Putting x = \(cos2\theta \left\{ 2tan^{ -1 }\sqrt { \frac { 1-x }{ 1+x } } \right\} \) we get 
\(2tan^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
i,e.,\(2tan^{ -1 }\sqrt { \frac { 2sin^{ 2 }\theta }{ 2cos^{ 2 }\theta } } \)
\(=2\quad tan^{ -1 }\left( tan\theta \right) \)
\(=2\theta =cos^{ -1 }x\)
Hence \(y=e^{ sin^{ 2 } }xcos^{ -1 }x\)
\(\Rightarrow logy=sin^{ 2 }x+log\left( cos^{ 1 }x \right) \)
\(\Rightarrow \frac { 1 }{ y } \times \frac { dy }{ dx } =2sinxcosx+\frac { 1 }{ cos^{ -1 }x } \times \frac { -1 }{ \sqrt { 1-x^{ 2 } } } \)
= sin 2x \(-\frac { 1 }{ cos^{ 1 }x\sqrt { 1-x^{ 2 } } } \)
\(\Rightarrow \frac { dy }{ dx } =e^{ sin^{ 2 } }xcos^{ -1 }x\left[ sin2x-\frac { 1 }{ cos^{ -1 }x\sqrt { 1-x^{ 2 } } } \right] \)

\(A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] \)
|A| = 2 x 0-(-3) x -2+5 x 1
= -6 + 5 = -1
a₁₁ = (-4 + 4) = 0
a₁₂ = -(-6 + 4) = 2
a₁₃ = (3 - 2) = 1
a₂₁ = -(6 - 5) = -1
a₂₂ = (-4 - 5) = -9
a₂₃ = -(2 + 3) = -5
a₃₁ = (12 - 10) = 2
a₃₂ = -(-8 - 15) = 23
a₃₃ = (4 + 9) = 13
\({ A }^{ -1 }=\frac { 1 }{ -1 } \left[ \begin{matrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{matrix} \right] \)
Given set of equations may be written as
\(AX=B,\)
where \(B=\left[ \begin{matrix} 11 \\ -5 \\ -3 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{matrix} \right] \left[ \begin{matrix} 11 \\ -5 \\ -3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0-5+6 \\ -22-45+69 \\ 11-25+39 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] \)
Hence x = 1, y = 2 and x = 3.

Assume, \(f(x)=x^{ 2 }+4x+7\) 
= \((x+2)^{ 2 }+3\)
\(\Rightarrow y-3=(x+2)^{ 2 }\)
\(\Rightarrow x+2=\pm \sqrt { y-3 } \)
\(\Rightarrow x=\sqrt { y-3 } -2\quad y>3\)
\(g:s\rightarrow N\)
\(g(y)=\sqrt { y-3 } -2\)
\(gof(x)=g[f(x)]=g(x^{ 2 }+4x+7)\)
\(=g[(x+2)^{ 2 }+3]\)
\(=\sqrt { (x+3)^{ 2 }+3-3-2 } \)
\(=x+2-2=x\)
\(fog(y)=f[g(y)]\)
\( =f\sqrt { y-3 } -2\)
\(=(\sqrt { y-3 } -2+2)^{ 3 }=y\)
\( \Rightarrow gof=I_{ N }\)
\(gog=I_{ s }\)
\(\Rightarrow f^{ -1 }=g=\sqrt { x-3 } -2\)
Yes, interest & knowledge have bijective relation Value Interest leads to knowledge

We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)

Getting \({ A }^{ 2 }=\left( \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right) \)
and \({ A }^{ 3 }=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) \)
\(\therefore\)  A³ - 4A² - 3A + 11I
\(=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) -\left( \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right) -\left( \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right) +\left( \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) =0\)
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \)
\({ A }^{ 3 }=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] \)
\(\therefore\)  A³ - 4A² - 4A + 11A
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -4\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -\left[ \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right] -\left[ \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
= 0

