\left|\begin{matrix} a^2+2a-3 & 2a-2 & 0 \\ 2a-2 & a-1 & 0 \\ 3 & 3 & 1 \end{matrix}\right|
\displaystyle \Rightarrow \Delta=\left|\begin{matrix} (a+3)(a-1) & 2(a-1) & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \end{matrix}\right|
\displaystyle \Rightarrow \Delta=(a-1)^2\left|\begin{matrix} a+3 & 2 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{matrix}\right|
\displaystyle \text{Expanding along } C_3
\displaystyle \Rightarrow \Delta=(a-1)^2[0-0+1(a+3-4)]
\displaystyle \Rightarrow \Delta=(a-1)^3
\displaystyle \text{OR}
\displaystyle \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}A=\begin{bmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{bmatrix}
\displaystyle \text{Here } 3\times 2 \text{ matrix is multiplied with } A \text{ and the result is } 3\times 2
\displaystyle \text{matrix, so } A \text{ is a } 2\times 2 \text{ matrix}
\displaystyle \text{Let } A=\begin{bmatrix} x & y \\ a & b \end{bmatrix}
\displaystyle \therefore \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} x & y \\ a & b \end{bmatrix}=\begin{bmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix} 2x-a & 2y-b \\ x & y \\ -3x+4a & -3y+4b \end{bmatrix}=\begin{bmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle 2x-a=-1
\displaystyle 2y-b=-8
\displaystyle x=1
\displaystyle y=-2
\displaystyle \text{On solving, we get } a=3 \text{ and } b=4
\displaystyle \therefore A=\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}

\displaystyle \textbf{15. } \text{If } x^y+y^x=a^b,\ \text{then find } \frac{dy}{dx}.
\displaystyle \text{OR}
\displaystyle \text{If } e^y(x+1)=1,\ \text{then show that } \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let } u=x^y \text{ and } v=y^x
\displaystyle \text{Then } u+v=a^b
\displaystyle \text{Differentiating w.r.t. } x
\displaystyle \frac{du}{dx}+\frac{dv}{dx}=0 \quad ...(i)
\displaystyle \text{Now } u=x^y \Rightarrow \log u=y\log x
\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx}=y\frac{1}{x}+\log x\frac{dy}{dx}
\displaystyle \Rightarrow \frac{du}{dx}=x^y\left(\frac{y}{x}+\log x\frac{dy}{dx}\right)
\displaystyle \Rightarrow \frac{du}{dx}=x^{y-1}\left(y+x\log x\frac{dy}{dx}\right) \quad ...(ii)
\displaystyle \text{Now } v=y^x \Rightarrow \log v=x\log y
\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx}=x\frac{1}{y}\frac{dy}{dx}+\log y
\displaystyle \Rightarrow \frac{dv}{dx}=y^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right)
\displaystyle \Rightarrow \frac{dv}{dx}=y^{x-1}\left(x\frac{dy}{dx}+y\log y\right) \quad ...(iii)
\displaystyle \text{From (i), (ii) and (iii)}
\displaystyle x^{y-1}\left(y+x\log x\frac{dy}{dx}\right)+y^{x-1}\left(x\frac{dy}{dx}+y\log y\right)=0
\displaystyle \Rightarrow x^{y-1}y+x^yx\log x\frac{dy}{dx}+y^{x-1}x\frac{dy}{dx}+y^x\log y=0
\displaystyle \Rightarrow \left(x^y\log x+xy^{x-1}\right)\frac{dy}{dx}=-(y^x\log y+x^{y-1}y)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-(y^x\log y+x^{y-1}y)}{x^y\log x+xy^{x-1}}
\displaystyle \text{OR}
\displaystyle \text{Since } e^y(x+1)=1
\displaystyle \Rightarrow x+1=e^{-y} \quad ...(i)
\displaystyle \text{Differentiating (i)}
\displaystyle 1=-e^{-y}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=-e^y \quad ...(ii)
\displaystyle \text{Differentiating (ii)}
\displaystyle \frac{d^2y}{dx^2}=-e^y\frac{dy}{dx}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\frac{dy}{dx}\cdot\frac{dy}{dx}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2

\displaystyle \textbf{16. } \text{Find : } \int \frac{\cos \theta}{(4+\sin^2 \theta)(5-4\cos^2 \theta)}\,d\theta
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\cos \theta}{(4+\sin^2\theta)(5-4\cos^2\theta)}\,d\theta
\displaystyle =\int \frac{\cos \theta}{(4+\sin^2\theta)(5-4(1-\sin^2\theta))}\,d\theta
\displaystyle =\int \frac{\cos \theta}{(4+\sin^2\theta)(1+4\sin^2\theta)}\,d\theta \quad ...(i)
\displaystyle \text{Let } \sin \theta=t \Rightarrow \cos \theta d\theta=dt
\displaystyle \Rightarrow I=\int \frac{dt}{(4+t^2)(1+4t^2)} \quad ...(ii)
\displaystyle \text{Let } t^2=y
\displaystyle \frac{1}{(4+y)(1+4y)}=\frac{A}{4+y}+\frac{B}{1+4y}
\displaystyle \Rightarrow 1=A(1+4y)+B(4+y)
\displaystyle \Rightarrow A=-\frac{1}{15},\ B=\frac{4}{15}
\displaystyle \Rightarrow I=\int \left(\frac{-1}{15(4+t^2)}+\frac{4}{15(1+4t^2)}\right)dt
\displaystyle =-\frac{1}{15}\int \frac{dt}{4+t^2}+\frac{4}{15}\int \frac{dt}{1+4t^2}
\displaystyle =-\frac{1}{15}\int \frac{dt}{4+t^2}+\frac{4}{15}\int \frac{dt}{4\left(\frac{1}{4}+t^2\right)}
\displaystyle =-\frac{1}{15}\int \frac{dt}{4+t^2}+\frac{1}{15}\int \frac{dt}{\frac{1}{4}+t^2}
\displaystyle =-\frac{1}{15}\left(\frac{1}{2}\tan^{-1}\frac{t}{2}\right)+\frac{1}{15}\left(2\tan^{-1}(2t)\right)+C
\displaystyle =-\frac{1}{30}\tan^{-1}\left(\frac{\sin\theta}{2}\right)+\frac{2}{15}\tan^{-1}(2\sin\theta)+C

