\(y={ tan }^{ -1 }\frac { 3x+2x }{ 1-3x2x } \)
= \({ tan }^{ -1 }3x+{ tan }^{ -1 }2x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { 3 }{ 1+9{ x }^{ 3 } } +\frac { 2 }{ 1+4{ x }^{ 2 } } \)

Let \(y=x^{1/3}\)
Let  \(x=125 \ and \ x+\Delta x=127\)
\(\Rightarrow \Delta x=2\)
\(y=(125)^{1/3}\Rightarrow5\)
\(y=x^{1/3}\)
\(\frac{dy}{dx}=\frac{1}{3}(x)^{-2/3}\)
\(\Rightarrow \Delta y=\frac{x^{-2/3}}{3}\times \Delta x \)
\(=\frac{2}{3(5)^2}=\frac{2}{75}=0.0266 \)
\(\therefore\ y+\Delta y=5+0.0266\Rightarrow5.0266 \)
\( \Rightarrow 3\sqrt{127}=5.0266\)

(i) f(g) = f(x)g(x)
\(fg=(\sqrt { x+4 } )(\sqrt { x-4 } )=\sqrt { x^{ 2 }-4 } \)
(ii) \(\frac { f }{ g } =\frac { f(x) }{ g(x) } =\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \times \frac { \sqrt { x-4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x^{ 2 }-16 } }{ x-4 } \)

\(L.H.S.={ tan }^{ -1 }\left( \frac { 3{ a }^{ 2 }x-{ x }^{ 3 } }{ { a }^{ 3 }-3a{ x }^{ 2 } } \right) \)
\(Divide\quad by\quad a^{ 3 },\)
\(={ tan }^{ -1 }\left( \frac { 3\frac { x }{ a } -\frac { { x }^{ 3 } }{ { a }^{ 3 } } }{ 1-3\left( \frac { { x }^{ 2 } }{ { a }^{ 2 } } \right) } \right) \)
\(={ tan }^{ -1 }\left[ \frac { 3\left( \frac { x }{ a } \right) -{ \left( \frac { x }{ a } \right) }^{ 3 } }{ 1-3{ \left( \frac { x }{ a } \right) }^{ 2 } } \right] \)
\(Put,\frac { x }{ a } =tan\theta ,\quad \theta ={ tan }^{ -1 }\frac { x }{ a } \)
\(={ tan }^{ -1 }\left[ \frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \right] \)
\(={ tan }^{ -1 }(tan\quad 3\theta )\)
\(3\theta =3{ tan }^{ -1 }\left( \frac { x }{ a } \right) =R.H.S.\)

We have, \(A=\left| \begin{matrix} 2 & 3 & -1 \\ 4 & 1 & 0 \\ 3 & 3 & 2 \end{matrix} \right| \)
\(M_{ 12 }\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =8-0=8\)
\(M_{ 21 }\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =6+3=9\)
\(C_{ 21 }=-\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =-(6+3)=-9\)
\(C_{ 12 }=-\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =-(8-0)=-8\)
\( M_{ 12 }\times M_{ 21 }+C_{ 21 }\times C_{ 12 }\)
\( 8\times (9)+(-9)(-8)=72+72\)
\(=144\)

6 sq.units

\(={2\over3}x^{3/2}-{2\over5}x^{5/2}+c\)

\(cos\alpha =1,sin\alpha =0\Rightarrow \alpha ={ 0 }^{ o }\)

For reflexive : As (1, 1), (2, 2), (3, 3) ∈ R. Hence, reflexive
For Symmetric: (1, 2) ∈ R but (2, 1) ∉ R. Hence, not symmetric
For transitive: (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.Hence, not transitive
Hence, R is not an equivalence relation

