0 (Using properties)

Under what conditions is the matrix equation
A² -B² = (A-B)(A+B)is true 
\(A=\left[ \begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix} \right] B=\left[ \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right] AB=\left[ \begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix} \right] \)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

We have y = log(sin x)
dy/dx = d/dx \(\left| log(sinx) \right| \)
= \(\frac { 1 }{ sinx } \times cosx\)
\(\Rightarrow\) dy/dx = cot x
\(\therefore\)  \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } =-{ cosec }^{ 2 }x\)

We have 2*(3*x) = 33
\(\Rightarrow 2*(3+x+3x)=33\)
\(\Rightarrow 2*(3+4x)=33\)
\(\Rightarrow 2+(3+4x)+2(3+4x)=33\)
\(\Rightarrow 2+3+4x+6+8x=33\)
\(\Rightarrow 12x+11=33\)
\(\Rightarrow 12x=22\Rightarrow x=\frac { 11 }{ 6 } \)

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.

Let x = cos 2\(\theta \)
\(=sin\left[ 2t{ an }^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \right] \)
\(=sin\left[ 2tan^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \right] \)
\(\left[ \because cos2\theta =1-2{ sin }^{ 2 }\theta \ and\ cos\ 2\theta =2{ cos }^{ 2 }\theta -1 \right] \)
\(=sin\left[ 2{ tan }^{ -1 }\left( tan\quad \theta \right) \right] \)
\(=sin(2\theta )=\sqrt { 1-{ cos }^{ 2 }2\theta } \)
\(=sin\quad 2\theta =\sqrt { 1-{ x }^{ 2 } } \)

\(LHS={ tan }^{ -1 }\frac { 1 }{ 4 } +{ tan }^{ -1 }\frac { 2 }{ 9 } \)
\(={ tan }^{ -1 }\left[ \frac { \frac { 1 }{ 4 } +\frac { 2 }{ 9 } }{ 1-\frac { 1\times 2 }{ 4\times 9 } } \right] \)
\(\left[ { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \frac { 9+8 }{ 36 } }{ \frac { 36-2 }{ 36 } } \right] \)
\(={ tan }^{ -1 }\left( \frac { 17 }{ 34 } \right) ={ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(=\frac { 1 }{ 2 } \left( 2{ tan }^{ -1 }\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 2\times \frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 4 } } \right) \)
\(\because \left[ 2{ tan }^{ -1 }={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right] \)
\(=\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 1 }{ 3/4 } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)

We have, \(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
\({ A }^{ \prime }=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right] \)
\({ A+A }^{ \prime }=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A+A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0\)
and \({ A-A }^{ \prime }=\left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A-A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)

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

\(\frac { d }{ dx } \left( { e }^{ m\quad tan^{ -1 } }x \right) ={ e }^{ m\quad tan^{ -1 }x }.\frac { m }{ 1+{ x }^{ 2 } } =\frac { me^{ m\quad tan^{ -1 }x } }{ 1+{ x }^{ 2 } } \)

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

(i) Let \(a,b\in S\) Then \(a\neq 1,b\neq 1\)
To Prove: \(a+b-ab\neq 1\)
If possible, let       a+b-ab=1
\(\Rightarrow a+b-ab-1=0\)
\(\Rightarrow ab-a-b+1=0\)
\(\Rightarrow a(b-1)-(b-1)=0\)
\(\Rightarrow (a-1)(b-1)=0\)
\(\Rightarrow a=1\ or\ b=1,\)
Which is contrary to hypothesis.
Thus \(a+b-ab\neq 1\) for all , \(a,b\in S\)
Hence, S is closed under given operation.
(ii) (a) For all \(a,b\in S\) ,
aob = a+b-ab = b+a-ba = boa.
Hence, the given operation is commutative.
(b) For \(a,b,c\in S,\)
(aob)oc = (a+b-ab)oc
= (a+b-ab)+c-(a+b-ab)c
= a+b+c-ab-bc-ac+abc
and  ao(boc) = ao(b+c-bc)
= a+(b+c-bc)-a(b+c-bc)
= a+b+c-ab-bc-ac+abc.
Thus  (aob)oc = a0(boc).
Hence, the given operation is associative.