\displaystyle \textbf{17. } \text{Evaluate : } \int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}\,dx
\displaystyle \text{OR}
\displaystyle \text{Evaluate : } \int_{1}^{4}\left\{|x-1|+|x-2|+|x-4|\right\}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}\,dx \quad ...(i)
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)}\,dx
\displaystyle \text{Using } \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{-(\pi-x)\tan x}{-(\sec x+\tan x)}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{(\pi-x)\tan x}{\sec x+\tan x}\,dx \quad ...(ii)
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi}\frac{\pi\tan x}{\sec x+\tan x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi}\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi}\frac{\sin x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi}1\,dx-\pi\int_{0}^{\pi}\frac{1}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi[x]_{0}^{\pi}-\pi\int_{0}^{\pi}\frac{1-\sin x}{\cos^2 x}\,dx
\displaystyle \Rightarrow 2I=\pi^2-\pi\int_{0}^{\pi}\left(\frac{1}{\cos^2 x}-\frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}\right)\,dx
\displaystyle \Rightarrow 2I=\pi^2-\pi\int_{0}^{\pi}(\sec^2 x-\tan x\sec x)\,dx
\displaystyle \Rightarrow 2I=\pi^2-\pi[\tan x-\sec x]_{0}^{\pi}
\displaystyle \Rightarrow 2I=\pi^2-\pi[\tan\pi-\sec\pi-\tan 0+\sec 0]
\displaystyle \Rightarrow 2I=\pi^2-\pi[0-(-1)-0+1]
\displaystyle \Rightarrow 2I=\pi^2-2\pi
\displaystyle \Rightarrow 2I=\pi(\pi-2)
\displaystyle \Rightarrow I=\frac{\pi}{2}(\pi-2)
\displaystyle \text{OR}
\displaystyle \text{Suppose } I=\int_{1}^{4}(|x-1|+|x-2|+|x-4|)\,dx
\displaystyle \Rightarrow I=\int_{1}^{4}|x-1|\,dx+\int_{1}^{4}|x-2|\,dx+\int_{1}^{4}|x-4|\,dx
\displaystyle \text{Now, } |x-1|=\begin{cases} -(x-1), & x\le 1 \\ x-1, & 1<x\le 4 \end{cases}
\displaystyle |x-2|=\begin{cases} -(x-2), & 1\le x\le 2 \\ x-2, & 2<x\le 4 \end{cases}
\displaystyle |x-4|=\begin{cases} -(x-4), & 1\le x\le 4 \\ x-4, & x>4 \end{cases}
\displaystyle \Rightarrow I=\int_{1}^{4}(x-1)\,dx-\int_{1}^{2}(x-2)\,dx+\int_{2}^{4}(x-2)\,dx-\int_{1}^{4}(x-4)\,dx
\displaystyle \Rightarrow I=\left[\frac{x^2}{2}-x\right]_{1}^{4}-\left[\frac{x^2}{2}-2x\right]_{1}^{2}+\left[\frac{x^2}{2}-2x\right]_{2}^{4}-\left[\frac{x^2}{2}-4x\right]_{1}^{4}
\displaystyle \Rightarrow I=\left[(8-4)-\left(\frac{1}{2}-1\right)\right]-\left[(2-4)-\left(\frac{1}{2}-2\right)\right]
\displaystyle \qquad +\left[(8-8)-(2-4)\right]-\left[(8-16)-\left(\frac{1}{2}-4\right)\right]
\displaystyle \Rightarrow I=\frac{9}{2}+\frac{1}{2}+2+\frac{9}{2}
\displaystyle \Rightarrow I=\frac{23}{2}

\displaystyle \textbf{18. } \text{Solve the differential equation } (\tan^{-1}x-y)\,dx=(1+x^2)\,dy.
\displaystyle \text{Answer:}
\displaystyle \text{Since } (\tan^{-1}x-y)\,dx=(1+x^2)\,dy
\displaystyle \Rightarrow (1+x^2)\frac{dy}{dx}+y=\tan^{-1}x
\displaystyle \Rightarrow \frac{dy}{dx}+\frac{y}{1+x^2}=\frac{\tan^{-1}x}{1+x^2} \quad ...(i)
\displaystyle \text{This is a linear differential equation of the form } \frac{dy}{dx}+Py=Q
\displaystyle \text{where } P=\frac{1}{1+x^2},\ Q=\frac{\tan^{-1}x}{1+x^2}
\displaystyle \text{I.F.}=e^{\int P\,dx}=e^{\int \frac{1}{1+x^2}\,dx}=e^{\tan^{-1}x}
\displaystyle \Rightarrow y\cdot e^{\tan^{-1}x}=\int \frac{\tan^{-1}x\ e^{\tan^{-1}x}}{1+x^2}\,dx+C \quad ...(ii)
\displaystyle \text{Let } I=\int \frac{\tan^{-1}x\ e^{\tan^{-1}x}}{1+x^2}\,dx
\displaystyle \text{Put } \tan^{-1}x=t \Rightarrow \frac{1}{1+x^2}\,dx=dt
\displaystyle \Rightarrow I=\int te^t\,dt
\displaystyle \Rightarrow I=t\int e^t\,dt-\int \frac{d}{dt}(t)\left(\int e^t\,dt\right)dt
\displaystyle \Rightarrow I=te^t-\int e^t\,dt
\displaystyle \Rightarrow I=(t-1)e^t
\displaystyle \Rightarrow I=(\tan^{-1}x-1)e^{\tan^{-1}x}
\displaystyle \text{Substituting in (ii), we get}
\displaystyle y\cdot e^{\tan^{-1}x}=(\tan^{-1}x-1)e^{\tan^{-1}x}+C
\displaystyle \Rightarrow y=\tan^{-1}x-1+Ce^{-\tan^{-1}x}
\displaystyle \text{Hence, the required solution is } y=\tan^{-1}x-1+Ce^{-\tan^{-1}x}