2Let D₁ denote the event that the patient has disease d₁. The events D₂ and D₃ are defined similarly.
Then, \(P({ D }_{ 1 })=\frac { 3,200 }{ 10,000 } =0.32\)
\(P({ D }_{ 2 })=\frac { 3,500 }{ 10,000 } =0.35\)
and \(P({ D }_{ 3 })=\frac { 3,300 }{ 10,000 } =0.33\)
Let S be the event that the patient shows the symptoms S,
Then, \(P(S/{ D }_{ 1 })=\frac { P(S\cap D)) }{ P({ D }_{ 1 }) } =\frac { 3,100 }{ 3,200 } \)
= 0.97 (approx.)
\(P(S/{ D }_{ 3 })=\frac { 3,000 }{ 3,500 } =0.94\quad (approx.)\quad \)
\(P(S/{ D }_{ 3 })=\frac { 3,000 }{ 3,300 } =0.91\quad (approx.)\)
Using Bayes' theorem, we get
P(D₁/S) = The probability that the patient has disease d₁ knowing that he/she has symtoms S₁, S₂ ,....,S₆
P(D₁/S)
\(=\frac { P({ D }_{ 1 })|P(S/{ D }_{ 1 }) }{ P({ D }_{ 1 })P(S/{ D }_{ 1 })+P({ D }_{ 2 })P(S/{ D }_{ 2 })+P({ D }_{ 3 })P(S/{ D }_{ 3 }) } \)
\(=\frac { 0.32\times 0.97 }{ 0.32\times 0.97+0.35\times 0.94+0.33\times 0.91 } \)
\(=\frac { 0.3104 }{ 0.3104+0.329+0.3003 } \)
\(=\frac { 0.3104 }{ 0.9397 } =0.33\quad approx\)
Similarly,
\(P({ D }_{ 2 }/S)=\frac { 0.329 }{ 0.9397 } =0.35\quad approx\)
and \(P({ D }_{ 3 }/S)=\frac { 0.3003 }{ 0.9397 } =0.32\quad approx\)
Thus, knowing that the patient has symtoms S₁,S₂, ..., S₃, the probability that he has disease d₁ is 0.33, the probability that he has disease d₂ is 0.35, the probability that he has disease d₃ is 0·32. Therefore, the doctor should conclude that the patient is most likely to have disease d₂. 

\(\therefore \ \frac{dx}{dt}=2\ cm/s\)
Area (A)=\(\frac{\sqrt3}{4}x^2\)
\(\Rightarrow \frac{dA}{dt}=\frac{\sqrt3}{2}x\frac{dx}{dt}\)
\(\Rightarrow \frac{dA}{dt}=\frac{\sqrt3}{2}(20)\times (2)\)
[∵ Side of triangle=20 cm]
\(=20\sqrt3\ cm^2/s\)
Hence its area increasing at the rate of \(20\sqrt3\ cm^2/s\)

The given equation is:
\(xy\quad log\left( \frac { y }{ x } \right) dx+\left( { y }^{ 2 }-{ x }^{ 2 }log\left( \frac { y }{ x } \right) \right) dy=0\)
\(\Rightarrow\)  \(\frac { dy }{ dx } =-\frac { xy\quad log\left( \frac { y }{ x } \right) }{ { y }^{ 2 }-{ x }^{ 2 }log\left( \frac { y }{ x } \right) } \)
\(\Rightarrow\)  \(\frac { dy }{ dx } =-\frac { \frac { y }{ x } \quad log\left( \frac { y }{ x } \right) }{ \frac { { y }^{ 2 } }{ { x }^{ 2 } } -log\left( \frac { y }{ x } \right) } \) ..(1)
Put \(\frac { y }{ x } =v\quad i.e.\)  \(y=vx\)    so that \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } .\)
\(\therefore\) (1) becomes:  \(v+x\frac { dv }{ dx } =-\frac { v\quad log\quad v }{ { v }^{ 2 }-\quad log\quad v } \)
\(\Rightarrow \)  \(x\frac { dv }{ dx } = -\frac { v\ log\ v }{ { v }^{ 2 }-\quad log\quad v } -v\\ \)
\(\Rightarrow \)  \(x\frac { dv }{ dx } = \frac { -v\quad log\quad v-{ v }^{ 3 }\quad +\quad v\quad log\quad v }{ { v }^{ 2 }\quad -\quad log\quad v\quad } \\ \)
\(\Rightarrow \)  \(x\frac { dv }{ dx } = -\frac { { v }^{ 3 } }{ { v }^{ 2 }\quad -logv } \\ \)
\(\Rightarrow \) \(\frac { { v }^{ 2 }-log\quad v }{ { v }^{ 3 } } dv=-\frac { dv }{ x } .\)
Integrating, \(\int { \frac { v^{ 2 }-log\quad v }{ { v }^{ 3 } } } dv=-\int { \frac { dx }{ x } } +c\) 

