Mathematics

\displaystyle \textbf{Time Allowed : 3 Hours \qquad 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 three Sections A, B and C.}
\displaystyle \text{Section A comprises of 10 questions of one mark each, Section B comprises of 12 questions}
\displaystyle \text{of four marks each and Section C comprises of 07 questions of six marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as per the}
\displaystyle \text{exact requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, internal choice has been provided in 04 questions}
\displaystyle \text{of four marks each and 02 questions 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 logarithmic tables, if required.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question numbers 1 to 10 carry 1 mark each.}

\displaystyle \textbf{1. } \text{Write the principal value of } \tan^{-1}(1)+\cos^{-1}\left(-\frac{1}{2}\right).
\displaystyle \text{Answer:}
\displaystyle  \tan^{-1}(1)+\cos^{-1}\left(-\frac{1}{2}\right)
\displaystyle =\tan^{-1}\left(\tan\frac{\pi}{4}\right)+\cos^{-1}\left(\cos\left(\pi-\frac{\pi}{3}\right)\right)
\displaystyle =\frac{\pi}{4}+\cos^{-1}\left(\cos\frac{2\pi}{3}\right) \ \ \ \ \ \ \ \ \ \  \left[ \because \frac{\pi}{4}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\text{ and }\frac{2\pi}{3}\in[0,\pi]\right]
\displaystyle =\frac{\pi}{4}+\frac{2\pi}{3}=\frac{3\pi+8\pi}{12}=\frac{11\pi}{12}

\displaystyle \textbf{2. } \text{Write the value of } \tan\left(2\tan^{-1}\frac{1}{5}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let } 2\tan^{-1}\frac{1}{5}=\theta
\displaystyle \tan^{-1}\frac{1}{5}=\frac{\theta}{2}\Rightarrow \tan\frac{\theta}{2}=\frac{1}{5}
\displaystyle \tan\left(2\tan^{-1}\frac{1}{5}\right)=\tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}
\displaystyle =\frac{2\cdot\frac{1}{5}}{1-\left(\frac{1}{5}\right)^2}=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{\frac{2}{5}}{\frac{24}{25}}=\frac{2}{5}\cdot\frac{25}{24}=\frac{5}{12}

\displaystyle \textbf{3. } \text{Find the value of } a \text{ if } \begin{bmatrix}a-b & 2a+c \\ 2a-b & 3c+d\end{bmatrix}=\begin{bmatrix}-1 & 5 \\ 0 & 13\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle  \begin{bmatrix}a-b & 2a+c \\ 2a-b & 3c+d\end{bmatrix}=\begin{bmatrix}-1 & 5 \\ 0 & 13\end{bmatrix}
\displaystyle \text{On comparing elements of both the sides}
\displaystyle a-b=-1 \qquad (i)
\displaystyle 2a+c=5 \qquad (ii)
\displaystyle 2a-b=0 \qquad (iii)
\displaystyle 3c+d=13 \qquad (iv)
\displaystyle \text{From (iii)}\ b=2a\Rightarrow a=\frac{b}{2}
\displaystyle \text{Putting value of } a \text{ in (i)}
\displaystyle \frac{b}{2}-b=-1\Rightarrow b=2\Rightarrow a=1
\displaystyle \text{From (ii)}\ 2a+c=5\Rightarrow c=5-2(1)=3
\displaystyle \text{From (iv)}\ d=13-3c=13-9=4

\displaystyle \textbf{4. } \text{If } \begin{vmatrix}x+1 & x-1 \\ x-3 & x+2\end{vmatrix}=\begin{vmatrix}4 & -1 \\ 1 & 3\end{vmatrix}, \text{ then write the value of } x.
\displaystyle \text{Answer:}
\displaystyle  \begin{vmatrix}x+1 & x-1 \\ x-3 & x+2\end{vmatrix}=\begin{vmatrix}4 & -1 \\ 1 & 3\end{vmatrix}
\displaystyle (x+1)(x+2)-(x-1)(x-3)=12+1
\displaystyle x^2+2x+x+2-(x^2-3x-x+3)=13
\displaystyle 7x-1=13\Rightarrow 7x=14\Rightarrow x=2

\displaystyle \textbf{5. } \text{If } \begin{bmatrix}9 & -1 & 4 \\ -2 & 1 & 3\end{bmatrix}=A+\begin{bmatrix}1 & 2 & -1 \\ 0 & 4 & 9\end{bmatrix}, \text{ then find the matrix } A.
\displaystyle \text{Answer:}
\displaystyle  \begin{bmatrix}9 & -1 & 4 \\ -2 & 1 & 3\end{bmatrix}=A+\begin{bmatrix}1 & 2 & -1 \\ 0 & 4 & 9\end{bmatrix}
\displaystyle A=\begin{bmatrix}9 & -1 & 4 \\ -2 & 1 & 3\end{bmatrix}-\begin{bmatrix}1 & 2 & -1 \\ 0 & 4 & 9\end{bmatrix}  =\begin{bmatrix}8 & -3 & 5 \\ -2 & -3 & -6\end{bmatrix}

\displaystyle \textbf{6. } \text{Write the degree of the differential equation } x^3\left(\frac{d^2y}{dx^2}\right)^2+x\left(\frac{dy}{dx}\right)^4=0.
\displaystyle \text{Answer:}
\displaystyle  x^3\left(\frac{d^2y}{dx^2}\right)^2+x\left(\frac{dy}{dx}\right)^4=0
\displaystyle \text{Degree }=2

\displaystyle \textbf{7. } \text{If } \overrightarrow{a}=x\widehat{i}+2\widehat{j}-z\widehat{k} \text{ and } \overrightarrow{b}=3\widehat{i}-y\widehat{j}+\widehat{k} \text{ are two equal vectors, then write the value of }
\displaystyle x+y+z.
\displaystyle \text{Answer:}
\displaystyle \text{ As } \overrightarrow{a}=\overrightarrow{b}
\displaystyle x\widehat{i}+2\widehat{j}-z\widehat{k}=3\widehat{i}-y\widehat{j}+\widehat{k}
\displaystyle \text{On equating both of the sides}
\displaystyle x=3,\ y=-2,\ z=-1
\displaystyle \therefore x+y+z=3-2-1=0

\displaystyle \textbf{8. } \text{If a unit vector } \overrightarrow{a} \text{ makes angles } \frac{\pi}{3} \text{ with } \widehat{i},\ \frac{\pi}{4} \text{ with } \widehat{j} \text{ and} \text{an acute angle } \theta \text{ with } \widehat{k},\
\displaystyle\text{then find the value of } \theta.
\displaystyle \text{Answer:}
\displaystyle \text{Let } l,m,n \text{ be direction cosines of } \overrightarrow{a}
\displaystyle l=\cos\frac{\pi}{3}\Rightarrow l=\frac{1}{2}
\displaystyle m=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}
\displaystyle n=\cos\theta
\displaystyle l^2+m^2+n^2=1
\displaystyle \Rightarrow \left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos^2\theta=1
\displaystyle \Rightarrow \frac{1}{4}+\frac{1}{2}+\cos^2\theta=1
\displaystyle \Rightarrow \cos^2\theta=\frac{1}{4}
\displaystyle \Rightarrow \cos\theta=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{3}