\(|A|=\left[\begin{matrix}a&b\\c&{1+bc\over a}\end{matrix}\right]\)
\(=a\left(1+bc\over a\right)-bc =1+bc-bc=1\neq0\)
Since A is non-singular matrix ∴  A^-1 exists
Now \(A_{11}={1+bc\over a}; A_{12}=-b; A_{22}=a\)
\(\therefore\ adj.A=\begin{bmatrix}A _{11}&A_{12}\\A_{21}&A_{22} \end{bmatrix}'=\begin{bmatrix}{1+bc\over a}&-c\\-b&a \end{bmatrix}\)
\(=\begin{bmatrix} {1+bc\over a}&-b\\-c&a\end{bmatrix}.\)
\(\therefore\ A^{-1}={adj.A\over |A|}={1\over1}\begin{bmatrix} {1+bc\over a}&-b\\-c&a\end{bmatrix}\)
\(\therefore\ aA^{-1}=a\begin{bmatrix}{1+bc\over a}&-b\\-c&a \end{bmatrix}=\begin{bmatrix} 1+bc&-ab\\-ac&a^2\end{bmatrix}\)....(1)
And (a¹+bc+1)I-aA
\(=(a^2+bc+1)\begin{bmatrix} 1&0\\0&1\end{bmatrix}-a\begin{bmatrix} a&b\\c&{1+bc\over a}\end{bmatrix}\)
\(=\begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1 \end{bmatrix}=\begin{bmatrix}-a^2&-ab\\-ac&-(1+bc) \end{bmatrix}\)
\(=\begin{bmatrix}a^2+bc+1-a^2&0-ab\\0-ac&a^2+bc+-1-1bc \end{bmatrix}\)
\(=\begin{bmatrix}1+bc&-ab\\-ac&a^2 \end{bmatrix}\)..(2)
From (1) and (2), aA^-1 = (a² + bc + 1)I-aA

\(\Rightarrow \left( { 1+x }^{ 2 } \right) ^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2x({ 1+x }^{ 2 })\frac { { d }y }{ d{ x } } =2\)

\(A^{2}=\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{cc} 16+6 & 12+15 \\ 8+10 & 6+25 \end{array}\right]=\left[\begin{array}{ll} 22 & 27 \\ 18 & 31 \end{array}\right]\)
\(A^{2}-x A+y I=O \)
\(\Rightarrow\left[\begin{array}{ll} 22 & 27 \\ 18 & 31 \end{array}\right]-\left[\begin{array}{ll} 4 x & 3 x \\ 2 x & 5 x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 0 & y \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
\(\Rightarrow\left[\begin{array}{ll} 22-4 x+y & 27-3 x+0 \\ 18-2 x+0 & 31-5 x+y \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
\(\Rightarrow 27-3 x=0 \Rightarrow x=9 \)
\(\text {Also } 22-4 x+y=0 \)
y = 14

k = 10/3

On differentiating e^y (x+1) = 1
e^y+(x+1)e^ydy/dx = 0
\(\Rightarrow { e }^{ y }+\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } ={ -e }^{ y }\)

We have, y = tan^-1\(\sqrt { \frac { 1-x }{ 1+x } } \)
Put x = cos2\(\theta\)
\(\Rightarrow\)2\(\theta\) = cos^-1x
\(\Rightarrow\)\(\theta\) = 1/2cos^-1x
y = tan^-1\(\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
y = \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \)
(\(\because\)cos2\(\theta\) = 2cos²\(\theta\)-1 = 1-2sin²\(\theta\)
y = tan^-1(tan \(\theta\))
y = \(\theta\) = 1/2cos^-1x
\(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)​​​​​​​

(i) Given a*b = 3a + b²
\(\Rightarrow 4*3=3(4)+(3)^{ 2 }\)
= 12 + 9
4*3 = 21
(ii) Given a*b = 3a + b²
\(\Rightarrow 5*2=3(5)+2^{ 2 }\)
\(\Rightarrow 5*2=15\)
\(\therefore 5*2=19\)