\displaystyle \textbf{19. } \text{Show that the points A, B, C with position vectors } 2\widehat{i}-\widehat{j}+\widehat{k}, \  \widehat{i}-3\widehat{j}-5\widehat{k}
\displaystyle \text{and }   3\widehat{i}-4\widehat{j}-4\widehat{k}\ \text{respectively, are the vertices of a right-angled triangle. }
\displaystyle \text{Hence find the area of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Since } \overrightarrow{OA}=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{OB}=\widehat{i}-3\widehat{j}-5\widehat{k} \text{ and }
\displaystyle \overrightarrow{OC}=3\widehat{i}-4\widehat{j}-4\widehat{k}
\displaystyle \text{Then } \overrightarrow{AB}=(1-2)\widehat{i}+(-3+1)\widehat{j}+(-5-1)\widehat{k}=-\widehat{i}-2\widehat{j}-6\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{AB}|=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}
\displaystyle \overrightarrow{BC}=(3-1)\widehat{i}+(-4+3)\widehat{j}+(-4+5)\widehat{k}=2\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{BC}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}
\displaystyle \overrightarrow{CA}=(2-3)\widehat{i}+(-1+4)\widehat{j}+(1+4)\widehat{k}=-\widehat{i}+3\widehat{j}+5\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{CA}|=\sqrt{(-1)^2+3^2+5^2}=\sqrt{35}
\displaystyle \text{Now } |\overrightarrow{BC}|^2+|\overrightarrow{CA}|^2=6+35=41=|\overrightarrow{AB}|^2
\displaystyle \text{Hence, } \triangle ABC \text{ is a right-angled triangle}
\displaystyle \text{Area of } \triangle ABC=\frac{1}{2}\times |\overrightarrow{BC}|\times |\overrightarrow{CA}|
\displaystyle =\frac{1}{2}\times \sqrt{6}\times \sqrt{35}=\sqrt{\frac{105}{2}}\ \text{sq. units}

\displaystyle \textbf{20. } \text{Find the value of } \lambda,\ \text{if four points with position vectors } 3\widehat{i}+6\widehat{j}+9\widehat{k},
\displaystyle \widehat{i}+2\widehat{j}+3\widehat{k},\ 2\widehat{i}+3\widehat{j}+\widehat{k}   \text{and } 4\widehat{i}+6\widehat{j}+\lambda \widehat{k}\ \text{are coplanar.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } \overrightarrow{OA}=3\widehat{i}+6\widehat{j}+9\widehat{k},\ \overrightarrow{OB}=\widehat{i}+2\widehat{j}+3\widehat{k},
\displaystyle \overrightarrow{OC}=2\widehat{i}+3\widehat{j}+\widehat{k},\ \overrightarrow{OD}=4\widehat{i}+6\widehat{j}+\lambda \widehat{k}
\displaystyle \overrightarrow{AB}=(1-3)\widehat{i}+(2-6)\widehat{j}+(3-9)\widehat{k}=-2\widehat{i}-4\widehat{j}-6\widehat{k}
\displaystyle \overrightarrow{AC}=(2-3)\widehat{i}+(3-6)\widehat{j}+(1-9)\widehat{k}=-\widehat{i}-3\widehat{j}-8\widehat{k}
\displaystyle \overrightarrow{AD}=(4-3)\widehat{i}+(6-6)\widehat{j}+(\lambda-9)\widehat{k}=\widehat{i}+(\lambda-9)\widehat{k}
\displaystyle \text{Since the four points are coplanar, } \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD} \text{ are coplanar}
\displaystyle \Rightarrow \left|\begin{matrix} -2 & -4 & -6 \\ -1 & -3 & -8 \\ 1 & 0 & \lambda-9 \end{matrix}\right|=0
\displaystyle \Rightarrow -2[-3(\lambda-9)-0]-(-4)[-\lambda+9+8]-6[3]=0
\displaystyle \Rightarrow 6\lambda-54-4\lambda+68-18=0
\displaystyle \Rightarrow 2\lambda-4=0
\displaystyle \Rightarrow \lambda=2