\displaystyle \textbf{9. } \text{Find the Cartesian equation of the line which passes through the point } (-2,4,-5)
\displaystyle \text{and is parallel to the line } \frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}.
\displaystyle \text{Answer:}
\displaystyle \text{Given line:}
\displaystyle \frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}
\displaystyle \Rightarrow \frac{x-(-3)}{3}=\frac{y-4}{-5}=\frac{z-(-8)}{6}
\displaystyle \text{Direction ratios are } (3,-5,6)
\displaystyle \text{Required line passes through } (-2,4,-5) \text{ and is parallel to given line}
\displaystyle \text{So, equation of required line is}
\displaystyle \frac{x-(-2)}{3}=\frac{y-4}{-5}=\frac{z-(-5)}{6}
\displaystyle \Rightarrow \frac{x+2}{3}=\frac{4-y}{5}=\frac{z+5}{6}

\displaystyle \textbf{10. } \text{The amount of pollution content added in air in a city due to x-diesel }
\displaystyle \text{vehicles is given by} P(x)=0.005x^3+0.02x^2+30x. \text{ Find the marginal increase }
\displaystyle \text{in pollution content when  3 diesel vehicles are added and write which value is}
\displaystyle \text{indicated in the above question.}
\displaystyle \text{Answer:}
\displaystyle  P(x)=0.005x^3+0.02x^2+30x
\displaystyle \text{On differentiating}
\displaystyle P'(x)=0.005(3x^2)+0.02(2x)+30
\displaystyle =0.015x^2+0.04x+30
\displaystyle P'(3)=0.015(9)+0.04(3)+30=30.255
\displaystyle \text{This question indicates how increase in number of diesel vehicles increases air pollution,}
\displaystyle \text{which is harmful for living beings.}

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question numbers 11 to 22 carry 4 marks each.}

\displaystyle \textbf{11. } \text{Show that the function } f \text{ in } A=R-\left\{\frac{2}{3}\right\} \text{ defined as } f(x)=\frac{4x+3}{6x-4}
\displaystyle \text{is one-one and onto. Hence find } f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Let } x_1,x_2\in A
\displaystyle \text{Now } f(x_1)=f(x_2)
\displaystyle \frac{4x_1+3}{6x_1-4}=\frac{4x_2+3}{6x_2-4}
\displaystyle \Rightarrow (4x_1+3)(6x_2-4)=(4x_2+3)(6x_1-4)
\displaystyle \Rightarrow 24x_1x_2+18x_2-16x_1-12=24x_1x_2+18x_1-16x_2-12
\displaystyle \Rightarrow -34x_1=-34x_2 \Rightarrow x_1=x_2
\displaystyle \text{Hence } f \text{ is one-one function.}
\displaystyle \text{For onto,}
\displaystyle \text{Let } y=\frac{4x+3}{6x-4}\Rightarrow 6xy-4y=4x+3
\displaystyle \Rightarrow x=\frac{4y+3}{6y-4}
\displaystyle \Rightarrow \forall y \text{ in codomain } \exists x \text{ in domain } \left[x\neq \frac{2}{3}\right]
\displaystyle \Rightarrow f \text{ is onto function.}
\displaystyle \text{Thus } f \text{ is one-one onto function.}
\displaystyle \text{Also } f^{-1}(x)=\frac{4x+3}{6x-4}

\displaystyle \textbf{12. } \text{Find the value of the following:}
\displaystyle \tan\left[\frac{1}{2}\left(\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right)\right],\ |x|<1,\ y>0 \text{ and } xy<1.
\displaystyle \text{OR}
\displaystyle \text{Prove that: } \tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{5}\right)+\tan^{-1}\left(\frac{1}{8}\right)=\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle  \tan\left[\frac{1}{2}\left(\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right)\right]
\displaystyle =\tan\left[\frac{1}{2}\left(2\tan^{-1}x+2\tan^{-1}y\right)\right]
\displaystyle \Rightarrow \tan(\tan^{-1}x+\tan^{-1}y)
\displaystyle =\tan\left(\tan^{-1}\frac{x+y}{1-xy}\right)
\displaystyle =\frac{x+y}{1-xy}
\displaystyle \text{OR}
\displaystyle \tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{5}\right)+\tan^{-1}\left(\frac{1}{8}\right)
\displaystyle =\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2}\cdot\frac{1}{5}}\right)+\tan^{-1}\left(\frac{1}{8}\right)
\displaystyle =\tan^{-1}\left(\frac{\frac{7}{10}}{\frac{9}{10}}\right)+\tan^{-1}\left(\frac{1}{8}\right)
\displaystyle =\tan^{-1}\left(\frac{7}{9}\right)+\tan^{-1}\left(\frac{1}{8}\right)
\displaystyle =\tan^{-1}\left(\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{9}\cdot\frac{1}{8}}\right)
\displaystyle =\tan^{-1}\left(\frac{\frac{65}{72}}{\frac{65}{72}}\right)=\tan^{-1}(1)=\frac{\pi}{4}

\displaystyle \textbf{13. } \text{Using properties of determinants, prove the following:} \\
\displaystyle \begin{vmatrix}1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1\end{vmatrix}=(1-x^3)^2.
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}
\displaystyle \text{Let } \Delta=\begin{vmatrix}1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1\end{vmatrix}
\displaystyle \text{Applying } C_1\rightarrow C_1+C_2+C_3
\displaystyle \Delta=\begin{vmatrix}1+x+x^2 & x & x^2 \\ 1+x+x^2 & 1 & x \\ 1+x+x^2 & x^2 & 1\end{vmatrix}
\displaystyle \text{Taking } (1+x+x^2) \text{ common from } C_1
\displaystyle \Delta=(1+x+x^2)\begin{vmatrix}1 & x & x^2 \\ 1 & 1 & x \\ 1 & x^2 & 1\end{vmatrix}
\displaystyle \text{Applying } R_2\rightarrow R_2-R_1 \text{ and } R_3\rightarrow R_3-R_1
\displaystyle \Delta=(1+x+x^2)\begin{vmatrix}1 & x & x^2 \\ 0 & 1-x & x(1-x) \\ 0 & x(x-1) & 1-x^2\end{vmatrix}
\displaystyle =(1+x+x^2)(1-x)^2\begin{vmatrix}1 & x & x^2 \\ 0 & 1 & x \\ 0 & -x & 1+x\end{vmatrix}
\displaystyle \text{Expanding by } C_1
\displaystyle \Delta=(1+x+x^2)(1-x)^2(1+x+x^2)
\displaystyle \Delta=[(1-x)(1+x+x^2)]^2
\displaystyle \Delta=(1-x^3)^2