For the function \(f\left( x \right) ={ x }^{ 3 }-{ 6x }^{ 2 }+ax+b\), it is given that f(1) = f(3) = 0.
Find the values of a and b and hence verify Rolle's Theorem on [1, 3].

We have: \({ x }^{ 16 }{ y }^{ 9 }={ \left( { x }^{ 2 }+y \right) }^{ 17 },\)
Taking logs., \(log({ x }^{ 16 }{ y }^{ 9 })=log{ \left( { x }^{ 2 }+y \right) }^{ 17 }\)
\(16logx+9logy=17log\left( { x }^{ 2 }+y \right) \)
\( \frac { 16 }{ x } +\frac { 9 }{ y } .\frac { dy }{ dx } =17\frac { 1 }{ { x }^{ 2 }+y } \left[ 2x+\frac { dy }{ dx } \right] \)
\(=\left[ \frac { 9 }{ y } -\frac { 17 }{ { x }^{ 2 }+y } \right] \frac { dy }{ dx } =\frac { 34x }{ { x }^{ 2 }+y } -\frac { 16 }{ x } \)
\(=\frac { { 9x }^{ 2 }+9y-17y }{ y\left( { x }^{ 2 }+y \right) } \frac { dy }{ dx } =\frac { { 34x }^{ 2 }-{ 16x }^{ 2 }-16y }{ x\left( { x }^{ 2 }+y \right) }\)
\(=\frac { 1 }{ y } \left( { 9x }^{ 2 }-8y \right) \frac { dy }{ dx } =\frac { 1 }{ x } \quad \left( { 18x }^{ 2 }-16y \right) \)
\(\frac { 1 }{ y } \quad \frac { dy }{ dx } =\frac { 2 }{ x } \)
\(\frac { dy }{ dx } =\quad \frac { 2y }{ x } \)

Let \(x_{ 1 },x_{ 2 }\in R-\{ 3\} \)
Now \(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(\frac { x_{ 1 }-2 }{ x_{ 1 }-3 } =\frac { x_{ 2 }-2 }{ x_{ 2 }-3 } \)
\(\Rightarrow \) \((​​x_{ 1 }-2)(x_{ 2 }-3)=(x_{ 2 }-2)(x_{ 1 }-3)\)
\(\Rightarrow \) \(x_{ 1 }x_{ 2 }-3x_{ 1 }-2x_{ 2 }+6=\) \(x_{ 2 }x_{ 1 }-3x_{ 2 }-2x_{ 1 }+6\)
\(\Rightarrow \) \(-3x_{ 1 }-2x_{ 2 }=3x_{ 2 }-2x_{ 1 }\)
\(\Rightarrow \) \(x_{ 1 }=x_{ 2 }\)
\(\Rightarrow \) \(f\) is one-one.
Let \(y\in R-\{ 1\} \).
Then  f(x)=y.
When \(\left( \frac { x-2 }{ x-3 } \right) =y,x\neq 3\)
\(\Rightarrow \) x - 2 = yx - 3y
\(\Rightarrow \) x - xy = 2 - 3y
\(\Rightarrow \) \(x=\frac { 2-3y }{ 1-y } \in A\)
\(\left[ \because \frac { 2-3y }{ 1-y } =3-\frac { 1 }{ 1-y } \neq 3\quad \right] \)
\(\therefore \) Corresponding to each \(y\in B\), there exists \(\frac { 2-3y }{ 1-y } \in A\)
such that \(f\left( \frac { 2-3y }{ 1-y } \right) =y\)
\(\Rightarrow \)f is onto.
Hence, 'f' is one-one and onto.