\displaystyle \textbf{21. } \text{There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards}
\displaystyle \text{are drawn at random without replacement. Let denote the sum of the numbers on } X
\displaystyle \text{the two drawn cards. Find the}  \text{mean and variance of } X.
\displaystyle \text{Answer:}
\displaystyle \text{Here } X \text{ can take the values } 4,6,8,10 \text{ and } 12
\displaystyle \text{Now, } P(X=4)=\text{Probability of getting 4 as sum}
\displaystyle =P[(\text{Getting } 1 \text{ in the first draw and } 3 \text{ in the second draw})
\displaystyle \text{or } (\text{Getting } 3 \text{ in the first draw and } 1 \text{ in the second draw})]
\displaystyle =\frac{1}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}=\frac{2}{12}=\frac{1}{6}
\displaystyle \text{Similarly,}
\displaystyle P(X=6)=\frac{1}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}=\frac{2}{12}=\frac{1}{6}
\displaystyle \text{Corresponding pairs are } (1,5),(5,1)
\displaystyle P(X=8)=\frac{1}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}=\frac{4}{12}=\frac{2}{6}
\displaystyle \text{Corresponding pairs are } (1,7),(7,1),(3,5),(5,3)
\displaystyle P(X=10)=\frac{1}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}=\frac{2}{12}=\frac{1}{6}
\displaystyle \text{Corresponding pairs are } (3,7),(7,3)
\displaystyle P(X=12)=\frac{1}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}=\frac{2}{12}=\frac{1}{6}
\displaystyle \text{Corresponding pairs are } (5,7),(7,5)
\displaystyle \text{Thus, the probability distribution of } X \text{ is given below.}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline X & 4 & 6 & 8 & 10 & 12 \\ \hline P(X) & \frac{1}{6} & \frac{1}{6} & \frac{2}{6} & \frac{1}{6} & \frac{1}{6} \\ \hline \end{array}
\displaystyle \sum p_i=\frac{1}{6}+\frac{1}{6}+\frac{2}{6}+\frac{1}{6}+\frac{1}{6}=\frac{6}{6}=1
\displaystyle \text{Now, } \sum p_i x_i=\frac{1}{6}\times 4+\frac{1}{6}\times 6+\frac{2}{6}\times 8+\frac{1}{6}\times 10+\frac{1}{6}\times 12
\displaystyle =\frac{1}{6}(4+6+16+10+12)
\displaystyle =\frac{1}{6}\times 48=8
\displaystyle \therefore \text{Mean } (E(X))=\frac{\sum p_i x_i}{\sum p_i}=\frac{8}{1}=8
\displaystyle \sum p_i x_i^2=\frac{1}{6}\times 16+\frac{1}{6}\times 36+\frac{2}{6}\times 64+\frac{1}{6}\times 100+\frac{1}{6}\times 144
\displaystyle =\frac{1}{6}(16+36+128+100+144)
\displaystyle =\frac{1}{6}\times 424=\frac{212}{3}
\displaystyle \text{Variance}=\sum p_i x_i^2-\left(\sum p_i x_i\right)^2=\sum p_i x_i^2-(\text{Mean})^2
\displaystyle =\frac{212}{3}-64=\frac{212-192}{3}=\frac{20}{3}

\displaystyle \textbf{22. } \text{Of the students in a school, it is known that 30\% have 100\% attendance and 70\% }
\displaystyle \text{students are irregular. Previous year results report that 70\% of all students who have}
\displaystyle \text{100\% attendance attain A grade and 10\% irregular students attain A grade in their }
\displaystyle \text{annual examination. At the end of the year, one student is chosen at random from the school }
\displaystyle \text{and he was  found to have an A grade. What is the probability that the student has }
\displaystyle \text{100\% attendance? Is regularity required only in school? Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1:\text{ the event that the student has 100\% attendance}
\displaystyle E_2:\text{ the event that the student does not have 100\% attendance}
\displaystyle A:\text{ the event that the student gets grade A}
\displaystyle \text{So, } P(E_1)=\frac{30}{100},\ P(E_2)=\frac{70}{100}
\displaystyle P\left(\frac{A}{E_1}\right)=\frac{70}{100},\ P\left(\frac{A}{E_2}\right)=\frac{10}{100}
\displaystyle \text{Using Bayes' theorem, required probability}
\displaystyle =\text{Probability that the student has 100\% attendance given that he was found to have grade A}
\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{30}{100}\times\frac{70}{100}}{\frac{30}{100}\times\frac{70}{100}+\frac{70}{100}\times\frac{10}{100}}=\frac{21}{28}=\frac{3}{4}
\displaystyle \text{No, regularity is not required only in school but also in colleges,}
\displaystyle \text{offices and even in day-to-day life. Suppose a person goes to work}
\displaystyle \text{every day and if he misses a day, he will feel lazy all day long.}
\displaystyle \text{Thus regularity is needed everywhere and in everything.}

\displaystyle \textbf{23. } \text{Maximise } Z=x+2y
\displaystyle \text{Subject to the constraints}
\displaystyle x+2y \ge 100
\displaystyle 2x-y \le 0
\displaystyle 2x+y \le 200
\displaystyle x,y \ge 0
\displaystyle \text{Solve the above LPP graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{Since } x+2y\geq 100,\ 2x+y\leq 200,\ 2x-y\leq 0,\ x\geq 0,\ y\geq 0\displaystyle \text{Converting the given inequalities into equations, we have}
\displaystyle x+2y=100,\ 2x+y=200,\ 2x-y=0,\ x=0,\ y=0
\displaystyle \text{The line } x+2y=100 \text{ meets the x-axis at } A_1(100,0)\ \text{and y-axis at } B_1(0,50)
\displaystyle \text{Join these points to obtain the line } x+2y=100
\displaystyle \text{It is clear that } (0,0) \text{ does not satisfy } x+2y\geq 100
\displaystyle \text{Thus, the region not containing the origin represents the solution set}
\displaystyle \text{of the inequality } x+2y\geq 100
\displaystyle \text{The line } 2x+y=200 \text{ meets the x-axis at } A_2(100,0)\ \text{and y-axis at } B_2(0,200)
\displaystyle \text{Join these points to obtain the line } 2x+y=200
\displaystyle \text{It is clear that } (0,0) \text{ satisfies } 2x+y\leq 200
\displaystyle \text{Thus, the region containing the origin represents the solution set}
\displaystyle \text{of the inequality } 2x+y\leq 200
\displaystyle \text{The line } 2x-y=0 \text{ is the line passing through the origin}
\displaystyle \text{Point of intersection of } 2x-y=0 \text{ and } 2x+y=200 \text{ is } R(50,100)
\displaystyle \text{Point of intersection of } 2x-y=0 \text{ and } x+2y=100 \text{ is } Q(20,40)
\displaystyle \text{For } x\geq 0,\ y\geq 0,\ \text{the first quadrant is the region represented}
\displaystyle \text{by } x\geq 0 \text{ and } y\geq 0
\displaystyle \text{The feasible region determined by the system of constraints is shown below}
\displaystyle \text{The corner points of the feasible region are } B_1(0,50),\ B_2(0,200),\ R(50,100)\ \text{and } Q(20,40)
\displaystyle \text{The values of } Z \text{ at these corner points are as follows}
\displaystyle \text{For } B_1(0,50):\ Z=x+2y=0+2\times 50=100
\displaystyle \text{For } B_2(0,200):\ Z=0+2\times 200=400\ (\text{maximum})
\displaystyle \text{For } R(50,100):\ Z=50+2\times 100=250
\displaystyle \text{For } Q(20,40):\ Z=20+2\times 40=100
\displaystyle \text{Thus, the maximum value of } Z \text{ is } 400\ \text{at } (0,200)

