{-1}

\(\left\{ \frac { a }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 }\left( a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } \right) } } \right\} \)

0

x = 1, 2 y = 3\(\pm\)3\(\sqrt { 2 } \)

We have: \(y={ e }^{ m{ sin }^{ -1 }x }\)
\(\left( 1-{ x }^{ 2 } \right) { y }_{ 2 }-x{ y }_{ 1 }={ m }^{ 2 }y\)
\({ y }_{ 1 }={ e }^{ m{ sin }^{ -1 }x }\frac { m }{ \sqrt { 1-{ x }^{ 2 } } } \)
\(\sqrt { 1-{ x }^{ 2 } } { y }_{ 1 }=m{ e }^{ m{ sin }^{ -1 }x }\)
squaring, \(\left( 1-{ x }^{ 2 } \right) { y }_{ 1 }^{ 2 }\quad ={ m }^{ 2 }{ e }^{ 2m{ sin }^{ -1 }x }\)
\(\left( 1-{ x }^{ 2 } \right) { y }_{ 1 }^{ 2 }={ m }^{ 2 }{ y }^{ 2 }\)
Again diff.w.r.t.x, \(x,\left( 1-{ x }^{ 2 } \right) 2{ y }_{ 1 }{ y }_{ 2 }+(0-2x){ y }_{ 1 }^{ 2 }=2{ m }^{ 2 }y{ y }_{ 1 }\)
\( \left( 1-{ x }^{ 2 } \right) { y }_{ 2 }-{ xy }_{ 1 }={ m }^{ 2 }y\)
\( \left( 1-{ x }^{ 2 } \right) { y }_{ 2 }-{ xy }_{ 1 }-{ m }^{ 2 }y=0\)
which is true.

\((I)\ A+B+C=\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ 1 & -3 & -1 \end{matrix} \right] +\left[ \begin{matrix} 4 & 5 & 6 \\ -1 & 0 & 1 \\ 2 & 1 & 2 \end{matrix} \right] +\left[ \begin{matrix} -1 & -2 & 1 \\ -1 & 2 & 3 \\ -1 & -2 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+4-1 & 2+5-2 & 3+6+1 \\ -1-1-1 & 0+0+2 & 2+1+3 \\ 1+2-1 & -3+1-2 & -1+2+2 \end{matrix} \right] =\left[ \begin{matrix} 4 & 5 & 10 \\ -3 & 2 & 6 \\ 2 & -4 & 3 \end{matrix} \right] \)
\(\therefore (A+B+C)\prime =\left[ \begin{matrix} 4 & -3 & 2 \\ 5 & 2 & -4 \\ 10 & 6 & 3 \end{matrix} \right] ...(1)\)
\((II) A\prime +B\prime +C\prime =\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 0 & -3 \\ 3 & 2 & -1 \end{matrix} \right] +\left[ \begin{matrix} 4 & -1 & 2 \\ 5 & 0 & 1 \\ 6 & 1 & 2 \end{matrix} \right] +\left[ \begin{matrix} -1 & -1 & -1 \\ -2 & 2 & -2 \\ 1 & 3 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+4-1 & -1-1-1 & 1+2-1 \\ 2+5-2 & 0+0+2 & -3+1-2 \\ 3+6+1 & 2+1+3 & -1+2+2 \end{matrix} \right] =\left[ \begin{matrix} 4 & -3 & 2 \\ 5 & 2 & -4 \\ 10 & 6 & 3 \end{matrix} \right] ...(2)\)
From (1) and (1), \((A+B+C)\prime =A\prime +B\prime +C\prime \) .

\(\Delta=\left| \begin{matrix} -a({ b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 }) & { 2b }^{ 3 } & { 2c }^{ 3 } \\ { 2 }a^{ 3 } & -b({ c }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 }) & { 2c }^{ 3 } \\ { 2a }^{ 3 } & { 2b }^{ 3 } & -c({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }) \end{matrix} \right|\)
Operating R₁-->R₁-R₂& R₂-->R₂-R₃, we get:
\(\Delta=\begin{vmatrix}-a(b^2+c^2-a^2)&b(a^2+b^2+c^2)&0\\0&-b(c^2+a^2+b^2)&c(a^2+b^2+c^2)\\2a^3&2b^3&-c(a^2+b^2-c^2) \end{vmatrix}\)
\(=(a^2+b^2+c^2)\begin{vmatrix} -a&b&0\\0&-b&c\\2a^3&2b^3&-c(a^2+b^2-c^2)\end{vmatrix}\)
\(=abc(a^2+b^2+c^2)^2\begin{vmatrix}-1&1&0\\0&-1&1\\2a^2&2b^2&-a^2-b^2+c^2 \end{vmatrix}\)
\(=abc(a^2+b^2+c^2)^2\begin{vmatrix}-1&1&0\\0&-1&1\\2a^2&2b^2&-a^2-b^2+c^2 \end{vmatrix}\) [Operating C₂--> C₂+C₁]
\(abc(a^2+b^2+c^2)^2(-1)[a^2+b^2-c^2-2a^2-2b^2]\) [Expanding by R₁]
\(=abc(a^2+b^2+c^2)^2(-1)(-a^2-b^2-c^2)\)
\(=abc(a^2+b^2+c^2)^3,\)
which is true

\(\Delta=\left| \begin{matrix} { yz }-x^{ 2 } & { zx }-y^{ 2 } & { xy-z }^{ 2 } \\ { zx-y }^{ 2 } & { xy }-z^{ 2 } & { yz-x }^{ 2 } \\ { xy-z }^{ 2 } & { yz }-x^{ 2 } & { zx-x }^{ 2 } \end{matrix} \right| \)
\(=\begin{vmatrix} -(x^2+y^2+z^2-xy-yz-zx)&zx-y^2&xy-z^2\\-(x^2+y^2+z^2-xy-yz-zx)&xy-z^2&yz-x^2\\-(x^2+y^2+z^2-xy-yz-zx)&yz-x^2&zx-y^2\end{vmatrix}\) [Operating C₁-->C₁+C₂+C₃]
\(= -(x^2+y^2+z^2-xy-yz-zx)\begin{vmatrix}1&zx-y^2&xy-z^2\\1&xy-z^2&yz-x^2\\1&yz-x^2&zx-y^2\end{vmatrix}\)  [Taking -(x²+y²+z²-xy-yz-zx) common from C₁]
=-(x²+y²+z²-xy-yz-zx)
\(\begin{vmatrix}0&(x-y)(x+y+z)&(y-z)(x+y+z)\\0&(x-y)(x+y+z )&(y-z)(x+y+z)\\1&yz-x^2&zx-y^2 \end{vmatrix}\) [Operating R₁-->R₁-R₃ and R₂+R₂-R₃]
\(=-(x^2+y^2+z^2-xy-yz-zx)(x+y +z)\)
\(=\begin{vmatrix} 0&x-y&y-z\\0&x-z&y-x\\1&yz-x^2&zx-y^2\end{vmatrix}\) [Taking (x+y+z)common from C₂and C₃]
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)\begin{vmatrix}0&x-y&y-z\\0&x-z&y-x \\1&yz-x^2&zx-y^2\end{vmatrix}\)
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(z-y)(y-x)-(x-z)(x-2)]\)
\(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(yz-zx-y^2+xy-x^2-zx+zx-z^2)]\) [Expanding by C₁]
 \(=-(x+y+z)(x^3+y^3+z^3-3xyz)[(x^2+y^2+z^2-xy-yz-zx)]\)
Hence, △ is divisible by (x+y+z) and quotient is (x³+y³+z³-3xyz)(x²+y²+z²-x-yz-zx)