f(x) = 5x - 3
LHL = \(\underset { x\rightarrow 3^{ - } }{ lim } \left( 5x-3 \right) =5(-3)-3\)
= -15 - 3
= -18
RHL \(\underset { x\rightarrow 3^{ + } }{ lim } \left( 5x-3 \right) =5(-3)-3\)
= -15-3
= - 18
\(\left[ f(x) \right] _{ atx=-3 }=5(-3)-3=-18\)
Hence F(x) is continuous at x = -3

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 shall prove the result by mathematical induction on n.
Step 1 : When n = 1, by the definition or integral powers of a matrix, we have
\({ A }^{ 1 }=\left[ \begin{matrix} 1+2\left( 1 \right) & -4n \\ n & 1-2\left( 1 \right) \end{matrix} \right] =\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \end{matrix} \right] \)
So, the result is true for n = 1.
Step 2 : Let the result be true for n = m. Then,
\({ A }^{ m }=\left[ \begin{matrix} 1+2m & -4m \\ m & 1-2m \end{matrix} \right] \)
Now, we will show that the result is true for n = m + 1, i.e.,
\({ A }^{ m+1 }=\left[ \begin{matrix} 1+2\left( m+1 \right) & -4\left( m+1 \right) \\ \left( m+1 \right) & 1-2\left( m+1 \right) \end{matrix} \right] \)
By the definition of integral powers of a square matrix, we have
\({ A }^{ m+1 }={ A }^{ m }.A\)
\(\Rightarrow { A }^{ m+1 }=\left[ \begin{matrix} 1+2m & -4m \\ m & 1-2m \end{matrix} \right] \left[ \begin{matrix} 3 & -4 \\ 1 & -1 \end{matrix} \right] \)
[by supposition (i)]
\(\Rightarrow { A }^{ m+1 }=\left[ \begin{matrix} 3+6m-4m & -4-8m+4m \\ 3m+1-2m & -4m-4+2m \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+2\left( m+1 \right) & -4\left( m+1 \right) \\ \left( m+1 \right) & 1-2\left( m+1 \right) \end{matrix} \right] \)
This shows that the result is true for n = m + 1, whenever it is true for n = m.
Hence, by the principle of mathematical induction, the result is true for any positive integer n.

\(R.H.S.=2{ sin }^{ -1 }\sqrt { \frac { 1-x }{ 2 } } \quad \quad \begin{cases} Let\quad x=cos\theta \\ \Rightarrow \theta ={ cos }^{ -1 }x \end{cases}\)
\(=2{ sin }^{ -1 }\left( \sqrt { \frac { 1-cos\theta }{ 2 } } \right) \)
\( \begin{cases} as\quad cos\theta =1=2{ sin }^{ 2 }\frac { \theta }{ 2 } \\ \Rightarrow 1-cos\theta =2sin^{ 2 }\frac { \theta }{ 2 } \end{cases}\)
\(=2{ sin }^{ -1 }\left( sin\frac { \theta }{ 2 } \right) \)
\(=2\left( \frac { \theta }{ 2 } \right) =\theta ={ cos }^{ -1 }x\)
L.H.S = R.H.S

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

\(f(x)=x^{ 2 },x\in Ra\)
\(f(1.1)=(1.1)^{ 2 }z\)
\(=1.21\)
\(f(1)=(1)^{ 2 }=1\)
\(\frac { f(1.1)-f(1) }{ 1.1-1 } =\frac { 1.21 }{ 1.1-1 } =\frac { 0.21 }{ 0.1 } \)
\(=2.1\)

