\(\left( AB \right) ^{ -1 }=\left[ \begin{matrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix} \right] \)

\({ A }^{ -1 }=\frac { 1 }{ 8 } \left[ \begin{matrix} 5 & -1 \\ -7 & 3 \end{matrix} \right] \)

\(xcos\left( \frac { y }{ x } \right) (ydx+xdy)\)
\(=ysin\left( \frac { y }{ x } \right) (xdy-ydx)\)
\(\Rightarrow \left( y+x\frac { dy }{ dx } \right) xcos\left( \frac { y }{ x } \right) \)
\(=\left( x\frac { dy }{ dx } -y \right) ysin\left( \frac { y }{ x } \right) \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} }{ x\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} } \)
Eqn (i) is homogeneous So, putting y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(v+x\frac { dv }{ dx } =\frac { vx[xcosv+vxsinv] }{ x[vxsinv-xcosv] } \)
\(\Rightarrow v+x\frac { dv }{ dx } =\frac { v[cosv+vsinv] }{ vsinv-cosv } \)
\(=\frac { 2vcosv }{ vsinv-cosv } \)
\(\Rightarrow \frac { 1-cosv-vsinv }{ vcosv } dv=-\frac { 2dx }{ x } \)
\(\Rightarrow \int { \frac { 1.cosv-vsinv }{ vcosv } dv=-2\int { \frac { dx }{ x } } } \)
\(\Rightarrow\) log |v cosv| = -2log x + log C
\(\Rightarrow\) x²v cosv = C
\(\Rightarrow\) xy cos(y/x) = C[as y = vx]

Equation of plane: 2x + y + 3z = 26
Normal to the plane = \(2\hat{i}+\hat{j}+3\hat{k}\)
\(\therefore \) < 2, 1, 3 > are direction ratios of the normal to the plane.

Equation of the line through P(2, 3, 4) and perpendicular to the give plane is
\({{x-2}\over{2}}={{y-3}\over{1}}={{z-4}\over{3}}=\lambda\)
\(\therefore\) Co-ordinates of point N are N \((2\lambda+2,\lambda+3,3\lambda+4)\)
Since N lies on the plane,
\(\Rightarrow\)  \(2(2\lambda+2)+(\lambda+3)+3(3\lambda+4)=26\)
\(\Rightarrow\)  \(4\lambda+4+\lambda+3+9\lambda+12=26\)
\(\Rightarrow\)  \(14\lambda=26-19\)
\(\Rightarrow\)  \(14\lambda=7\)
\(\Rightarrow\)  \(\lambda={{1}\over{2}}\)
\(\therefore\) Co-ordinates of the foot of the perpendicular i.e., N are
\(\left( 2\left({{1}\over{2}} \right)+2,{{1}\over{2}}+3,3\left({{1}\over{2}} \right)+4 \right)\) i.e., \(\left( 3,{{7}\over{2}},{{11}\over{2}} \right)\)
\(\therefore\) The length of perpendicular form P to given plane is
\(|NP|=\sqrt{{2-3}^{2}+{\left(3-{{7}\over{2}} \right)}^{2}+\left(4-{{11}\over{2}}\right)^{2}}\)
\(=\sqrt{1+{{1}\over{4}}+{{9}\over{4}}}\)
\(=\sqrt{{{14}\over{1}}}\)
\(=\sqrt{{7}\over{2}}\) units
Now, N is the mid-point of PP', where P'\((\alpha,\beta,\gamma)\) is the image of point P.
\(\therefore\)   \(\left(3,{{7}\over{2}},{{11}\over{2}} \right)=\left({{2+\alpha}\over{2}} ,{{3+\beta}\over{2}},{{4+\gamma}\over{2}}\right)\)
\(\Rightarrow \) \(3={{2+\alpha}\over{2}};{{7}\over{4}}={{3+\beta}\over{2}};{{11}\over{2}}={{4+\gamma}\over{2}}\)
\(\Rightarrow\)  \(6=2+\alpha;7=3+\beta;11=4+\gamma\)
\(\Rightarrow\)  \(\alpha=4;\beta=4;\gamma=7\)
\(\therefore\) Image of point P is P'(4, 4, 7).