\displaystyle \textbf{14. } \text{Differentiate the following function with respect to } x:  \  (\log x)^x+x^{\log x}.
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=(\log x)^x+x^{\log x}
\displaystyle y=u+v \qquad \ldots (i)
\displaystyle u=(\log x)^x
\displaystyle \text{Taking log on both sides}
\displaystyle \log u=x\log(\log x)
\displaystyle \text{Differentiating on both sides w.r.t. } x
\displaystyle \frac{1}{u}\frac{du}{dx}=x\cdot\frac{1}{\log x}\cdot\frac{1}{x}+\log(\log x)
\displaystyle \Rightarrow \frac{du}{dx}=u\left[\frac{1}{\log x}+\log(\log x)\right]
\displaystyle \Rightarrow \frac{du}{dx}=(\log x)^x\left[\frac{1}{\log x}+\log(\log x)\right] \qquad \ldots (ii)
\displaystyle v=x^{\log x}
\displaystyle \text{Taking log on both sides}
\displaystyle \log v=\log x^{\log x}
\displaystyle \Rightarrow \log v=\log x\cdot\log x
\displaystyle \Rightarrow \log v=(\log x)^2
\displaystyle \text{Differentiating on both sides w.r.t. } x
\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx}=2\log x\cdot\frac{1}{x}
\displaystyle \Rightarrow \frac{dv}{dx}=2x^{\log x}\frac{\log x}{x} \qquad \ldots (iii)
\displaystyle \text{From (i)}
\displaystyle y=u+v
\displaystyle \text{Differentiating on both sides w.r.t. } x
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
\displaystyle \text{From (ii) and (iii)}
\displaystyle \frac{dy}{dx}=(\log x)^x\left[\frac{1}{\log x}+\log(\log x)\right]+\frac{2\log x\cdot x^{\log x}}{x}

\displaystyle \textbf{15. } \text{If } y=\log\left[x+\sqrt{x^2+a^2}\right], \text{ show that } (x^2+a^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0.
\displaystyle \text{Answer:}
\displaystyle y=\log\left[x+\sqrt{x^2+a^2}\right]
\displaystyle \text{Differentiating w.r.t. } x
\displaystyle \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\times\left[1+\frac{1}{2}(x^2+a^2)^{-1/2}(2x)\right]
\displaystyle =\frac{1}{x+\sqrt{x^2+a^2}}\times\left[1+\frac{x}{\sqrt{x^2+a^2}}\right]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\times\left[\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{x^2+a^2}}
\displaystyle \Rightarrow \sqrt{x^2+a^2}\frac{dy}{dx}=1
\displaystyle \text{Again differentiating both sides w.r.t. } x
\displaystyle \sqrt{x^2+a^2}\frac{d^2y}{dx^2}+\frac{1}{2}(x^2+a^2)^{-1/2}(2x)\frac{dy}{dx}=0
\displaystyle \Rightarrow \sqrt{x^2+a^2}\frac{d^2y}{dx^2}+\frac{x}{\sqrt{x^2+a^2}}\frac{dy}{dx}=0
\displaystyle \Rightarrow (x^2+a^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0

\displaystyle \textbf{16. } \text{Show that the function } f(x)=|x-3|,\ x\in R, \text{ is continuous but not } \\ \text{differentiable at } x=3.
\displaystyle \text{OR}
\displaystyle \text{If } x=a\sin t \text{ and } y=a\left(\cos t+\log\tan\frac{t}{2}\right), \text{ find } \frac{d^2y}{dx^2}.
\displaystyle \text{Answer:}
\displaystyle  f(x)=|x-3|,\ x\in R
\displaystyle f(x)=\begin{cases}-(x-3), & x<3 \\ 0, & x=3 \\ x-3, & x>3 \end{cases}
\displaystyle \text{Now, } \lim_{x\to 3^+}f(x)=\lim_{h\to 0}f(3+h)
\displaystyle \left[\text{Let } x=3+h \text{ and } x\to 3^+ \Rightarrow h\to 0\right]
\displaystyle =\lim_{h\to 0}(3+h-3)=\lim_{h\to 0}h=0
\displaystyle \therefore\ \lim_{x\to 3^+}f(x)=0 \qquad \ldots (i)
\displaystyle \lim_{x\to 3^-}f(x)=\lim_{h\to 0}f(3-h)
\displaystyle \left[\text{Let } x=3-h \text{ and } x\to 3^- \Rightarrow h\to 0\right]
\displaystyle =\lim_{h\to 0}-(3-h-3)=\lim_{h\to 0}h=0
\displaystyle \therefore\ \lim_{x\to 3^-}f(x)=0 \qquad \ldots (ii)
\displaystyle \text{Also } f(3)=0 \qquad \ldots (iii)
\displaystyle \text{From equations (i), (ii) and (iii)}
\displaystyle \lim_{x\to 3^+}f(x)=\lim_{x\to 3^-}f(x)=f(3)
\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=3
\displaystyle \text{At } x=3
\displaystyle \text{R.H.D.}=\lim_{h\to 0}\frac{f(3+h)-f(3)}{h}=\lim_{h\to 0}\frac{|3+h-3|-0}{h}
\displaystyle =\lim_{h\to 0}\frac{|h|}{h}=1 \qquad \left[\because\ |h|=h,\ |0|=0\right]
\displaystyle \therefore\ \text{R.H.D.}=1 \qquad \ldots (iv)
\displaystyle \text{L.H.D.}=\lim_{h\to 0}\frac{f(3-h)-f(3)}{-h}=\lim_{h\to 0}\frac{|3-h-3|-0}{-h}
\displaystyle =\lim_{h\to 0}\frac{|h|}{-h}=-1 \qquad \left[\because\ |h|=h\right] \qquad \ldots (v)
\displaystyle \text{Equations (iv) and (v)}\Rightarrow \text{L.H.D.}\neq \text{R.H.D. at } x=3
\displaystyle \therefore\ f(x) \text{ is not differentiable at } x=3
\displaystyle \therefore\ f(x)=|x-3|,\ x\in R \text{ is continuous but not differentiable at } x=3