\(\Rightarrow \int { \frac { 1 }{ 2 } } dv-\int { logv.{ v }^{ -3 } } dv =-\int { \frac { dx }{ x } } +c\) 
\( \Rightarrow log|v|-\left[ log\quad v.\frac { { v }^{ -2 } }{ -2 } -\int { \frac { 1 }{ v } . } \frac { { v }^{ -2 } }{ -2 } dv \right] =- log|x|+c\)
\(\Rightarrow log|v|-\left[ -\frac { log\quad v }{ 2{ v }^{ 2 } } +\frac { 1 }{ 2 } \int { { v }^{ -3 }\quad dv } \right] =- log|x|+c\)
\(\Rightarrow log|v|+\frac { log\quad v }{ 2{ v }^{ 2 } } -\frac { 1 }{ 2 } \frac { { v }^{ -2 } }{ -2 } =- log|x|+c\)
\(\Rightarrow log|\frac { y }{ x } |+\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } log\frac { y }{ x } +\frac { 1 }{ 4 } \frac { { x }^{ 2 } }{ { y }^{ 2 } } =- log|x|+c\)
\(\Rightarrow log|y|-log|x|+\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } \left( log\frac { y }{ x } +\frac { 1 }{ 2 } \right) =- log|x|+c\)
\(\Rightarrow log|y|+\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } \left( log\frac { y }{ x } +\frac { 1 }{ 2 } \right) =- log|x|+c\)
which is the required solution

We have \(y=log\left( x+\sqrt { { x }^{ 2 }+1 } \right) \)
\(\frac { dy }{ dx } =\frac { 1 }{ x+\sqrt { { x }^{ 2 }+1 } } \left( 1+\frac { 1 }{ 2\sqrt { { x }^{ 2 }+1 } } (2x+0) \right) \)
\( =\frac { 1 }{ x+\sqrt { { x }^{ 2 }+1 } } \left( 1+\frac { x }{ \sqrt { { x }^{ 2 }+1 } } \right) \)
\(=\frac { 1 }{ x+\sqrt { { x }^{ 2 }+1 } } \left( \frac { \sqrt { { x }^{ 2 }+1 } +x }{ \sqrt { { x }^{ 2 }+1 } } \right) =\frac { 1 }{ \sqrt { { x }^{ 2 }+1 } } \)
\(\sqrt { { x }^{ 2 }+1 } \frac { dy }{ dx } =1\)
Squaring, \(\left( { x }^{ 2 }+1 \right) { \left( \frac { dy }{ dx } \right) }^{ 2 }=1\)
Diff.w.r.t.x, \(\left( { x }^{ 2 }+1 \right) 2\left( \frac { dy }{ dx } \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +(2x+0){ \left( \frac { dy }{ dx } \right) }^{ 2 }=0\)
\(\left( { x }^{ 2 }+1 \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +x\frac { dy }{ dx } =0\)
which is true.

\(\text { Let } \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x\)
Then, by P₄
\(\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x=\int_{0}^{\frac{\pi}{2}} d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}\)
\(\text { Hence } \ \mathrm{I}=\frac{\pi}{4}\)

(A) is the correct answer.
Reason: Since A and B are symmetric matrices,
\(\therefore \ A\prime =A\quad and\quad B\prime =B ...(1)\)
\(Now\ (AB-BA)\prime =(AB)\prime -(BA)\prime \)
\(=B\prime A\prime -A\prime B\prime\)
\(=BA-AB\)
\(=-(AB-BA)\)
\(\Rightarrow (AB-BA)\) is skew-symmetric.

let \(\overset { \rightarrow }{ c } \) = \(\left| \overrightarrow { a } \right| \overrightarrow { b } +\left| \overrightarrow { b } \right| \overrightarrow { a } \) 
and  \(\overset { \rightarrow }{ d } \) = \(\left| \overrightarrow { a } \right| \overrightarrow { b } -\left| \overrightarrow { b } \right| \overrightarrow { a } \)
\(\therefore \vec{c} \cdot \vec{d}=|| \vec{a}|\vec{b}+| \vec{b} \mid \vec{a}) \cdot(|\vec{a}| \vec{b}-|\vec{b}| \vec{a}\rangle \mid \)
\(=\left|\overrightarrow{a \mid}^{2} \vec{b} \cdot \vec{b}-\vec{a}\right||b| \vec{b} \cdot \vec{a}+\overrightarrow{\mid}|\vec{b}| \overrightarrow{|a|} \mid \vec{a} \cdot \vec{b}-\overrightarrow{b^{2}} \vec{a} \cdot \vec{a} \)
\(=\overrightarrow{a^{2}} b^{2}-\vec{a}|| b|\vec{b} \cdot \vec{a}+| \vec{a}|| \vec{b}|\vec{a} \cdot \vec{b}-| \overrightarrow{b \mid}^{2} \mid \overrightarrow{a \mid}^{2}[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}] \)
\(=0\) 
Hence, is perpendicular to \(\overset { \rightarrow }{ d } \).