Let, \(I=\int _{ 0 }^{ \pi }{ \frac { x\sin { x } }{ 1+\cos ^{ 2 }{ x } } } dx\)
Apply the property \(\int _{ 0 }^{ a }{ f\left( x \right) } =\int _{ 0 }^{ a }{ f\left( a-x \right) } \)
\(=\int _{ 0 }^{ \pi }{ \frac { \left( \pi -x \right) \sin { \left( \pi -x \right) } }{ 1+\cos ^{ 2 }{ \left( \pi -x \right) } } } dx\)
\(=\int _{ 0 }^{ \pi }{ \frac { \left( \pi -x \right) \sin { x } }{ 1+\cos ^{ 2 }{ x } } } dx\)
\(\therefore \ 2I=\int _{ 0 }^{ \pi }{ \frac { \sin { x } }{ 1+\cos ^{ 2 }{ x } } } dx\)
\(\pi \int _{ -1 }^{ 1 }{ \frac { 1 }{ 1+{ t }^{ 2 } } dx } \)  (where t = cosx)
As it is an even function
\(=2\pi \int _{ 0 }^{ 1 }{ \frac { dt }{ 1+{ t }^{ 2 } } } \)
\(=2\pi \left[ \tan ^{ -1 }{ t } \right] _{ 0 }^{ 1 }\)
\(\Rightarrow 2I=2\pi \frac { \pi }{ 4 } =\frac { 2{ \pi }^{ 2 } }{ 4 } \)
\(\Rightarrow I=\frac { { \pi }^{ 2 } }{ 4 } \)

Let the award money for the values of honesty, regularity and hard work be Rs. x Rs. y and Rs. z respectively. Then the. system of equations is
x + y + z = 6,000 
x + 3z = 11,000
x - 2y + z = 0
Matrix equation is
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 6,000 \\ 11,000 \\ 0 \end{matrix} \right] \)
i.e., AX = B
\(|A|=\quad \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{matrix} \right| \)
= 1(0+6)-1(1-3)+1(-2-0)
= 6 \(\ne\)0
\(\therefore\) A^-1 exists.
adj A = \({ \left[ \begin{matrix} 6 & 2 & -2 \\ -3 & 0 & 3 \\ 3 & -2 & -1 \end{matrix} \right] }^{ T }\)
\(=\left[ \begin{matrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{matrix} \right] \)
\(X={ A }^{ -1 }B\)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{matrix} \right] \left[ \begin{matrix} 6,000 \\ 11,000 \\ 0 \end{matrix} \right] \quad \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 500 \\ 2,000 \\ 3,500 \end{matrix} \right] \)
x = 500, y = 2,000, z = 3,500 
Hence, award money given for the value of Honesty = Rs. 500, award money given for the value of Regularity = Rs. 2,000 and, award money given. for the value of Hard-work = Rs. 3,500.
Value: The school must include the value of Obedience for the awards.

Let E₁: urn has 2 white balls
E₂: urn has 3 white balls
E₃: urn has 4 white balls
A: 2 balls drawn are white
P(E₁) = P(E₂) = P(E₃) = \(\frac { 1 }{ 3 } \)
P(A/E₁)=\(\frac { { 2 }_{ { C }_{ 2 } } }{ { 4 }_{ { C }_{ 2 } } } =\frac { 1 }{ 6, } \)
\(P(A/{ E }_{ 2 })=\frac { { 2 }_{ { C }_{ 2 } } }{ { 4 }_{ C_{ 2 } } } =\frac { 1 }{ 6 } ,\)
\(P(A/{ E }_{ 2 })=\frac { { 3 }_{ { C }_{ 2 } } }{ { 4 }_{ C_{ 2 } } } =\frac { 1 }{ 2 } ,\)
\(P(A/{ E }_{ 3 })=\frac { { 4 }_{ { C }_{ 2 } } }{ { 4 }_{ { C }_{ 2 } } } =1\)
\(\therefore P({ E }_{ 3 }/A)\)
\(=\frac { { P(E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .1 }{ \frac { 1 }{ 3 } .\frac { 1 }{ 6 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } +\frac { 1 }{ 3 } .1 } \)
\(=\frac { 6 }{ 10 } =0.6\)

A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \);  A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
L.H.S = A³ - 6A² + 7A + 2I
\(=\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) -\left( \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right) +\left( \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right) +\left( \begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) =0\)
= RHS