\displaystyle \textbf{SECTION - D}
\displaystyle \text{Questions number 24 to 29 carry 6 Mark each.}

\displaystyle \textbf{24. } \text{Determine the product } \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \text{and use it to solve the}
\displaystyle \text{system of equations } x-y+z=4, \  x-2y-2z=9,\ 2x+y+3z=1.
\displaystyle \text{Answer:}
\displaystyle \textbf{Let } A=\begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix},\ B=\begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}
\displaystyle BA=\begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix}
\displaystyle BA=\begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix}=8I_3
\displaystyle \Rightarrow \frac{1}{8}BA=I_3 \Rightarrow \left(\frac{1}{8}B\right)A=I_3
\displaystyle \Rightarrow A^{-1}=\frac{1}{8}B
\displaystyle A^{-1}=\frac{1}{8}\begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}
\displaystyle \text{The given system can be written as } AX=D
\displaystyle A=\begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix},\ X=\begin{bmatrix} x \\ y \\ z \end{bmatrix},\ D=\begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix}
\displaystyle \Rightarrow X=A^{-1}D=\frac{1}{8} \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix}
\displaystyle X=\frac{1}{8} \begin{bmatrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{bmatrix} =\frac{1}{8}\begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix}
\displaystyle X=\begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix}
\displaystyle \therefore x=3,\ y=-2,\ z=-1

\displaystyle \textbf{25. } \text{Consider } f:R-\left\{-\frac{4}{3}\right\}\rightarrow R-\left\{\frac{4}{3}\right\} \text{ given by } f(x)=\frac{4x+3}{3x+4}. \text{ Show that }
\displaystyle f \text{ is bijective. Find the inverse of } f \text{ and hence find } f^{-1}(0) \text{ and } x \text{ such that }
\displaystyle f^{-1}(x)=2.
\displaystyle \text{OR}
\displaystyle \text{Let } A=Q \times Q \text{ and let } * \text{ be a binary operation on } A \text{ defined by}
\displaystyle  (a,b)*(c,d)=(ac,b+ad) \text{ for } (a,b),\ (c,d) \in A. \text{ Determine, } \text{whether } *
\displaystyle \text{ is commutative and associative. Then, with respect to}   * \text{ on } A
\displaystyle \text{(i) find the identity element in } A
\displaystyle \text{(ii) find the invertible elements of } A
\displaystyle \text{Answer:}
\displaystyle \text{The function } f:R-\left\{-\frac{4}{3}\right\}\rightarrow R-\left\{\frac{4}{3}\right\} \text{ is given by}
\displaystyle f(x)=\frac{4x+3}{3x+4}
\displaystyle \text{Injectivity: Let } x,y\in R-\left\{-\frac{4}{3}\right\} \text{ be such that } f(x)=f(y)
\displaystyle \Rightarrow \frac{4x+3}{3x+4}=\frac{4y+3}{3y+4}
\displaystyle \Rightarrow (4x+3)(3y+4)=(4y+3)(3x+4)
\displaystyle \Rightarrow 12xy+16x+9y+12=12xy+16y+9x+12
\displaystyle \Rightarrow 7x=7y
\displaystyle \Rightarrow x=y
\displaystyle \text{Hence, } f \text{ is one-one}
\displaystyle \text{Surjectivity: Let } y\in R-\left\{\frac{4}{3}\right\}
\displaystyle \text{Then } \frac{4x+3}{3x+4}=y
\displaystyle \Rightarrow 4x+3=3xy+4y
\displaystyle \Rightarrow 4x-3xy=4y-3
\displaystyle \Rightarrow x(4-3y)=4y-3
\displaystyle \Rightarrow x=\frac{4y-3}{4-3y}
\displaystyle \text{Now } x\in R-\left\{-\frac{4}{3}\right\}
\displaystyle \text{Also } \frac{4y-3}{4-3y}\ne -\frac{4}{3} \text{, otherwise}
\displaystyle \frac{4y-3}{4-3y}=-\frac{4}{3}
\displaystyle \Rightarrow 12y-9=-16+12y
\displaystyle \Rightarrow 9=16,\ \text{which is impossible}
\displaystyle \text{Thus } x=\frac{4y-3}{4-3y}\in R-\left\{-\frac{4}{3}\right\} \text{ such that}
\displaystyle f(x)=f\left(\frac{4y-3}{4-3y}\right)=\frac{4\left(\frac{4y-3}{4-3y}\right)+3}{3\left(\frac{4y-3}{4-3y}\right)+4}
\displaystyle =\frac{16y-12+12-9y}{12y-9+16-12y}=\frac{7y}{7}=y
\displaystyle \text{So every element of } R-\left\{\frac{4}{3}\right\} \text{ has a pre-image in } R-\left\{-\frac{4}{3}\right\}
\displaystyle \text{Hence, } f \text{ is onto}
\displaystyle \text{Thus } f \text{ is bijective}
\displaystyle \text{Now, } x=\frac{4y-3}{4-3y}
\displaystyle \text{Replacing } x \text{ by } f^{-1}(x) \text{ and } y \text{ by } x,\ \text{we have}
\displaystyle f^{-1}(x)=\frac{4x-3}{4-3x}
\displaystyle \therefore f^{-1}(0)=\frac{4\times 0-3}{4-3\times 0}=-\frac{3}{4}
\displaystyle \text{Now, } f^{-1}(x)=2
\displaystyle \Rightarrow \frac{4x-3}{4-3x}=2
\displaystyle \Rightarrow 4x-3=8-6x
\displaystyle \Rightarrow 10x=11
\displaystyle \Rightarrow x=\frac{11}{10}
\displaystyle \text{OR}
\displaystyle \text{Since } A=Q\times Q \text{ and } * \text{ is defined by } (a,b)*(c,d)=(ac,b+ad)
\displaystyle \text{Commutativity: Let } X=(a,b),\ Y=(c,d)\in A,\ \forall a,c\in Q \text{ and } b,d\in Q
\displaystyle X*Y=(ac,b+ad)
\displaystyle Y*X=(ca,d+cb)
\displaystyle \text{Since } b+ad\ne d+cb \text{ in general, } X*Y\ne Y*X
\displaystyle \text{Therefore, } * \text{ is not commutative on } A
\displaystyle \text{Associativity: Let } X=(a,b),\ Y=(c,d),\ Z=(e,f)
\displaystyle X*(Y*Z)=(a,b)*(ce,d+cf)=(ace,b+ad+acf)
\displaystyle (X*Y)*Z=(ac,b+ad)*(e,f)=(ace,b+ad+acf)
\displaystyle \therefore X*(Y*Z)=(X*Y)*Z,\ \forall X,Y,Z\in A
\displaystyle \text{Thus, } * \text{ is associative on } A
\displaystyle \text{(i) Let } E=(x,y) \text{ be the identity element in } A \text{ with respect to } *
\displaystyle \text{Then } X*E=X \text{ and } E*X=X,\ \forall X=(a,b)\in A
\displaystyle \Rightarrow (a,b)*(x,y)=(a,b) \text{ and } (x,y)*(a,b)=(a,b)
\displaystyle \Rightarrow (ax,b+ay)=(a,b) \text{ and } (xa,y+xb)=(a,b)
\displaystyle \text{From } (ax,b+ay)=(a,b),\ ax=a \text{ and } b+ay=b
\displaystyle \Rightarrow x=1,\ y=0
\displaystyle \text{Also from } (xa,y+xb)=(a,b),\ xa=a \text{ and } y+xb=b
\displaystyle \Rightarrow x=1,\ y=0
\displaystyle \therefore (1,0) \text{ is the identity element in } A \text{ with respect to } *
\displaystyle \text{(ii) Let } F=(m,n) \text{ be the inverse of } (a,b)\in A
\displaystyle \text{Then } (a,b)*F=E \text{ and } F*(a,b)=E,\ \text{where } E=(1,0)
\displaystyle \Rightarrow (a,b)*(m,n)=(1,0) \text{ and } (m,n)*(a,b)=(1,0)
\displaystyle \Rightarrow (am,b+an)=(1,0) \text{ and } (ma,n+mb)=(1,0)
\displaystyle \text{From } (am,b+an)=(1,0),\ am=1 \text{ and } b+an=0
\displaystyle \Rightarrow m=\frac{1}{a},\ n=-\frac{b}{a}
\displaystyle \text{Hence, the inverse of } (a,b)\in A \text{ with respect to } * \text{ is } \left(\frac{1}{a},-\frac{b}{a}\right)