Let 'x' units of goods A and 'y' units of goods B be produced.
Then LLP is as below:
Maximise:  Z=100x + 120y...(1)
Subject to:  \(2x+3y\le 30\) ..(2)
\(3x+y\le 17\) ...(3)
and  \(x\ge 0,y\ge 0\)...(4)

The shaded portion represents the feasible region OCEB, which is bounded.
Its vertices are O(0,0), \(C\left( \frac { 17 }{ 3 } ,0 \right) \) , B(0,10) and E(3,8).
[Solving 2x + 3y = 30 and 3x + y = 17; x = 3, y = 8]
Applying Corner Point Method, we have:

Hence, the maximum total revenue is obtained when 3 units of goods A and 8 units of goods B be produced.
Comments: We agree with the views of the manufacturer because men and women have equal rights according to the constitution of the country.

Corresponding equations are  y = |x - 1|
\(\text { and } y=\sqrt{5-x^{2}}\)
\(\text { Eliminating } y \text { , we get }|x-1|=\sqrt{5-x^{2}}\)
\(\Rightarrow x^{2}+1-2 x=5-x^{2} \Rightarrow x=2,-1\)
\(\therefore \text { area }=\int_{-1}^{2} \sqrt{5-x^{2}} d x-\int_{-1}^{1}-(x-1) d x-\int_{1}^{2}(x-1) d x \)
\(=\left[\frac{x}{2} \sqrt{5-x^{2}}+\left.\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right|_{-1} ^{2}-\left[-\frac{x^{2}}{2}+x\right]_{-1}^{1}\right.\)
\(-\left[\frac{x^{2}}{2}-x\right] \mid\)
\(=\left(1+\frac{5}{2} \sin ^{-1} \frac{2}{\sqrt{5}}\right)-\left[-1+\frac{5}{2} \sin ^{-1}\left(-\frac{1}{\sqrt{5}}\right)\right] \)
\(-\left[-\frac{1}{2}+1\right]+\left[-\frac{1}{2}-1\right]-[2-2]+\left(\frac{1}{2}-1\right) \)
\(=1+\frac{5}{2} \sin ^{-1} \frac{2}{\sqrt{5}}+1+\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}-\frac{1}{2}-\frac{3}{2}-\frac{1}{2}\)

\(=\frac{5}{2} \sin ^{-1}\left[\frac{2}{\sqrt{5}} \sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}} \sqrt{1-\frac{4}{5}}\right]-\frac{1}{2} \)
\(=\frac{5}{2} \sin ^{-1}\left[\frac{4}{5}+\frac{1}{5}\right]-\frac{1}{2}=\frac{5}{2} \cdot \frac{\pi}{2}-\frac{1}{2} \)
\(=\frac{1}{4}(5 \pi-2) \text { sq units } \)

Equation of plane through the intersection of given two planes is:
\((x+y+z-1)+\lambda(2x+3y+4z-5)=0\)
\(\Rightarrow\)  \((1+2\lambda)x+(1+3\lambda)y+(1+4\lambda)z-1-5\lambda=0\) ...(i)
Plane (i) is perpendicular to the plane
x - y + z = 0
So, \(1(1+2\lambda)-1(1+3\lambda)+1(1+4\lambda=0)\)
\(\Rightarrow\)  \(3\lambda=-1\)
\(\therefore\)  \(\lambda=-{{1}\over{3}}\)
\(\therefore\)  Equation of plane is
\(\left(1-{{2}\over{3}} \right)x+(1-1)y+\left( 1-{{4}\over{3}}\right)z-1+{{5}\over{3}}=0\)
i.e., x - z + 2 = 0
Distance of above plane from origin
\(={{2}\over{\sqrt{2}}}=\sqrt{2}\) units.

Let the young man drives x km and y km at 25kmh and 40 km!h speed respectively, then the LPP is Maximise distance:
Z = x + y
Subject to
2x + 5y \(\le\)100
\(x \geq 0, y \geq 0, \frac{x}{25}+\frac{y}{40} \leq 1\)
x, y \(\ge\)0
Plotting the inequations as graph we notice shaded portion is feasible solution.