Let A be any square matrix. Then, 
\(A=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) +\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) \)
= P + Q (say),
where, \(P=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) \)
and \(Q=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) \)
Now, \({ P }^{ T }=\left[ \frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) \right] ^{ T }\)
\(=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) ^{ T }\quad \left[ \because \quad \left( KA \right) ^{ T }=K.{ A }^{ T } \right] \)
\(\Rightarrow \ { P }^{ T }=\frac { 1 }{ 2 } \left[ { A }^{ T }+\left( { A }^{ T } \right) ^{ T } \right] \) \(\left[ \because \ \left( A+B \right) ^{ T }={ A }^{ T }+{ B }^{ T } \right] \)
\(\Rightarrow \ { P }^{ T }=\frac { 1 }{ 2 } \left( { A }^{ T }+A \right) \)  \(\left[ \because \ \left( { A }^{ I } \right) ^{ T }=A \right] \)
\(\Rightarrow { P }^{ T }=\frac { 1 }{ 2 } \left( A{ +A }^{ T } \right) =P\)
\(\therefore \)  P is symmetric matrix.
Also, \({ Q }^{ T }=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) ^{ T }\)
\(=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) ^{ T }\)
\(=\frac { 1 }{ 2 } \left[ { A }^{ T }-\left( { A }^{ T } \right) ^{ T } \right] \)
\(=\frac { 1 }{ 2 } \left[ { A }^{ T }-A \right] \)
\({ \Rightarrow Q }^{ T }=\frac { 1 }{ 2 } \left[ A-{ A }^{ T } \right] =-Q\)
\(\therefore \)  Q is skew symmetric matrix.
Thus, A = P + Q, where P is a symmetric matrix and Q is a skew symmetric matrix.
Hence, A is expressible as the sum of a symmetric and a skew symmetric matrix.
Uniqueness : If possible, let A = R + S, where R is symmetric and S is skew symmetric, then
A^T = (R + S)^T = R^T + S^T
\(\Rightarrow \)  A^T = R - S  (\(\because \)  R^T = R and S^T = - S)
Now, A = R + S and A^T = R - S
\(\Rightarrow R=\frac { 1 }{ 2 } \left[ A+{ A }^{ T } \right] =P\)
\(S=\frac { 1 }{ 2 } \left[ A-{ A }^{ T } \right] =Q\)
Hence, A is uniquely expressible as the sum of a symmetric and a skew symmetric matrix.

\({ sin }^{ -1 }\frac { 2a }{ 1+{ a }^{ 2 } } =2{ tan }^{ -1 }a\)
\(and\quad { sin }^{ -1 }\frac { 2b }{ 1+{ b }^{ 2 } } =2{ tan }^{ -1 }b\)
\(as\left[ 2{ tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) \right] \)
\(2{ tan }^{ -1 }a+2{ tan }^{ -1 }b=2ta{ n }^{ -1 }x\)
\({ tan }^{ -1 }a+{ tan }^{ -1 }b={ tan }^{ -1 }x\)
\({ tan }^{ -1 }\left( \frac { a+b }{ 1-ab } \right) =ta{ n }^{ -1 }x\)
\(x=\frac { a+b }{ 1-ab } \)  
Hence Proved.

\(4\quad { tan }^{ -1 }\frac { 1 }{ 5 } =2\left[ 2{ tan }^{ -1 }\frac { 1 }{ 5 } \right] \)
\(=2\left[ { tan }^{ -1 }\left( \frac { 2\times \frac { 1 }{ 5 } }{ 1-\frac { 1 }{ 25 } } \right) \right] \)
\(\left[ \because 2{ tan }^{ -1 }x={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right] \)
\(=2{ tan }^{ -1 }\left( \frac { \frac { 2 }{ 5 } }{ \frac { 25-1 }{ 25 } } \right) \)
\(=2{ tan }^{ -1 }\left( \frac { 2\times 25 }{ 24\times 5 } \right) =2{ tan }^{ -1 }\left( \frac { 5 }{ 12 } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 2\times 5 }{ 12 } }{ 1-\frac { 25 }{ 144 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 10 }{ 12 } }{ \frac { 144-25 }{ 144 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 10\times 144 }{ 119\times 12 } \right) \)
\(={ tan }^{ -1 }\left( \frac { 120 }{ 119 } \right) \)