\displaystyle \textbf{26. } \text{Show that the surface area of a closed cuboid with square base and given}
\displaystyle \text{volume is minimum, when it is a cube.}
\displaystyle \text{Answer:}
\displaystyle \text{Let length = breadth = } x,\ \text{height = } y
\displaystyle \text{Surface area }=S \text{ and fixed volume }=V \text{ of a closed cuboid}
\displaystyle V=x^2y \quad ...(i)
\displaystyle \text{and } S=2(x^2+xy+xy)=2x^2+4xy \quad ...(ii)
\displaystyle \text{Now } S=2x^2+4xy
\displaystyle \Rightarrow S=2x^2+4x\left(\frac{V}{x^2}\right)
\displaystyle \Rightarrow S=2x^2+\frac{4V}{x}
\displaystyle \Rightarrow \frac{dS}{dx}=4x-\frac{4V}{x^2} \quad ...(iii)
\displaystyle \text{For maximum or minimum value, we have}
\displaystyle \Rightarrow \frac{dS}{dx}=0
\displaystyle \Rightarrow 4x-\frac{4V}{x^2}=0
\displaystyle \Rightarrow V=x^3
\displaystyle \Rightarrow x^2y=x^3
\displaystyle \Rightarrow x=y
\displaystyle \text{Differentiating equation (iii) with respect to } x
\displaystyle \Rightarrow \frac{d^2S}{dx^2}=4+\frac{8V}{x^3}=4+\frac{8x^2y}{x^3}=4+\frac{8y}{x}
\displaystyle \Rightarrow \left[\frac{d^2S}{dx^2}\right]_{y=x}=12>0
\displaystyle \text{Hence, } S \text{ is minimum when length = } x,\ \text{breadth = } x \text{ and height = } x,
\displaystyle \text{i.e., when it is a cube}