Possible points for maximum Z are A(25, 0), \(B\left(\frac{50}{3}, \frac{40}{3}\right)\) and C(0, 20).

Z  is maximum at \( B\left(\frac{50}{3}, \frac{40}{3}\right) \text {, i.e. } x=\frac{50}{3}, y=\frac{40}{3} \text {. }\)
Hence, he must travel \( \frac{50}{3} \mathrm{~km} \)at a speed of  25 km h
\(\text { and } \frac{40}{3} \mathrm{~km} \text { at a speed of } 40 \mathrm{~km} / \mathrm{h} \)for a maximum  distance of 30 km.

Here a = 1, b = 3, nh = 2, f(x) = 3x² + 1
\(\int _{ 1 }^{ 3 }{ \left( { 3x }^{ 2 }+1 \right) } dx\)
\(=\lim _{ h\rightarrow 0 }{ h\left[ { 3h }^{ 2 }\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+...+\left( n-1 \right) ^{ 2 } \right) +6h\left( 1+2+...+\left( n-1 \right) \right) +4n \right] } \)
\(=\lim _{ h\rightarrow 0 }{ h\left[ \frac { 3\left( nh-h \right) \left( nh \right) \left( 2nh-h \right) }{ 6 } +\frac { 6\left( nh-h \right) \left( nh \right) }{ 2 } +4nh \right] } \)
\(=\lim _{ h\rightarrow 0 }{ h\left[ \frac { 3\left( 2-h \right) .2.\left( 4-h \right) }{ 6 } +\frac { 6\left( 2-h \right) .2 }{ 2 } +8 \right] } \)
= 8 + 12 + 8 = 28

\(\frac{dx}{d\theta}=ae^{ \theta }(\sin\theta-\cos\theta)+ae^{ \theta }(\sin\theta+\cos\theta)\)
\(=2ae^{ \theta}\sin\theta\)
\(\frac{dy}{d\theta}=ae^{ \theta }(\sin\theta+\cos\theta)+ae^{ \theta }(\cos\theta-\sin\theta)\)
\(=2ae^{ \theta}\cos\theta\)
\(\therefore \frac{dy}{dx}=\frac{2ae^{ \theta}\cos\theta}{2ae^{ \theta}\sin\theta}=\cot\theta\)
\( \frac { dy }{ dx } |_{at\theta=\frac{\pi}{4}}=\cot\frac{\pi}{4}=1\)

The given differential equation is :
\(x\frac { dy }{ dx } \sin { \left( \frac { y }{ x } \right) } +x-y\sin { \left( \frac { y }{ x } \right) } =0\)
\(\Rightarrow y\sin { \left( \frac { y }{ x } \right) } -x=\frac { xdy }{ dx } \sin { \left( \frac { y }{ x } \right) } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y\sin { \left( \frac { y }{ x } \right) } -x }{ x\sin { \left( \frac { y }{ x } \right) } } \) ... (i)
\(f\left( x,y \right) =\frac { y }{ x } -cosec\left( \frac { y }{ x } \right) \)
Now, put \(x=\lambda x,y=\lambda y\), then
\(f\left( \lambda x,\lambda y \right) =\frac { \lambda y }{ \lambda x } -cosec\left( \frac { \lambda y }{ \lambda x } \right) \)
\(=\frac { y }{ x } -cosec\left( \frac { y }{ x } \right) \)
= f (x, y)
So, it is homogeneous
Now, put y = vx in equation (i).
On differentiating both sides w.r.t. x, we get
\(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
Now from (i), \(v+x\frac { dv }{ dx } =\frac { vx }{ x } -\frac { 1 }{ \sin { \left( \frac { vx }{ x } \right) } } \)
\(\Rightarrow x\frac { dv }{ dx } =-\frac { 1 }{ \sin { v } } \)
\(\Rightarrow -\int { \sin { v } dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow \cos { v } =\log { \left| x \right| } +C\)
\(\Rightarrow \cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } +C\)
It is given that x = 1 and y = \(y={ \pi }/{ 2 }\)
So, \(\cos { \frac { \left( \frac { \pi }{ 2 } \right) }{ 1 } } =C+\log { \left| 1 \right| } \)
\(\Rightarrow C=0\)
\(\Rightarrow \cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } \)
Hence, required solution is
\(\cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } \)