\displaystyle \textbf{17. } \text{Evaluate: } \int \frac{\sin(x-a)}{\sin(x+a)}\,dx
\displaystyle \text{OR}
\displaystyle \text{Evaluate: } \int \frac{5x-2}{1+2x+3x^2}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int\frac{\sin(x-a)}{\sin(x+a)}\,dx=\int\frac{\sin(x+a-2a)}{\sin(x+a)}\,dx
\displaystyle =\int\frac{\sin(x+a)\cos 2a-\sin 2a\cos(x+a)}{\sin(x+a)}\,dx
\displaystyle \left[\because\ \sin(A-B)=\sin A\cos B-\cos A\sin B\right]
\displaystyle I=\int\left[\frac{\sin(x+a)\cos 2a}{\sin(x+a)}-\frac{\sin 2a\cos(x+a)}{\sin(x+a)}\right]dx
\displaystyle I=\int\cos 2a\,dx-\int\sin 2a\cot(x+a)\,dx
\displaystyle I=x\cos 2a-\sin 2a\log\sin(x+a)+C
\displaystyle \left[\because\ \int \cot x\,dx=\log\sin x+C\right]
\displaystyle \text{OR}
\displaystyle \text{Let } I=\int\frac{5x-2}{1+2x+3x^2}\,dx
\displaystyle \text{Let } 5x-2=A+B\frac{d}{dx}(1+2x+3x^2)
\displaystyle 5x-2=A+B(6x+2)
\displaystyle 5x-2=(6B)x+A+2B
\displaystyle A=-\frac{11}{3} \text{ and } B=\frac{5}{6}
\displaystyle I=\int\frac{A\,dx}{3x^2+2x+1}+B\int\frac{(6x+2)\,dx}{3x^2+2x+1}
\displaystyle I=-\frac{11}{3}\int\frac{dx}{3x^2+2x+1}+\frac{5}{6}\int\frac{6x+2}{3x^2+2x+1}\,dx
\displaystyle I=I_1+I_2
\displaystyle I_2=\frac{5}{6}\int\frac{6x+2}{3x^2+2x+1}\,dx
\displaystyle \text{Let } 3x^2+2x+1=t
\displaystyle (6x+2)\,dx=dt
\displaystyle I_2=\frac{5}{6}\int\frac{dt}{t}=\frac{5}{6}\log|t|+C_1
\displaystyle =\frac{5}{6}\log|3x^2+2x+1|+C_1
\displaystyle I_1=-\frac{11}{3}\int\frac{dx}{3x^2+2x+1}=-\frac{11}{9}\int\frac{dx}{x^2+\frac{2}{3}x+\frac{1}{3}}
\displaystyle =-\frac{11}{9}\int\frac{dx}{x^2+\frac{2}{3}x+\frac{1}{3}-\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2}
\displaystyle =-\frac{11}{9}\int\frac{dx}{\left(x+\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2}
\displaystyle =-\frac{11}{9}\left(\frac{1}{\frac{\sqrt{2}}{3}}\right)\tan^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)+C_2
\displaystyle =-\frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C_2
\displaystyle \text{Putting values of } I_1 \text{ and } I_2 \text{ in } I
\displaystyle I=\frac{5}{6}\log|3x^2+2x+1|-\frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C
\displaystyle \left[\text{where } C=C_1+C_2\right]

\displaystyle \textbf{18. } \text{Evaluate: } \int \frac{x^2}{(x^2+4)(x^2+9)}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int\frac{x^2}{(x^2+4)(x^2+9)}\,dx
\displaystyle \text{Let } x^2=t
\displaystyle \Rightarrow \frac{x^2}{(x^2+4)(x^2+9)}=\frac{t}{(t+4)(t+9)}
\displaystyle \Rightarrow \frac{t}{(t+4)(t+9)}=\frac{A}{t+4}+\frac{B}{t+9}
\displaystyle \Rightarrow t=A(t+9)+B(t+4)
\displaystyle \text{Put } t=-9,\ B=\frac{9}{5}
\displaystyle \text{Put } t=-4,\ A=-\frac{4}{5}
\displaystyle \therefore\ \frac{x^2}{(x^2+4)(x^2+9)}=-\frac{4}{5(x^2+4)}+\frac{9}{5(x^2+9)}
\displaystyle I=-\frac{4}{5}\int\frac{dx}{x^2+2^2}+\frac{9}{5}\int\frac{dx}{x^2+3^2}
\displaystyle I=-\frac{4}{5}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}+\frac{9}{5}\cdot\frac{1}{3}\tan^{-1}\frac{x}{3}+C
\displaystyle I=-\frac{2}{5}\tan^{-1}\frac{x}{2}+\frac{3}{5}\tan^{-1}\frac{x}{3}+C

\displaystyle \textbf{19. Evaluate: } \int_{0}^{4}(|x|+|x-2|+|x-4|)\,dx
\displaystyle \text{Answer:}
\displaystyle  \text{Let } I=\int_{0}^{4}(|x|+|x-2|+|x-4|)\,dx
\displaystyle I=\int_{0}^{4}|x|\,dx+\int_{0}^{4}|x-2|\,dx+\int_{0}^{4}|x-4|\,dx
\displaystyle =\int_{0}^{4}x\,dx+\int_{0}^{4}|x-2|\,dx+\int_{0}^{4}(4-x)\,dx
\displaystyle |x-2|=\begin{cases}-(x-2), & \text{if } 0\leq x<2 \\ x-2, & \text{if } 2\leq x\leq 4 \end{cases}
\displaystyle |x|=x \text{ if } 0\leq x\leq 4
\displaystyle |x-4|=-(x-4) \text{ if } 0\leq x\leq 4
\displaystyle I=\int_{0}^{4}x\,dx+\int_{0}^{2}-(x-2)\,dx+\int_{2}^{4}(x-2)\,dx+\int_{0}^{4}-(x-4)\,dx
\displaystyle =\left[\frac{x^2}{2}\right]_{0}^{4}-\left[\frac{x^2}{2}-2x\right]_{0}^{2}+\left[\frac{x^2}{2}-2x\right]_{2}^{4}-\left[\frac{x^2}{2}-4x\right]_{0}^{4}
\displaystyle =\frac{1}{2}[16]-\left[\frac{1}{2}\times 4-4\right]+\left[\frac{1}{2}\times 16-8-\frac{4}{2}+4\right]+\left[\frac{16}{2}-16\right]
\displaystyle =8+2+2+8=20

\displaystyle \textbf{20. } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are two vectors such that } |\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|, \text{ then prove that vector }
\displaystyle 2\overrightarrow{a}+\overrightarrow{b} \text{ is perpendicular to vector } \overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } |\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}|
\displaystyle \text{Squaring on both sides}
\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}|^2
\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=|\overrightarrow{a}|^2
\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}=|\overrightarrow{a}|^2
\displaystyle \Rightarrow |\overrightarrow{a}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2=|\overrightarrow{a}|^2
\displaystyle \Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^2=0
\displaystyle \Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{b}=0
\displaystyle \Rightarrow (2\overrightarrow{a}+\overrightarrow{b})\cdot\overrightarrow{b}=0
\displaystyle \text{Hence } (2\overrightarrow{a}+\overrightarrow{b}) \text{ is perpendicular to } \overrightarrow{b}