We have: \(x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix}+2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix}=2\begin{bmatrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{bmatrix}.\) 
\(x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix}+2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix}=2\begin{bmatrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{bmatrix}.\)
\(\Rightarrow \ \begin{bmatrix} { 2x }^{ 2 } & 2x \\ 3x & { x }^{ 2 } \end{bmatrix}+\begin{bmatrix} 16 & 10x \\ 8 & 8x \end{bmatrix}=\begin{bmatrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 12x \end{bmatrix}\)
\(\Rightarrow \ \begin{bmatrix} { 2x }^{ 2 }+16 & 12x \\ 3x+8 & { x }^{ 2 }+8x \end{bmatrix}=\begin{bmatrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 12x \end{bmatrix}\)
Comparing, \(12x=48,\quad 3x+8=20,\quad and\quad { x }^{ 2 }+8x=12x.\) 
All these give \(x=4\)

\(\left| \begin{matrix} x & { x }^{ 2 } & yz \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
= \(\left| \begin{matrix} x-y & { x }^{ 2 }-y^2 & yz-zx \\ y-z & { y }^{ 2 }-z^2 & zx-xy \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
[Operating R\(\rightarrow\)R₁-R₂ & R\(\rightarrow\)R₂-R₃] =
\(\left| \begin{matrix} x-y & (x-y)(x+y) & z(y-x) \\ y-z &(y-z)(y+z)& x(z-y)\\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
= \((x-y)(y-z)\left| \begin{matrix}0 &x-z&-z+x \\ 1 & y+z &-x \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
[Taking (z-x)common from R₁]
\((x-y)(y-z)(z-x)\left| \begin{matrix} 0 &-1 & -1 \\ 1 & y+z & -x \\ z & z^2 & xy \end{matrix} \right| \)
\((x-y)(y-z)(z-x)\left| \begin{matrix} 0 & 0 & -1 \\ 1 & x+y+z & -x \\ z & z^2-xy & xy \end{matrix} \right| \)
\((x-y)(y-z)(z-x)(-1)\left| \begin{matrix} 1 & x+y+z \\z &z^2-xy\end{matrix} \right| \)
= (x-y) (y-z) (z-x) (-1) [z²-xy-zx-yz-z²]
= (x-y) (y-z) (z-x) (1) (-xy-yz-zx)
= (x-y) (y-z) (z-x) (xy+yz+zx)

\(\Delta=\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| \)
=\(\left| \begin{matrix} 0 & -a & b \\a & 0 &c \\- b & -c & 0 \end{matrix} \right| \)
=\((-1)^3=\left| \begin{matrix} 0 & a &- b \\-a & 0 &-c \\ b & c & 0 \end{matrix} \right| \)
\(=-\left|\begin{matrix}0&a&-b\\ -a&0&-c\\b&c&0\end{matrix}\right|\)
\(\Rightarrow2\Delta=0\ \Rightarrow \Delta=0\)

Here,
\(f(x)f(y)=\left[ \begin{matrix} cosx & -sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} cosxcosy-sinxsiny & -sinycosx-sinxcosy & 0 \\ sinxcosy+cosxsiny & -sinxsiny+cosxcosy & 0 \\ 0 & 0 & 1 \end{matrix} \right]\)
\( =\left[ \begin{matrix} cos(x+y) & -sin(x+y) & 0 \\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=f(x+y).\)

Take x = cos θ, then proceeding as above, we get, \({ \sin }^{ -1 }(2x\sqrt { 1-{ x }^{ 2 } } )\) = 2 cos^–1 x

Let \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) =y\), Then \(\sin y=\frac{1}{\sqrt{2}}\)
We know that the range of the principal value branch of \(\sin ^{-1} \text { is }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) and \(\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\)
Therefore, principal value of \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) is \ \frac { \pi }{ 4 } \)

Let cosec⁻¹ (2) = y. Then,\(\cos e c y=2=\cos e c\left(\frac{\pi}{6}\right)\)
We know that the range of the principal value branch of cosec⁻¹ is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\) 
Therefore, the principal value of \(\cos e c^{-1}(2) \text { is } \frac{\pi}{6}\)