\(={ (x+y+z) }^{ 3 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
LHS = \(\left| \begin{matrix} { (y+z) }^{ 2 } & xy & zx \\ xy & { (x+z) }^{ 2 } & yz \\ xz & yz & { (x+y) }^{ 2 } \end{matrix} \right| \)
Apply \(({ R }_{ 1 }\rightarrow x{ R }_{ 1 },\quad { R }_{ 2 }y\rightarrow { R }_{ 2 },{ R }_{ 3 }z\rightarrow { R }_{ 3 })\)
\(=\frac { 1 }{ xyz } \left| \begin{matrix} { x(y+z) }^{ 2 } & { x }^{ 2 }y & z{ x }^{ 2 } \\ x{ y }^{ 2 } & { y(x+z) }^{ 2 } & { y }^{ 2 }z \\ x{ z }^{ 2 } & y{ z }^{ 2 } & { z(x+y) }^{ 2 } \end{matrix} \right| \)
\(=\left| \begin{matrix} { (y+z) }^{ 2 } & { x }^{ 2 } & { x }^{ 2 } \\ { y }^{ 2 } & { (x+z) }^{ 2 } & { y }^{ 2 } \\ x{ z }^{ 2 } & { z }^{ 2 } & { (x+y) }^{ 2 } \end{matrix} \right| \)
Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(=\left| \begin{matrix} { (y+z) }^{ 2 } & { x }^{ 2 }-{ (y+z) }^{ 2 } & { x }^{ 2 }-{ (y+z) }^{ 2 } \\ { y }^{ 2 } & { (x+z) }^{ 2 }-{ y }^{ 2 } & 0 \\ { z }^{ 2 } & 0 & { (x+y) }^{ 2 }-{ z }^{ 2 } \end{matrix} \right| \)
Taking (x + y + z) common from C₂ and C₃ . 
\(={ (x+y+z) }^{ 3 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
Apply \({ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }-{ R }_{ 3 }\)
\(={ (x+y+z) }^{ 2 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
Apply, \({ C }_{ 2 }\rightarrow y{ C }_{ 2 },{ C }_{ 3 }\rightarrow z{ C }_{ 3 }\)
\(=\frac { { (x+y+z) }^{ 2 } }{ yz } \left| \begin{matrix} 2yz & 0 & 0 \\ { y }^{ 2 } & x+z-y & { y }^{ 2 } \\ { z }^{ 2 } & { z }^{ 2 } & zx+yz \end{matrix} \right| \)
\(=\frac { { (x+y+z) }^{ 2 } }{ yz } \left| (2yz).({ x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 }-{ y }^{ 2 }{ z }^{ 2 } \right| \)
\(=2xyz{ (x+y+z) }^{ 3 }=RHS\)

(i) \(A+B=\left( \begin{matrix} 15000 & 30000 & 36000 \\ 70000 & 40000 & 20000 \end{matrix} \right) \)
(ii) 2 % of B = \(A+B=\left( \begin{matrix} 100 & 200 & 120 \\ 400 & 200 & 200 \end{matrix} \right) \)
Amount paid by X = Rs. 420
Amount paid Y = Rs. 800
(iii) Kindness, Helpfulness, Service to the society etc.

For getting A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \)
For getting A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
Simplifying A³ - 6A² + 7A + kI³ as
\(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
Equating  \(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \)
\(\Rightarrow\)  k - 2 = 0
\(\Rightarrow\)  k = 2
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)
A³ = A² . A = \(\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] \)
Now A³ - 6A² + 7A + kP₃ = 0
\(\Rightarrow \left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] -\left[ \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right] +\left[ \begin{matrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow\)   21 - 30 + 7 + k = 0
\(\Rightarrow\)  k = 2

Given, \(\left[ \begin{matrix} 2x & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 \\ -3 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ 3 \end{matrix} \right] =0\)
\(\Rightarrow \quad \left[ \begin{matrix} 2x-9 & 4x \end{matrix} \right] \left[ \begin{matrix} x \\ 3 \end{matrix} \right] =0\)
\(\Rightarrow \quad \left[ { 2x }^{ 2 }-9x+12x \right] =0\)
\(\Rightarrow \quad { 2x }^{ 2 }+3x=0\)
\(\Rightarrow \quad x\left( 2x+3 \right) =0\)
\(\therefore\)  x = 0 or x = \(-\frac{3}{2}\)