\displaystyle \textbf{21. } \text{Find the coordinates of the point, where the line } \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}
\displaystyle \text{intersects the plane } x-y+z-5=0. \text{ Also find the angle between the line and the plane.}
\displaystyle \text{OR}
\displaystyle \text{Find the vector equation of the plane which contains the line of intersection of the planes}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0 \text{ and } \overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}-\widehat{k})+5=0 \text{and which is perpendicular to the plane}
\displaystyle \overrightarrow{r}\cdot(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let the given line }
\displaystyle \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2} \qquad \ldots (i)
\displaystyle \text{intersect the plane } x-y+z-5=0 \text{ at point } P(\alpha,\beta,\gamma)
\displaystyle \therefore\ P(\alpha,\beta,\gamma) \text{ lies on line (i)}
\displaystyle \frac{\alpha-2}{3}=\frac{\beta+1}{4}=\frac{\gamma-2}{2}=\lambda
\displaystyle \text{Let } \frac{\alpha-2}{3}=\frac{\beta+1}{4}=\frac{\gamma-2}{2}=\lambda
\displaystyle \Rightarrow \alpha=3\lambda+2,\ \beta=4\lambda-1,\ \gamma=2\lambda+2
\displaystyle \text{Also } P(\alpha,\beta,\gamma) \text{ lies on plane}
\displaystyle \Rightarrow (3\lambda+2)-(4\lambda-1)+(2\lambda+2)-5=0
\displaystyle \Rightarrow 3\lambda+2-4\lambda+1+2\lambda+2-5=0 \Rightarrow \lambda=0
\displaystyle \therefore\ \alpha=2,\ \beta=-1,\ \gamma=2
\displaystyle \text{Hence, coordinates of required point } (2,-1,2)
\displaystyle \text{Angle between line and plane is given as}
\displaystyle \sin\theta=\left|\frac{\overrightarrow{b}\cdot\overrightarrow{n}}{|\overrightarrow{b}|\cdot|\overrightarrow{n}|}\right|
\displaystyle \text{where } \theta \text{ is required angle}
\displaystyle \left[\because\ \overrightarrow{b}=3\widehat{i}+4\widehat{j}+2\widehat{k},\ \overrightarrow{n}=\widehat{i}-\widehat{j}+\widehat{k}\right]
\displaystyle \therefore\ \overrightarrow{b}\cdot\overrightarrow{n}=3-4+2=1
\displaystyle \Rightarrow \sin\theta=\left|\frac{1}{\sqrt{9+16+4}\sqrt{1^2+(-1)^2+1^2}}\right|
\displaystyle =\frac{1}{\sqrt{29}\sqrt{3}}=\frac{1}{\sqrt{87}}
\displaystyle \Rightarrow \theta=\sin^{-1}\frac{1}{\sqrt{87}}
\displaystyle \text{OR}
\displaystyle \text{Equation of the given planes are}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0 \qquad \ldots (i)
\displaystyle \text{and } \overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}-\widehat{k})+5=0 \qquad \ldots (ii)
\displaystyle \text{Let equation of the required plane be}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})-4+\lambda\left[\overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}-\widehat{k})+5\right]=0
\displaystyle \Rightarrow \overrightarrow{r}\cdot[(1+2\lambda)\widehat{i}+(2+\lambda)\widehat{j}+(3-\lambda)\widehat{k}]-4+5\lambda=0 \qquad \ldots (iii)
\displaystyle \text{Now, plane (iii) is perpendicular to plane}
\displaystyle \overrightarrow{r}\cdot(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0
\displaystyle \Rightarrow 5(1+2\lambda)+3(2+\lambda)-6(3-\lambda)=0 \Rightarrow \lambda=\frac{7}{19}
\displaystyle \text{From (iii), we have}
\displaystyle \overrightarrow{r}\cdot\left[\left(1+\frac{14}{19}\right)\widehat{i}+\left(2+\frac{7}{19}\right)\widehat{j}+\left(3-\frac{7}{19}\right)\widehat{k}\right]-4+\frac{35}{19}=0
\displaystyle \Rightarrow \overrightarrow{r}\cdot(33\widehat{i}+45\widehat{j}+50\widehat{k})-41=0
\displaystyle \text{Which is required equation of plane.}

\displaystyle \textbf{22. } \text{A speaks truth in 60\% of the cases, while B in 90\% of the cases. In what percent}
\displaystyle \text{of cases are they likely to contradict each other in stating the same fact? In the cases }
\displaystyle \text{of contradiction do you think, the statement of B will carry more weight as he speaks}
\displaystyle \text{truth in more number of cases than A?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E \text{ be the event of A speaking truth and } F \text{ be the event of B speaking truth}
\displaystyle \therefore\ P(E)=\frac{60}{100}=\frac{3}{5}
\displaystyle P(F)=\frac{90}{100}=\frac{9}{10}
\displaystyle \text{Probability of A and B likely to contradict each other in stating the same fact}
\displaystyle =P(E\overline{F})+P(\overline{E}F)
\displaystyle =P(E)\cdot P(\overline{F})+P(\overline{E})\cdot P(F)
\displaystyle =\frac{3}{5}\left(1-\frac{9}{10}\right)+\left(1-\frac{3}{5}\right)\frac{9}{10}
\displaystyle =\frac{3}{5}\cdot\frac{1}{10}+\frac{2}{5}\cdot\frac{9}{10}=\frac{3}{50}+\frac{18}{50}=\frac{21}{50}
\displaystyle \text{Thus, A and B are likely to contradict each other in }42\%\text{ cases.}
\displaystyle \text{Yes, statement of B will carry more weight.}

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 23 to 29 carry 6 marks each.}

\displaystyle \textbf{23. } \text{A school wants to award its students for the values of honesty, regularity and}
\displaystyle \text{hardwork with a total cash award of Rs 6000. Three times the award money for}
\displaystyle \text{hardwork added to that given for honesty amounts to Rs 11000. The award money}
\displaystyle \text{given for honesty and hardwork together is double the one given for regularity. }
\displaystyle \text{Represent the above situation algebraically and find the award money for each value }
\displaystyle \text{using matrix method. Apart from these values, namely honesty, regularity and}
\displaystyle \text{hardwork, suggest one more value which the school must include for awards.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } x,y \text{ and } z \text{ be the awarded money for honesty, regularity }  \text{and hardwork.}
\displaystyle x+y+z=6000 \qquad \ldots (i)
\displaystyle x+3z=11000 \qquad \ldots (ii)
\displaystyle x+z=2y \Rightarrow x-2y+z=0 \qquad \ldots (iii)
\displaystyle \text{The above system of three equations may be written in }  \text{matrix form as}
\displaystyle AX=B
\displaystyle \text{where } A=\begin{bmatrix}1&1&1\\1&0&3\\1&-2&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}6000\\11000\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\1&0&3\\1&-2&1\end{vmatrix}
\displaystyle =1(0+6)-1(1-3)+1(-2-0)
\displaystyle =6+2-2=6\neq 0,\ \text{Hence } A^{-1} \text{ exist}
\displaystyle A_{11}=(-1)^{1+1}\begin{vmatrix}0&3\\-2&1\end{vmatrix}=0+6=6
\displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}1&3\\1&1\end{vmatrix}=2,\ A_{13}=(-1)^{1+3}\begin{vmatrix}1&0\\1&-2\end{vmatrix}=-2
\displaystyle A_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\-2&1\end{vmatrix}=-3,\ A_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0
\displaystyle A_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\1&-2\end{vmatrix}=3,\ A_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\0&3\end{vmatrix}=3
\displaystyle A_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\1&3\end{vmatrix}=-2,\ A_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\1&0\end{vmatrix}=-1
\displaystyle \text{adj }A=\begin{bmatrix}6&2&-2\\-3&0&3\\3&-2&-1\end{bmatrix}^{\prime}=\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{|A|}\text{adj }A=\frac{1}{6}\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}
\displaystyle \therefore\ AX=B \Rightarrow X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}\begin{bmatrix}6000\\11000\\0\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}36000-33000+0\\12000+0+0\\-12000+33000+0\end{bmatrix}=\frac{1}{6}\begin{bmatrix}3000\\12000\\21000\end{bmatrix}=\begin{bmatrix}500\\2000\\3500\end{bmatrix}
\displaystyle \Rightarrow x=500,\ y=2000,\ z=3500
\displaystyle \text{Except above three values, school must include discipline for award as discipline has }
\displaystyle \text{great importance in student's life}