0

x =8 y = 8

\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right) \)
\(\text { Put } x^{2}=\cos 2 \theta, \text { then } \)
\(\mathrm{LHS}=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)\) 
\( \tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right)\) 
\(\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)\)
[divide numerator and denominator inside the bracket by \(cos \theta]\)

The given system of equations can be written as 
AX = B   ..(i)
where \(A=\left[ \begin{matrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{matrix} \right] ,\)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] and\quad B=\left[ \begin{matrix} 3 \\ 10 \\ 1 \end{matrix} \right] \)
Now, \(|A|=\left| \begin{matrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{matrix} \right| \)
= 1(-21+1) -1(-14-3) -1(-2-9)
= -20 + 17 + 11 = 8 \(\ne\)0
The given system of equations is consistent and have unique solution.
Let C_ij be the cofactor of the element a_ij in the determinant A, then
C₁₁ = -20, C₁₂ = 17, C₁₃ = -11
C₂₁ = 8, C₂₂ = -4, C₂₃ = 4
C₃₁ = 4, C₃₂ = -3, C₃₃ = 1
\(\therefore \ { A }^{ -1 }=\frac { 1 }{ \left| A \right| } (adj\quad A)=\frac { 1 }{ 8 } { [{ C }_{ ij }] }^{ T }\)
\(=\frac { 1 }{ 8 } { \left[ \begin{matrix} -20 & 17 & -11 \\ 8 & -4 & 4 \\ 4 & -3 & 1 \end{matrix} \right] }^{ T }\)
\(=\frac { 1 }{ 8 } { \left[ \begin{matrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{matrix} \right] }\)
From (i), \(X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 8 } { \left[ \begin{matrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{matrix} \right] }\left[ \begin{matrix} 3 \\ 10 \\ 1 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 8 } { \left[ \begin{matrix} -60+80+4 \\ 51-40-3 \\ -33+40+1 \end{matrix} \right] }=\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ 8 \\ 8 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 1 \\ 1 \end{matrix} \right] \)
\(\therefore\) x = 3, y = 1 and z = 1.

LHL = a + 3
RHL = \(\frac{b}{2}\)
f(x) is continuous at x = 0,
So, a + 3 = 2 = \(\frac{b}{2}\)
a = -1 and b = 4
a = 3 and b = -2
Alterantive method
We have given that f(x) is continuous at x = 0
\(\therefore \lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } \)
= f(0) = 2...(i)
Now,
\(\therefore \lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ h\rightarrow { 0 }^{ } }{ f\left( 0-h \right) } \)
Consider x = 0 - h \(\Rightarrow\) x \(\rightarrow\) 0 ^- \(\Rightarrow\) h \(\rightarrow\) 0
Thus,
\(\\ \lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ h\rightarrow { 0 }^{ } }{ f\left( h \right) } \)
\(=\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { -sin\left( a+1 \right) h-2sinh }{ -h } } \)
\(=\lim _{ h\rightarrow { 0 }^{ - } }{ \frac { -sin\left( a+1 \right) h-2sinh }{ -h } } \)
\(=\lim _{ h\rightarrow { 0 }^{ - } }{ \frac { sin\left( a+1 \right) h }{ -h } + } \lim _{ h\rightarrow { 0 }^{ - } }{ \frac { 2sinh }{ h } } \)
\(=\lim _{ h\rightarrow { 0 }^{ - } }{ \frac { sin\left( a+1 \right) h }{ -h } + } \lim _{ h\rightarrow { 0 }^{ - } }{ \frac { 2sinh }{ h } } \)  
= a + 1 + 2 
= a + 3....(ii)
Now using eqn.(ii)
\(\lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =2\)
\(\Rightarrow\) a + 3 = 2
\(\Rightarrow\) a = -1
\(\\ \lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ h\rightarrow { 0 }^{ } }{ f\left( 0+h \right) } \)
\(=\lim _{ h\rightarrow { 0 }^{ } }{ f\left( h \right) } \)
Considering x = 0 + h, x\(\rightarrow\)0⁺\(\Rightarrow\) h\(\rightarrow\)0
Now,
\(=\lim _{ h\rightarrow { 0 }^{ + } }{ \frac { \sqrt { 1+bx } -1 }{ x } } \)
\(=\lim _{ h\rightarrow { 0 }^{ } }{ \frac { \sqrt { 1+bh } -1 }{ h } } \)
\(=\lim _{ h\rightarrow { 0 }^{ } }{ \frac { \sqrt { 1+bh } -1 }{ h } } \times \frac { \sqrt { 1+bh } +1 }{ \sqrt { 1+bh } +1 } \)
\(​​​​=\lim _{ h\rightarrow { 0 }^{ } }{ \frac { \left( \sqrt { 1+bh } \right) ^{ 2 }-\left( 1 \right) ^{ 2 } }{ h\left( \sqrt { 1+bh } +1 \right) } } \)
\(​​​​=\lim _{h\rightarrow { 0 }^{ } }{ \frac { 1+bh-1 }{ h\left( \sqrt { 1+bh } +1 \right) } } \)
\(=\lim _{ h\rightarrow { 0 }^{ } }{ \frac { bh }{ h\left( \sqrt { 1+bh } +1 \right) } } (h\neq0)\)
\(​​​​=\lim _{ h\rightarrow { 0 }^{ } }{ \frac { b }{ h\left( \sqrt { 1+bh } +1 \right) } } \)
\(=\frac { b }{ \left( \sqrt { 1+b\times 0 } +1 \right) } \)
\(=\frac { b }{ 1+1 } =\frac { b }{ 2 } \)
Now, using eqn. (i)
\(​​​​=\lim _{ x\rightarrow { 0 }^{ + } }{ f } \left( x \right) =f\left( 0 \right) =2\)
\(\Rightarrow \frac{b}{2}=2\)
\(\Rightarrow\) b = 4