(i) Let
A = (1,2,3)
A x A = (1, 1) (2, 2) (3, 3) (1, 2) (2, 1) (3, 2) (2, 3) (1, 3), (3, 1)
\(\therefore\) R = {(1, 1) (1, 2) (2, 1) (2, 2) (2, 3) (1,3) (3, 3) (3,2) (3, 1)}.
(ii) R = {(1, 1) (1,2) (2,1) (2,2) (2,3), (3, 3) (3,2)}

We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

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

f(x) = [x]  is not continuous for integers.Hence not continuous at x = ±2, ±1, 0

\(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(\frac { x_{ 1 } }{ x_{ 1 }+2 } =\frac { x_{ 2 } }{ x_{ 2 }+2 } \)
\(\Rightarrow \) \(x_{ 1 },x_{ 2 }+2x_{ 1 }=x_{ 1 },x_{ 2 }+2x_{ 2 }\)
\(\Rightarrow \) \(2x_{ 1 }=2x_{ 2 }\Rightarrow x_{ 1 }=x_{ 2 }\)
\(\Rightarrow \) \(f\) is one-one.
Let \(y=f(x)=\frac { x }{ x+2 } \Rightarrow xy+2y=x\)
\(\Rightarrow \) \(x(y-1)=-2y\Rightarrow x=\frac { 2y }{ 1-y } \)
\(\Rightarrow \) \(f^{ 1 }(y)=\frac { 2y }{ 1-y } \)
\(\Rightarrow \) \(f^{ 1 }(x)=\frac { 2x }{ 1-x } ,x\neq 1.\)

\(We\ have:\ sin(A+B)\)
\(=sinAcosB+cosAsinB\)
Diff. w.r.t. x, we have:
\(cos\left( A+B \right) \left[ \frac { dA }{ dx } +0 \right] \)
\(=cosB\quad cos\quad A\frac { dA }{ dx } -\left( sin\quad A\frac { dA }{ dx } \right) sin\quad B\)
\(cos\left( A+B \right) \frac { dA }{ dx } \)
\(=cos\left( A+B \right) =cosAcosB-sinAsinB.\)

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\(\Rightarrow \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}A \quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}A \quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}A.\quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ 2R }_{ 2 }]\)
\(Hence,\quad { A }^{ -1 }=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}.\)

Since A and B are both symmetric matrices, therefore A′ = A and B′ = B.
Let AB be symmetric, then (AB)′ = AB
But (AB)′ = B′A′= BA 
Therefore BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric.
Now (AB)′ = B′A′
= BA (as A and B are symmetric)
= AB
Hence AB is symmetric.

We have:
\(A=\left| \begin{matrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{matrix} \right| \)
\(\therefore\ B=A'=\begin{bmatrix} 1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}\)
Now \(|B|=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1 \end{bmatrix}\)
= (1)(-1-8)-0-2(-8+3)
= -9 + 10 = 1 ≠ 0
= B^-1 Exsits
Now \(B_{11}=\begin{vmatrix}-1&2\\4&1 \end{vmatrix}=-1-8=-9;\)
\(B_{12}=-\begin{vmatrix}-2&2\\3&1 \end{vmatrix}=-(-2-6)=8;\)
\(B_{13}=\begin{vmatrix}-2&-1\\3&4 \end{vmatrix}=-8+3=-5\)
\(B_{21}=-\begin{vmatrix} 0&-2\\4&1\end{vmatrix}=-(0+8)=-8;\)
\(B_{22}=\begin{vmatrix} 1&-2\\3&1\end{vmatrix}=1+6=7;\)
\(B_{23}=-\begin{vmatrix} 1&0\\3&4\end{vmatrix}=-(4-0)=-4\)
\(B_{31}=\begin{vmatrix}0&-2\\-1&2 \end{vmatrix}=0-2=-2;\)
\(B_{32}=-\begin{vmatrix} 1&-2\\-2&2\end{vmatrix}=-(2-4)=2;\)
\(B_{33}=\begin{vmatrix} 1&0\\-2&-1\end{vmatrix}=-1+0=-1\)
\(\therefore\ adj.B=\begin{bmatrix}B_{11 }&B_{12}&B_{13}\\B_{21}&B_{22}&B_{23}\\B_{31}&B_{32}&B_{33}\end{bmatrix}=\begin{bmatrix} -9&8&-5\\-8&7&-4\\-2&2/&-1\end{bmatrix}\)
\(=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
\(\therefore\ B^{-1}={adj.B\over |B|}={1\over1}\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
\(=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
Hence, \((A')^{-1}=B^{-1}=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)