\displaystyle \textbf{24. } \text{Show that the height of the cylinder of maximum volume, that can be inscribed }
\displaystyle \text{in a sphere of radius } R \text{ is } \frac{2R}{\sqrt{3}}. \text{ Also find the maximum volume.}
\displaystyle \text{OR}
\displaystyle \text{Find the equation of the normal at a point on the curve } x^2=4y \text{ which passes }
\displaystyle \text{through the point } (1,2). \text{ Also find the equation of the corresponding tangent.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } r \text{ be the radius and } 2h \text{ be the height of cylinder}
\displaystyle \therefore\ OC=h
\displaystyle \text{In } \triangle OBC
\displaystyle r^2+h^2=R^2 \Rightarrow r^2=R^2-h^2 \qquad \ldots (i)
\displaystyle \text{Let } V \text{ be the volume of cylinder}
\displaystyle V=\pi r^2(2h)
\displaystyle V=2\pi h(R^2-h^2) \Rightarrow V=2\pi(R^2h-h^3)
\displaystyle \frac{dV}{dh}=2\pi(R^2-3h^2)
\displaystyle \text{For maxima or minima } \frac{dV}{dh}=0
\displaystyle 2\pi(R^2-3h^2)=0 \Rightarrow R^2=3h^2
\displaystyle \Rightarrow R=\sqrt{3}\,h \Rightarrow h=\frac{R}{\sqrt{3}}
\displaystyle \frac{d^2V}{dh^2}=2\pi(0-6h)
\displaystyle \text{At } h=\frac{R}{\sqrt{3}},\ \frac{d^2V}{dh^2}=-12\pi\frac{R}{\sqrt{3}}<0
\displaystyle \therefore\ V \text{ is maximum when } h=\frac{R}{\sqrt{3}}
\displaystyle \text{Height of cylinder, } 2h=\frac{2R}{\sqrt{3}}
\displaystyle \text{From (i)}
\displaystyle r^2=R^2-\frac{R^2}{3}=\frac{2R^2}{3}
\displaystyle \therefore\ \text{Maximum volume }=\pi r^2(2h)
\displaystyle =\pi\cdot\frac{2R^2}{3}\cdot\frac{2R}{\sqrt{3}}=\frac{4\pi R^3}{3\sqrt{3}}
\displaystyle \text{OR}
\displaystyle \text{Let the point of contact of tangent to the given curve be } (x_0,y_0)
\displaystyle \text{Now the given curve is } x^2=4y
\displaystyle \text{Differentiating on both sides w.r.t. } x
\displaystyle 2x=4\frac{dy}{dx} \Rightarrow \frac{dy}{dx}=\frac{x}{2}
\displaystyle \text{Now slope of tangent to the given curve at } (x_0,y_0)
\displaystyle =\left[\frac{dy}{dx}\right]_{(x_0,y_0)}=\frac{x_0}{2}
\displaystyle \text{Slope of normal to the given curve at } (x_0,y_0)
\displaystyle =-\frac{1}{\text{Slope of tangent at }(x_0,y_0)}=-\frac{1}{x_0/2}=-\frac{2}{x_0}
\displaystyle \text{Hence equation of required normal is}
\displaystyle (y-y_0)=-\frac{2}{x_0}(x-x_0) \qquad \ldots (i)
\displaystyle \because\ (i) \text{ passes through } (1,2)
\displaystyle \therefore\ (2-y_0)=-\frac{2}{x_0}(1-x_0)
\displaystyle \Rightarrow x_0y_0-x_0^2=2-2x_0
\displaystyle \Rightarrow x_0y_0=2 \qquad \ldots (ii)
\displaystyle \text{Also, } (x_0,y_0) \text{ lie on given curve } x^2=4y
\displaystyle \Rightarrow x_0^2=4y_0 \Rightarrow y_0=\frac{x_0^2}{4} \qquad \ldots (iii)
\displaystyle \text{Putting value of } y_0 \text{ in (ii)}
\displaystyle x_0\cdot\frac{x_0^2}{4}=2 \Rightarrow x_0^3=8 \Rightarrow x_0=2
\displaystyle \Rightarrow y_0=\frac{x_0^2}{4}=\frac{2^2}{4}=1
\displaystyle \text{Equation of normal}
\displaystyle (y-1)=-\frac{2}{2}(x-2) \Rightarrow y-1=-x+2
\displaystyle \Rightarrow x+y-3=0
\displaystyle \text{Also, equation of required tangent is}
\displaystyle (y-1)=\frac{2}{2}(x-2) \Rightarrow y-1=x-2
\displaystyle \Rightarrow x-y-1=0