The given function is f: X → Y and relation on X is R={(a, b): f(a) = f (b)}
Reflexive Since, for every x ∈ X, we have
f'(x) = f(x)
⇒ (xx) ∈ R, ∀ ∈ X Therefore, R is reflexive.
Symmetric Let (x, y) ∈ R
Then, f(x)=f(y)
⇒ f(y)=f(x)
⇒ (y, x) ∈ R
Thus, (x, y) ∈ R ⇒  (y, x)∈ R, ∀x, y∈ X
Therefore, R is symmetric.
Transitive Let x, y, z∈ X such that
(x, y) ∈ R and (y, z) ∈ R
Given a relation 5 in \(N \times N\), defined as
(a, b) S(c, d), if a+d=b+c.
Reflexive Let (a, b) be any arbitrary element of \(N \times N\)
i.e. \((a, b) \in N \times N\), where \(a, b \in N\)
Now, as a+b=b+a
[∴ addition is commutative ]
Therefore \quad(a, b) S(a, b)
So, S is reflexive.
Symmetric \(\operatorname{Let}(a, b),(c, d) \in N \times N\), such that (a, b)
S(c, d). Then, a+d=b+c
\( \Rightarrow  b+c=a+d \Rightarrow c+b=d+a \)
\(\Rightarrow (c, d) S(a, b)\)
So, S is symmetric.
Transitive Let \((a, b),(c, d),(e, f) \in N \times N\) such that (a, b) S(c, d) and (c, d) S(e, f).
Then, a+d=b+c and c+f=d+e
On adding the above equations, we get
a+d+c+f=b+c+d+e 
\( \Rightarrow  a+f=b+e \Rightarrow(a, b) S(e, f)\)
So, S is transitive.
Thus, S is reflexive, symmetric and transitive. Hence, S is an equivalence

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

We have, \(\left[ \begin{matrix} 2x+y & x-y \\ x-z & x+y+z \end{matrix} \right] =\left[ \begin{matrix} 10 & -1 \\ 2 & 8 \end{matrix} \right] \)
\(\Rightarrow\) 2x + y = 10, x - y = - 1
x - z = 2 and x + y + z = 8
\(\therefore\) 2(y - 1) + y = 10  \(\Rightarrow\)  2y + y + 2 = 10
\(\Rightarrow\)3y = 12  \(\Rightarrow\)  y = 4
\(\therefore\)  x = 3
3 - z,  \(\Rightarrow\)  z = 1
\(\therefore\)  x = 3, y = 4, z = 1

\(A^1=\left| \begin{matrix} 3 & 10 \\ 2 & 7 \end{matrix} \right|\)
Alternative Method:
\(A=\left| \begin{matrix} 3 & 10 \\ 2 & 7 \end{matrix} \right| \)
\(A^{ 1 }=\frac { 1 }{ \left| A \right| } (adjA) \)
\(\left| A \right| =\left| \begin{matrix} 3 & 10 \\ 2 & 7 \end{matrix} \right| =21-20=1 \)
\(adjA=\left| \begin{matrix} 7 & -10 \\ -2 & 3 \end{matrix} \right| \Rightarrow A^{ 1 }=\left| \begin{matrix} 7 & -10 \\ -2 & 3 \end{matrix} \right|\)

\({ tan }^{ -1 }\sqrt { 3 } -{ sec }^{ -1 }(-2)=-\frac { \pi }{ 3 } \)
Alternative Method :
\({ tan }^{ -1 }\sqrt { 3 } -{ sec }^{ -1 }(-2)\)
\(=tan^{ -1 }\left[ tan\left( \frac { \pi }{ 3 } \right) \right] -{ sec }^{ -1 }\left[ -sec\left( \frac { \pi }{ 3 } \right) \right] \)
\(=\frac { \pi }{ 3 } -\left( \pi -\frac { \pi }{ 3 } \right) \)
\(\left[ \because \quad { tan }^{ -1 }(tan\quad \theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \quad and\quad { sec }^{ -1 }(sec\quad \theta )=\theta \forall \theta \in (0,\quad \pi )-\left( \frac { \pi }{ 2 } \right) \right] \)
\(=\frac { 2\pi }{ 3 } -\pi =-\frac { \pi }{ 3 } \)

Given \(A=\left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{matrix} \right]  \)
For a skew symmetric matrix, a_ij = -a_ji ∀∀ i, j
\(\therefore \ x=-(-2)=2\)

Let numbers be x, y, z then
x + y + z = -1
2y + 3z = 5
x + y – z = -1
\(x=-\frac { 7 }{ 2 } \), y = \(\frac { 5 }{ 2 } \), z = 0