\(\forall x\in [0,\infty ),y=9x^{ 2 }+6x-5\)
\(=(3x+1)^{ 2 }-6\ge -5\prec \)
\( f=\ [-5,\infty )\)
Co-domain f, hence f is not onto and hence not invertible
Let us take the modified co-domain
\(f=\ [-5,\infty )\)
Let us now check whether f is one-one
Let \(x_{ 1 },x_{ 2 }\neq [0,\infty )\)
\( f(x_{ 1 })=f(x_{ 2 })\)
\( \Rightarrow (3x_{ 1 }+1)^{ 2 }-6=(3x_{ 2 }+1)^{ 2 }-6\)
\( \Rightarrow 3x_{ 1 }+1=3x_{ 2 }+1\)
\(\Rightarrow x_{ 1 }=x_{ 2 }\)
Hence f is oe-one
Since with the modified co-domain = the range f, f is both o0ne-one and onto hence invertible
From (i) above for any
\(y\neq [-5,\infty )\)
\(x=\frac { \sqrt { y+6 } -1 }{ 3 } \)
\(f^{ -1 }[-5,\infty )\rightarrow [0,\infty ),f^{ -1 }(y)\)
\(=\frac { \sqrt { y+6 } -1 }{ 3 } \)

Find AB, A', B', B'A', compare.Proceed.

As y = xy ⇒ log y = y log x
⇒ \(\frac { 1 }{ y } .{ y }^{ ' }=\frac { y }{ x } +logx.{ y }^{ ' }\)
\(\Rightarrow { y }^{ ' }\left[ \frac { 1 }{ y } -log \ x \right] \)
\(=\frac { y }{ x } \Rightarrow x(1-y \ log \ x){ y }^{ ' }={ y }^{ 2 }\)

as sec^-1x + sec^-1 y = \(\frac{\pi}{2}\)
⇒ \(\frac{\pi}{2}\) - cosec^-1 x + \(\frac{\pi}{2}\) - cosec^-1 y = \(\frac{\pi}{2}\)
⇒ cosec^-1 x + cosec^-1 y = \(\frac{\pi}{2}\)