\displaystyle \textbf{25. } \text{Using integration, find the area bounded by the curve } x^2=4y \\ \text{and the line } x=4y-2.
\displaystyle \text{OR}
\displaystyle \text{Using integration, find the area of the region enclosed between the two circles } \\ x^2+y^2=4 \text{ and } (x-2)^2+y^2=4.
\displaystyle \text{Answer:}
\displaystyle \text{The equation } x^2=4y \qquad \ldots (1) \text{ represents an upward parabola }  \text{with vertex at } (0,0)
\displaystyle \text{The equation } x=4y-2 \qquad \ldots (2) \text{ represents a straight line}
\displaystyle \text{Intersection of parabola and line}
\displaystyle x^2=4y,\ x=4y-2
\displaystyle x=x^2-2,\ x^2-x-2=0
\displaystyle x=2,-1
\displaystyle \text{When } x=2,\ y=\frac{x^2}{4}=\frac{4}{4}=1
\displaystyle \text{When } x=-1,\ y=\frac{1}{4}
\displaystyle \text{Intersection points are } (2,1) \text{ and } \left(-1,\frac{1}{4}\right)
\displaystyle \text{Required area }=\int_{-1}^{2}(\text{Line})\,dx-\int_{-1}^{2}(\text{Parabola})\,dx
\displaystyle =\int_{-1}^{2}\left(\frac{x+2}{4}\right)dx-\int_{-1}^{2}\frac{x^2}{4}\,dx
\displaystyle =\frac{1}{4}\left[\frac{x^2}{2}+2x\right]_{-1}^{2}-\frac{1}{4}\left[\frac{x^3}{3}\right]_{-1}^{2}
\displaystyle =\frac{1}{4}\left[\left(2+4\right)-\left(\frac{1}{2}-2\right)\right]-\frac{1}{12}\left[2^3-(-1)^3\right]
\displaystyle =\frac{1}{4}\left(6+\frac{3}{2}\right)-\frac{1}{12}\times 9
\displaystyle =\frac{15}{8}-\frac{3}{4}
\displaystyle =\frac{9}{8} \text{ square units}
\displaystyle \text{OR}
\displaystyle \text{Given: Equation of two circles are}
\displaystyle x^2+y^2=4 \qquad \ldots (1)
\displaystyle (x-2)^2+y^2=4 \qquad \ldots (2)
\displaystyle \text{Circle in eq. (1), Centre } (0,0),\ r=2
\displaystyle \text{Circle in eq. (2), Centre } (2,0),\ r=2\displaystyle \text{Now, we find the point of intersection of the two circles.}
\displaystyle \text{Subtracting (2) - (1)}
\displaystyle (x-2)^2-x^2=0 \Rightarrow x^2+4-4x-x^2=0
\displaystyle 4-4x=0 \Rightarrow x=1
\displaystyle 1+y^2=4 \Rightarrow y=\pm\sqrt{3}
\displaystyle \text{Point of intersection } (1,\sqrt{3}) \text{ and } (1,-\sqrt{3})
\displaystyle \text{Area of shaded region }=2\left[\int_{0}^{1}y\,dx+\int_{1}^{2}y\,dx\right]
\displaystyle \text{Area }=2\left[\int_{0}^{1}\sqrt{4-(x-2)^2}\,dx+\int_{1}^{2}\sqrt{4-x^2}\,dx\right]
\displaystyle \text{As } \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}
\displaystyle \text{Area }=2\left[\left(\frac{x-2}{2}\sqrt{4-(x-2)^2}+\frac{4}{2}\sin^{-1}\frac{x-2}{2}\right)\right]_{0}^{1}
\displaystyle \qquad +2\left[\left(\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}\right)\right]_{1}^{2}
\displaystyle =2\left[\left(\frac{x-2}{2}\sqrt{4-(x-2)^2}+2\sin^{-1}\frac{x-2}{2}\right)\right]_{0}^{1}
\displaystyle \qquad +2\left[\left(\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\frac{x}{2}\right)\right]_{1}^{2}
\displaystyle =2\left[\left(-\frac{1}{2}\sqrt{3}+2\sin^{-1}\left(-\frac{1}{2}\right)-2\sin^{-1}(-1)\right)\right.
\displaystyle \left.+\left(2\sin^{-1}1-\frac{1}{2}\sqrt{3}-2\sin^{-1}\frac{1}{2}\right)\right]
\displaystyle =2\left[-\frac{\sqrt{3}}{2}+2\sin^{-1}\left(-\frac{\pi}{6}\right)-2\sin^{-1}\left(-\frac{\pi}{2}\right)\right.
\displaystyle \left.+2\sin^{-1}\frac{\pi}{2}-\frac{\sqrt{3}}{2}-2\sin^{-1}\frac{\pi}{6}\right]
\displaystyle =2\left[-\frac{\sqrt{3}}{2}+2\left(-\frac{\pi}{6}\right)-2\left(-\frac{\pi}{2}\right)+2\left(\frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2\left(\frac{\pi}{6}\right)\right]
\displaystyle =2\left[-\sqrt{3}+\frac{2\pi}{3}+\pi-\frac{\pi}{3}\right]
\displaystyle =2\left[\frac{4\pi}{3}-\sqrt{3}\right]
\displaystyle =\left[\frac{8\pi}{3}-2\sqrt{3}\right] \text{ sq. units}
\displaystyle \text{Hence required area }=\left[\frac{8\pi}{3}-2\sqrt{3}\right] \text{ sq. units.}

\displaystyle \textbf{26. } \text{Show that the differential equation } 2ye^{x/y}dx+(y-2xe^{x/y})dy=0
\displaystyle \text{is homogeneous. Find the particular solution of this differential equation, given} \\ \text{that } x=0 \text{ when } y=1.
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle 2y\cdot e^{x/y}\,dx+(y-2xe^{x/y})\,dy=0
\displaystyle \frac{dx}{dy}=-\frac{(y-2xe^{x/y})}{2ye^{x/y}}=\frac{2xe^{x/y}-y}{2ye^{x/y}}
\displaystyle \text{Let } F(x,y)=\frac{2xe^{x/y}-y}{2ye^{x/y}}
\displaystyle \therefore\ F(\lambda x,\lambda y)=\frac{2\lambda xe^{\lambda x/\lambda y}-\lambda y}{2\lambda ye^{\lambda x/\lambda y}}
\displaystyle =\lambda^0\frac{2xe^{x/y}-y}{2ye^{x/y}}=\lambda^0F(x,y)
\displaystyle \text{Hence, given differential equation is homogeneous.}
\displaystyle \text{Now, } \frac{dx}{dy}=\frac{2xe^{x/y}-y}{2ye^{x/y}} \qquad \ldots (1)
\displaystyle \text{Let } x=vy \Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}
\displaystyle \text{From (1)}
\displaystyle \Rightarrow v+y\frac{dv}{dy}=\frac{2vye^{vy/y}-y}{2ye^{vy/y}}
\displaystyle \Rightarrow v+y\frac{dv}{dy}=\frac{2ve^v-1}{2e^v}
\displaystyle \Rightarrow y\frac{dv}{dy}=\frac{2ve^v-1}{2e^v}-v=-\frac{1}{2e^v}
\displaystyle \Rightarrow 2ye^v\,dv=-dy
\displaystyle \Rightarrow 2\int e^v\,dv=-\int\frac{dy}{y}
\displaystyle \Rightarrow 2e^v=-\log y+C
\displaystyle \Rightarrow 2e^{x/y}+\log y=C
\displaystyle \text{When } x=0,\ y=1
\displaystyle \Rightarrow 2e^0+\log 1=C
\displaystyle \Rightarrow C=2
\displaystyle \text{Hence required solution is}
\displaystyle 2e^{x/y}+\log y=2