\(\Delta =\left| \begin{matrix} ({ x+y) }^{ 2 } & zx & zy \\ zx & (z+y)^{ 2 } & xy \\ zy & xy & (z+x)^{ 2 } \end{matrix} \right| \)
R₁⟶ zR₁, R₂⟶ xR₂, R₃ ⟶ yR₃
\(\Delta =\frac { 1 }{ xyz } \left| \begin{matrix} z(x+y)^{ 2 } & { z }^{ 2 }x & { z }^{ 2 }y \\ { x }^{ 2 }z & x(x+y)^{ 2 } & { x }^{ 2 }y \\ { y }^{ 2 }z & { y }^{ 2 }x & y(z+x)^{ 2 } \end{matrix} \right| \)
\(\Delta =\left| \begin{matrix} (x+y)^{ 2 }-{ z }^{ 2 } & 0 & { z }^{ 2 } \\ 0 & (z+y)^{ 2 }-x^{ 2 } & { x }^{ 2 } \\ { y }^{ 2 }-(z+x){ x }^{ 2 } & { y }^{ 2 }-(z+x)^{ 2 } & (z+x)^{ 2 } \end{matrix} \right| \)
\((x+y+z)^{ 2 }\left| \begin{matrix} x+y-z & 0 & { z }^{ 2 } \\ 0 & z+y-x & x^{ 2 } \\ -2x & -2z & 2xz \end{matrix} \right| \)
R₃⟶ R₃-R₁-R₂
= \((x+y+z)^{ 2 }\left| \begin{matrix} x+y-z & 0 & { z }^{ 2 } \\ 0 & z+y-x & { x }^{ 2 } \\ -2x & -2z & 2xz \end{matrix} \right| \)
C₁⟶ C₁+C₃/z, C₂⟶ C₂+C₃/x, we get
\(\Delta =(x+y+z)^{ 2 }\left| \begin{matrix} x+y & \frac { { z }^{ 2 } }{ x } & { z }^{ 2 } \\ \frac { { x }^{ 2 } }{ z } & z+y & { x }^{ 2 } \\ 0 & 0 & 2xz \end{matrix} \right| \)
= 2xyz(x + y + z)³

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 \(-\frac { 1 }{ cos^{ 1 }x\sqrt { 1-x^{ 2 } } } \)
\(\Rightarrow \frac { dy }{ dx } =e^{ sin^{ 2 } }xcos^{ -1 }x\left[ sin2x-\frac { 1 }{ cos^{ -1 }x\sqrt { 1-x^{ 2 } } } \right] \)

Given, 
\(f(t)=\left| \begin{matrix} \cos { t } & t & 1 \\ 2\sin { t } & t & 2t \\ \sin { t } & t & t \end{matrix} \right| =f(t)=\left| \begin{matrix} \cos { t } & t & 1 \\ 0 & -t & 0 \\ \sin { t } & t & t \end{matrix} \right|\)
\( [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }]\)
\(=t\left| \begin{matrix} \cos { t } & 1 & 1 \\ 0 & -1 & 0 \\ \sin { t } & 1 & 1 \end{matrix} \right| \)
Expanding an along R₂, we get
t[(-1)(tcost-sint)]
= -t²cost+sint
\(\lim _{ t\rightarrow 0 }{ \frac { f(t) }{ { t }^{ 2 } } } =\lim _{ t\rightarrow 0 }{ \frac { -{ t }^{ 2 }cost+tsint }{ { t }^{ 2 } } } \)
\(=\lim _{ t\rightarrow 0 }{ \left( \frac { -{ t }^{ 2 }cost }{ { t }^{ 2 } } +\frac { tsint }{ { t }^{ 2 } } \right) } \)
\(=\lim _{ t\rightarrow 0 }{ \left( -cost+\frac { sint }{ t } \right) } \)
\(=-1+\lim _{ t\rightarrow 0 }{ \frac { sint }{ t } } \)
= -1+1 = 0