\displaystyle \textbf{27. } \text{Using the method of integration, find the area of the triangle} ABC,\ \text{ coordinates}
\displaystyle  \text{of whose vertices are } A(4,1),\ B(6,6) \text{ and } C(8,4).
\displaystyle \text{OR}
\displaystyle \text{Find the area enclosed between the parabola } 4y=3x^2 \text{ and } \text{the straight line }
\displaystyle 3x-2y+12=0.
\displaystyle \text{Answer:}
\displaystyle \text{The vertices of } \triangle ABC \text{ are } A(4,1),\ B(6,6) \text{ and } C(8,4)
\displaystyle \text{The equation of } AB \text{ is } y-1=\left(\frac{6-1}{6-4}\right)(x-4)
\displaystyle \Rightarrow 2y-2=5x-20
\displaystyle \Rightarrow 2y=5x-18
\displaystyle \Rightarrow y=\frac{5x-18}{2}=\frac{5x}{2}-9 \quad ...(i)
\displaystyle \text{The equation of } BC \text{ is } y-6=\left(\frac{4-6}{8-6}\right)(x-6)
\displaystyle \Rightarrow y-6=-(x-6)
\displaystyle \Rightarrow y=12-x \quad ...(ii)
\displaystyle \text{The equation of } CA \text{ is } y-4=\left(\frac{1-4}{4-8}\right)(x-8)
\displaystyle \Rightarrow 4y-16=3x-24
\displaystyle \Rightarrow 4y=3x-8
\displaystyle \Rightarrow y=\frac{3x-8}{4}=\frac{3x}{4}-2 \quad ...(iii)\displaystyle \text{Area of } \triangle ABC
\displaystyle =\text{Area of } ABFE+\text{Area of } BCGF-\text{Area of } ACGF
\displaystyle =\int_{4}^{6}\left(\frac{5x}{2}-9\right)dx+\int_{6}^{8}(12-x)\,dx-\int_{4}^{8}\left(\frac{3x}{4}-2\right)dx
\displaystyle =\left[\frac{5x^2}{4}-9x\right]_{4}^{6}+\left[12x-\frac{x^2}{2}\right]_{6}^{8}-\left[\frac{3x^2}{8}-2x\right]_{4}^{8}
\displaystyle =[(45-54)-(20-36)]+[(96-32)-(72-18)]-[(24-16)-(6-8)]
\displaystyle =(-9+16)+(64-54)-(8+2)
\displaystyle =7+10-10=7\ \text{square units}
\displaystyle \text{OR}
\displaystyle \text{The equations of the curve are}
\displaystyle 4y=3x^2 \quad ...(i)
\displaystyle 3x-2y+12=0 \quad ...(ii)
\displaystyle \text{The curve (i) represents a parabola having vertex at the origin,}
\displaystyle \text{axis along the positive direction of Y-axis and opening upwards}
\displaystyle \text{The curve (ii) represents a straight line}
\displaystyle \text{This straight line meets the x-axis at } (-4,0) \text{ and y-axis at } (0,6)
\displaystyle \text{Solving (i) and (ii), we get}
\displaystyle 3x-2\left(\frac{3x^2}{4}\right)+12=0
\displaystyle \Rightarrow 6x-3x^2+24=0
\displaystyle \Rightarrow x^2-2x-8=0
\displaystyle \Rightarrow (x+2)(x-4)=0
\displaystyle \Rightarrow x=-2,\ 4
\displaystyle \text{When } x=-2,\ y=3 \text{ and when } x=4,\ y=12\displaystyle \text{Therefore, the points of intersection of the given curves are } (-2,3) \text{ and } (4,12)
\displaystyle \text{Required area }=\text{Area of the shaded region}
\displaystyle =\int_{-2}^{4}\left(\frac{3x+12}{2}-\frac{3}{4}x^2\right)dx
\displaystyle =\left[\frac{3}{4}x^2+6x-\frac{x^3}{4}\right]_{-2}^{4}
\displaystyle =\left(\frac{3}{4}\times 16+6\times 4-\frac{64}{4}\right)-\left(\frac{3}{4}\times 4+6\times (-2)+\frac{8}{4}\right)
\displaystyle =20-(-7)
\displaystyle =27\ \text{square units}

\displaystyle \textbf{28. } \text{Find the particular solution of the differential equation}
\displaystyle (x-y)\frac{dy}{dx}=(x+2y),\ \text{given that } y=0 \text{ when } x=1.
\displaystyle \text{Answer:}
\displaystyle \text{Since } (x-y)\frac{dy}{dx}=x+2y
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-y}
\displaystyle \text{This is a homogeneous differential equation}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\ \text{we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x+2vx}{x-vx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1+2v}{1-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+2v}{1-v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+2v-v+v^2}{1-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v+v^2}{1-v}
\displaystyle \Rightarrow \frac{1-v}{1+v+v^2}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1-v}{1+v+v^2}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \frac{1}{2}\int \frac{2-2v}{v^2+v+1}\,dv=\int \frac{dx}{x}
\displaystyle \Rightarrow \int \frac{(2v+1)-3}{v^2+v+1}\,dv=-\int \frac{2\,dx}{x} \quad ...(i)
\displaystyle \text{Let } I_1=\int \frac{2v+1}{v^2+v+1}\,dv \text{ and } I_2=\int \frac{3}{v^2+v+1}\,dv
\displaystyle I_1=\int \frac{2v+1}{v^2+v+1}\,dv
\displaystyle \text{Let } v^2+v+1=t \Rightarrow (2v+1)\,dv=dt
\displaystyle \Rightarrow I_1=\int \frac{dt}{t}=\log |t|=\log |v^2+v+1|
\displaystyle \text{Also, } I_2=\int \frac{3}{v^2+v+1}\,dv
\displaystyle =\int \frac{3\,dv}{v^2+2v\times \frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1}
\displaystyle =3\int \frac{dv}{\left(v+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}
\displaystyle =3\left(\frac{2}{\sqrt{3}}\right)\tan^{-1}\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)
\displaystyle =2\sqrt{3}\tan^{-1}\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)
\displaystyle \text{Therefore from eqn (i), we have}
\displaystyle \log |v^2+v+1|-2\sqrt{3}\tan^{-1}\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=-2\log |x|+C
\displaystyle \text{Putting } v=\frac{y}{x} \text{ in the above equation, we get}
\displaystyle \log |x^2+y^2+xy|=2\sqrt{3}\tan^{-1}\left(\frac{x+2y}{x\sqrt{3}}\right)+C \quad ...(ii)
\displaystyle \text{At } y=0 \text{ and } x=1,\ \text{eqn. (ii) becomes}
\displaystyle \log |1|=2\sqrt{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)+C
\displaystyle \Rightarrow 0=2\sqrt{3}\times \frac{\pi}{6}+C
\displaystyle \Rightarrow C=-\frac{\pi}{\sqrt{3}}
\displaystyle \text{Putting } C=-\frac{\pi}{\sqrt{3}} \text{ in eqn. (ii), we have}
\displaystyle \log |x^2+y^2+xy|=2\sqrt{3}\tan^{-1}\left(\frac{x+2y}{x\sqrt{3}}\right)-\frac{\pi}{\sqrt{3}} \quad ...(iii)
\displaystyle \text{Hence eqn. (iii) is the required solution}