\displaystyle \textbf{27. } \text{Find the vector equation of the plane passing through three points with}
\displaystyle \text{position vectors } \widehat{i}+\widehat{j}-2\widehat{k},\ 2\widehat{i}-\widehat{j}+\widehat{k} \text{ and } \widehat{i}+2\widehat{j}+\widehat{k}. \text{ Also find the coordinates of }
\displaystyle \text{the point of intersection of this plane and the line}
\displaystyle \overrightarrow{r}=3\widehat{i}-\widehat{j}-\widehat{k}+\lambda(2\widehat{i}-2\widehat{j}+\widehat{k}).
\displaystyle \text{Answer:}
\displaystyle \text{Plane passes through three points}
\displaystyle (1,1,-2),\ (2,-1,1) \text{ and } (1,2,1)
\displaystyle \text{Equation of plane is given as}
\displaystyle \begin{vmatrix}x-1 & y-1 & z+2 \\ 2-1 & -1-1 & 1+2 \\ 1-1 & 2-1 & 1+2\end{vmatrix}=0
\displaystyle \begin{vmatrix}x-1 & y-1 & z+2 \\ 1 & -2 & 3 \\ 0 & 1 & 3\end{vmatrix}=0
\displaystyle \Rightarrow (x-1)(-6-3)-(y-1)(3-0)+(z+2)(1-0)=0
\displaystyle \Rightarrow -9x+9-3y+3+z+2=0
\displaystyle \Rightarrow 9x+3y-z=14
\displaystyle \text{Vector form: } \overrightarrow{r}\cdot(9\widehat{i}+3\widehat{j}-\widehat{k})=14
\displaystyle \text{The given line is}
\displaystyle \overrightarrow{r}=(3\widehat{i}-\widehat{j}+\widehat{k})+\lambda(2\widehat{i}-2\widehat{j}+\widehat{k})
\displaystyle \text{Cartesian form of line}
\displaystyle \frac{x-3}{2}=\frac{y+1}{-2}=\frac{z-1}{1}=\lambda
\displaystyle \Rightarrow \alpha=2\lambda+3,\ \beta=-2\lambda-1,\ \gamma=\lambda+1
\displaystyle \text{Let line intersect plane at } (\alpha,\beta,\gamma)
\displaystyle \therefore\ (\alpha,\beta,\gamma) \text{ lie on plane }
\displaystyle 9\alpha+3\beta-\gamma=14
\displaystyle \Rightarrow 9(2\lambda+3)+3(-2\lambda-1)-(\lambda+1)=14
\displaystyle \Rightarrow 18\lambda+27-6\lambda-3-\lambda-1=14
\displaystyle \Rightarrow 11\lambda+23=14 \Rightarrow 11\lambda=-9 \Rightarrow \lambda=-\frac{9}{11}
\displaystyle \therefore\ \text{Point of intersection is } \left(\frac{15}{11},\frac{7}{11},\frac{2}{11}\right)

\displaystyle \textbf{28. } \text{A cooperative society of farmers has 50 hectares of land to grow two crops A and B. }
\displaystyle \text{The profits from crops A and B per hectare are estimated as Rs }10500 \text{ and Rs }9000
\displaystyle \text{ respectively. To control weeds, a liquid herbicide has to be used for crops A and B at the rate }
\displaystyle \text{of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide }
\displaystyle \text{should be used in order to protect fish and wildlife using a pond which collects drainage from }
\displaystyle \text{this land. Keeping in mind that the protection of fish and other wildlife is more important than }
\displaystyle \text{earning profit, how much land should be allocated to each crop so as to maximize the total profit? }
\displaystyle \text{Form an LPP from the above and solve it graphically. Do you agree with the message the }
\displaystyle \text{protection of wildlife is utmost necessary to preserve the balance in environment?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ and } y \text{ hectare of land be allocated to crop } A \text{ and } B  \text{ respectively}
\displaystyle \text{If } Z \text{ is profit}
\displaystyle \text{Maximize } Z=10500x+9000y
\displaystyle \text{Subject to constraints}
\displaystyle x+y\leq 50
\displaystyle 20x+10y\leq 800 \Rightarrow 2x+y\leq 80
\displaystyle x\geq 0,\ y\geq 0
\displaystyle x+y=50
\displaystyle \text{When } x=0,\ y=50
\displaystyle \text{When } y=0,\ x=50
\displaystyle 2x+y=80
\displaystyle \text{When } x=0,\ y=80
\displaystyle \text{When } y=0,\ x=40
\displaystyle \text{Intersection point } y=50-x
\displaystyle 2x+y=80 \Rightarrow 2x+50-x=80
\displaystyle \Rightarrow y=50-30=20,\ \text{Point } (30,20)
\displaystyle OABC \text{ is feasible region with corner points } O(0,0),\ A(40,0),
\displaystyle B(30,20),\ C(0,50)
\displaystyle \text{At } O(0,0),\ Z=10500(0)+9000(0)=0
\displaystyle \text{At } A(40,0),\ Z=10500(40)+9000(0)=420000
\displaystyle \text{At } B(30,20),\ Z=10500(30)+9000(20)=495000
\displaystyle \text{At } C(0,50),\ Z=10500(0)+9000(50)=450000
\displaystyle \text{Hence cooperative society of farmers will get the maximum }\text{profit of Rs }495000
\displaystyle \text{ by allocating } 30 \text{ hectares for crop } A \text{ and } 20 \text{ hectares for crop } B.
\displaystyle \text{Yes, because excess use of herbicide can make drainage water poisonous and thus it }
\displaystyle \text{harm the life of water living creatures and wildlife.}
\displaystyle \text{}

\displaystyle \textbf{29. } \text{Assume that the chances of a patient having a heart attack is 40\%. Assuming }
\displaystyle \text{that a meditation and yoga course reduces the risk of heart attack by 30\% and prescription }
\displaystyle \text{of certain drug reduces its chance by 25\%. At a time a patient can choose any one}
\displaystyle \text{ of the two options with equal probabilities. It is given that after going through one }
\displaystyle \text{of the two options, the patient selected at random suffers a heart attack. Find the }
\displaystyle \text{probability that the patient followed a course of meditation and yoga. Interpret the }
\displaystyle \text{result and state which of the above stated methods is more beneficial for the patient.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2,A \text{ be events defined as}
\displaystyle E_1=\text{treatment of heart attack with yoga and meditation.}
\displaystyle E_2=\text{treatment of heart attack with certain drugs}
\displaystyle A=\text{Person getting heart attack.}
\displaystyle P(E_1)=\frac{1}{2},\ P(E_2)=\frac{1}{2}
\displaystyle \text{Now } P\left(\frac{A}{E_1}\right)=40\%-40\times\frac{30}{100}
\displaystyle =40\%-12\%=28\%=\frac{28}{100}
\displaystyle P\left(\frac{A}{E_2}\right)=40\%-\left(40\times\frac{25}{100}\right)
\displaystyle =40\%-10\%=30\%=\frac{30}{100}
\displaystyle P(E_1/A)=\text{Person getting heart attack treated with yoga }  \text{and meditation}
\displaystyle \text{By Baye's theorem}
\displaystyle P\left(\frac{E_1}{A}\right)=\frac{P(E_1)\cdot P\left(\frac{A}{E_1}\right)}{P(E_1)\cdot P\left(\frac{A}{E_1}\right)+P(E_2)\cdot P\left(\frac{A}{E_2}\right)}
\displaystyle =\frac{\frac{1}{2}\times\frac{28}{100}}{\frac{1}{2}\times\frac{28}{100}+\frac{1}{2}\times\frac{30}{100}}=\frac{28}{58}=\frac{14}{29}
\displaystyle \text{The problem emphasises the importance of yoga and }  \text{meditation.}
\displaystyle \text{Treatment with yoga and meditation is more beneficial for }  \text{the heart patient.}

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.