The given system of equations is:
5x + 2y = 3
3x + 2y = 5
These can be written as AX = B
\(\Rightarrow X=A^{-1}B\)
where \(A=\begin{bmatrix}5&2\\3&2 \end{bmatrix}, x=\begin{bmatrix}X\\Y \end{bmatrix}and\ B=\begin{bmatrix}3\\5 \end{bmatrix}\)
\(\therefore|A|=\begin{bmatrix} 5&2\\3&2\end{bmatrix}=10-6=4\neq0\Rightarrow A^{-1}\) exists.
\(\therefore A^{-1}={1\over|A|}(adj\ A)={1\over4}\begin{bmatrix} 2&-2\\-3&5\end{bmatrix}\)
From (1), \(X={1\over4} \begin{bmatrix} 2&-2\\-3&5\end{bmatrix}\begin{bmatrix} 3\\5\end{bmatrix}\)
\(={1\over4}\begin{bmatrix}6-10\\-9+25 \end{bmatrix}={1\over4}\begin{bmatrix}-4\\16 \end{bmatrix}\)
\(\Rightarrow\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} -1\\4\end{bmatrix}\)
Hence, x = 1, y = 4.

\( \text {Given, } \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right) \)
\(=\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right) ;\left[\because \cos \frac{\pi}{3}=\frac{1}{2}, \sin \frac{\pi}{6}=\frac{1}{2}\right] \)
\(=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right) \)
\(=\frac{\pi}{3}+\frac{\pi}{3} \)
\(=\frac{2 \pi}{3} \)

\( f(x)=\begin{cases} -x,if\quad x<0 \\ x,\quad if\quad x\ge 0. \end{cases}\)
Clearly the function is defined at 0 and f(0) = 0. Left hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (-x) } =0\)
Similarly, the right hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (x) } =0\)
Thus, the left hand limit, right hand limit and the value of the function coincide at x = 0.
Hence, f is continuous at x = 0.

Let \({ cos }^{ -1 }\left( -\frac { 1 }{ \sqrt { 2 } } \right) =y\), where  \(y\in [0,\pi ]\)
\(\Rightarrow cosy=\frac { 1 }{ \sqrt { 2 } } \)
\(\Rightarrow cosy=-cos\frac { \pi }{ 4 } =cos\left( \pi -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow cosy=cos\frac { 3\pi }{ 4 } \Rightarrow y=\frac { 3\pi }{ 4 } \)
Hence, the required principal value = \(\frac { 3\pi }{ 4 } \)

Let  \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =y\), Then \(\cot y=\frac{-1}{\sqrt{3}}=-\cot \left(\frac{\pi}{3}\right)=\cot \left(\pi-\frac{\pi}{3}\right)=\cot \left(\frac{2 \pi}{3}\right)\)
We know that the range of principal value branch of cot^–1 is (0, π) and \(\cot \left(\frac{2 \pi}{3}\right)=\frac{-1}{\sqrt{3}}\)
Hence, principal value of \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =\frac { 2\pi }{ 3 } .\)

As \(\lim _{ x\rightarrow 0 }{ \left( \frac { sinx }{ x } +cosx \right) } \) 
= 1 + 1 = 2 2k
⇒ k = 1

As sin^-1 x = \(\frac { \pi }{ 2 } \), sin^-1 y = \(\frac { \pi }{ 2 } \), sin^-1z = \(\frac { \pi }{ 2 } \)
⇒ x = 1, y = 1, z = 1
∴ x + y² + z³ = 1 + 1 + 1 = 3