\displaystyle \textbf{29. } \text{Find the coordinates of the point where the line through the } \text{points } (3,-4,-5)
\displaystyle \text{ and } (2,-3,1) \text{ crosses the plane determined } \text{by the points } (1,2,3),\ (4,2,-3) \text{ and }
\displaystyle (0,4,3).
\displaystyle \text{OR}
\displaystyle \text{A variable plane which remains at a constant distance } 3p \text{ from the origin cuts the }
\displaystyle \text{coordinate axes at } A,\ B,\ C. \text{ Show } \text{that the locus of the centroid of triangle } ABC \text{ is}
\displaystyle \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}.
\displaystyle \text{Answer:}
\displaystyle \text{The cartesian equation of a line passing through two points } (x_1,y_1,z_1) \text{ and } (x_2,y_2,z_2) \\ \text{ is given as}
\displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}
\displaystyle \text{Therefore the equation of the line passing through } (3,-4,-5) \text{ and } (2,-3,1) \text{ is}
\displaystyle \frac{x-3}{2-3}=\frac{y-(-4)}{-3-(-4)}=\frac{z-(-5)}{1-(-5)}
\displaystyle \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}
\displaystyle \text{Now the coordinates of any point on this line are given by}
\displaystyle \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k
\displaystyle \Rightarrow x=3-k,\ y=k-4,\ z=6k-5,\ \text{where } k \text{ is a constant}
\displaystyle \text{Let } (3-k,k-4,6k-5) \text{ be the required point of intersection}
\displaystyle \text{Now, let the equation of a plane passing through } (1,2,3) \text{ be}
\displaystyle a(x-1)+b(y-2)+c(z-3)=0 \quad ...(i)
\displaystyle \text{where } a,b,c \text{ are the direction ratios of the normal to the plane}
\displaystyle \text{Since the plane (i) passes through } (4,2,-3) \text{ and } (0,4,3)
\displaystyle \therefore a(4-1)+b(2-2)+c(-3-3)=0
\displaystyle \Rightarrow 3a-6c=0 \quad ...(ii)
\displaystyle \text{and } a(0-1)+b(4-2)+c(3-3)=0
\displaystyle \Rightarrow -a+2b=0 \quad ...(iii)
\displaystyle \text{We can solve (ii) and (iii) using the method of cross multiplication}
\displaystyle \frac{a}{0+12}=\frac{b}{6-0}=\frac{c}{6+0}
\displaystyle \Rightarrow \frac{a}{12}=\frac{b}{6}=\frac{c}{6}
\displaystyle \Rightarrow \frac{a}{2}=b=c=\lambda \text{ (say)}
\displaystyle \Rightarrow a=2\lambda,\ b=\lambda,\ c=\lambda
\displaystyle \text{From (i), we get}
\displaystyle 2\lambda(x-1)+\lambda(y-2)+\lambda(z-3)=0
\displaystyle \Rightarrow 2x+y+z-7=0 \quad ...(iv)
\displaystyle \text{Putting } x=3-k,\ y=k-4,\ z=6k-5 \text{ in (iv), we get}
\displaystyle 2(3-k)+(k-4)+(6k-5)-7=0
\displaystyle \Rightarrow 5k-10=0
\displaystyle \Rightarrow k=2
\displaystyle \therefore \text{Required point of intersection is}
\displaystyle (3-k,k-4,6k-5)=(3-2,2-4,6\times 2-5)=(1,-2,7)
\displaystyle \text{OR}
\displaystyle \text{Suppose this plane meets the X, Y and Z axes at } A(a,0,0),\ B(0,b,0) \text{ and } C(0,0,c)
\displaystyle \text{Equation of plane is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \quad ...(i)
\displaystyle \text{Let the coordinates of the centroid of triangle } ABC \text{ be } (\alpha,\beta,\gamma)
\displaystyle \therefore \alpha=\frac{a+0+0}{3}=\frac{a}{3},\ \beta=\frac{0+b+0}{3}=\frac{b}{3},\ \gamma=\frac{0+0+c}{3}=\frac{c}{3}
\displaystyle \Rightarrow a=3\alpha,\ b=3\beta,\ c=3\gamma
\displaystyle \text{Since required plane is at a distance of } 3p \text{ from the origin}
\displaystyle \therefore 3p=\text{length of perpendicular from } (0,0,0) \text{ to plane (i)}
\displaystyle \Rightarrow 3p=\frac{\left|0+0+0-1\right|}{\sqrt{\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2}}
\displaystyle \Rightarrow 3p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}
\displaystyle \Rightarrow \sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\frac{1}{3p}
\displaystyle \text{Squaring both sides, we get}
\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{9p^2} \quad ...(ii)
\displaystyle \text{Substituting the values of } a,b \text{ and } c \text{ in (ii), we get}
\displaystyle \frac{1}{9\alpha^2}+\frac{1}{9\beta^2}+\frac{1}{9\gamma^2}=\frac{1}{9p^2}
\displaystyle \Rightarrow \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{1}{p^2}
\displaystyle \text{or } \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}
\displaystyle \text{Which is the required locus of the centroid of the triangle } ABC

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