Let (a, b), (c, d) E Q x Q. Then b + ad may not be mequal to d + cd. We find that (1, 2) * (2, 3) = (2, 5), (2,3) * (1,2) = (2,7) '*(2, 5) Hence, * is not commutative.
Let, (a, b), (c, d), (e,f> E Q x Q, {(a, b) * (c, d) * (e,f) 
= (ace, b + ad + acf)
= (a, b) * {(c, d) * (e, f)
Hence * is associative. 1
(x, y) Q x Q is the identity element for * if 2
(x, y)  Q x Q is the inverse of (a, b) E Q x Q if (c, d) * (a, b) = (a, b) * (c, d) = (1,0),
i.e., (ac, b + ad) = (ca, d + cb) = (1,0)
\(\Rightarrow c=\frac { 1 }{ a } ,d=\frac { -b }{ a } \)
The inverse of (a,b) \(\in Q\) XQ \(a\neq 0\quad \left( \frac { 1 }{ a } ,\frac { -b }{ a } \right) \)

The given equation can be written as:
\(\tan ^{ -1 }{ \left( \frac { x-1 }{ x-2 } \right) } =\tan ^{ -1 }{ 1 } -\tan ^{ -1 }{ \left( \frac { x+1 }{ x+2 } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { 1-\frac { x+1 }{ x+2 } }{ 1+\frac { x+1 }{ x+2 } } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { x+2-x-1 }{ 2x+3 } \right) } \)
\(\therefore \quad \frac { x-1 }{ x-2 } =\frac { 1 }{ 2x+3 } \)
\(\Rightarrow (x-1)(2x+3)=x-2\)
\(\Rightarrow { 2x }^{ 2 }+x-3=x-2\)
\(\Rightarrow { x }^{ 2 }=\frac { 1 }{ 2 } \)
\(\Rightarrow x=\pm \frac { 1 }{ \sqrt { 2 } } \)

= (sin x)^{tan x}{sec²x.log (sin x) +1} + (cos x)^{sec x} {sec x tan x log (cos) - sec x tan x}.

\({\left[\begin{array}{ll} x-2 & -3 \end{array}\right]\left[\begin{array}{l} x \\ 3 \end{array}\right]=0} \)
\(\Rightarrow\left[x^{2}-2 x-9\right]=[0] \Rightarrow x^{2}-2 x-9=0 \)
\(x=\frac { 2\pm \sqrt { 4+36 } }{ 2 } =1\pm \sqrt { 10 } \)

\(We\quad have\quad :\quad x=a(\theta -sin\theta ),\quad y=a(1+cos\theta )\)
\(\frac { dx }{ d\theta } =a(1-cos\theta ),\quad \frac { dy }{ d\theta } =a(0-sin\theta )=-a\quad sin\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } \)
\(=\frac { -a\quad sin\theta }{ a(1-cos\theta ) } =\frac { -sin\theta }{ 1-cos\theta } \)
\(=\frac { -2sin\theta /2cos\theta /2 }{ 2{ sin }^{ 2 }\theta /2 } =-\frac { cos\theta /2 }{ sin\theta /2 } =-cot\frac { \theta }{ 2 } \)

\(We\quad have\quad :\quad x=cos\quad \theta -cos2\theta ,\quad y=sin\theta -sin2\theta \)
\(\frac { dx }{ d\theta } =-sin\theta +2sin2\theta ,\quad \frac { dy }{ dt } =cos\theta -2cos\quad 2\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { cos\quad \theta -cos2\theta }{ -sin\theta +2sin2\theta } =-\frac { cos\quad \theta -cos2\theta }{ -sin\theta -2sin2\theta } \)

The matrix A has 2 columns which is equal to the number of rows of B.
Hence AB is defined. Now
\(\mathrm{AB}=\left[\begin{array}{lll} 6(2)+9(7) & 6(6)+9(9) & 6(0)+9(8) \\ 2(2)+3(7) & 2(6)+3(9) & 2(0)+3(8) \end{array}\right]\)
\(=\left[\begin{array}{ccc} 12+63 & 36+81 & 0+72 \\ 4+21 & 12+27 & 0+24 \end{array}\right]=\left[\begin{array}{ccc} 75 & 117 & 72 \\ 25 & 39 & 24 \end{array}\right]\)

LHS:\(\begin{vmatrix} 2y&y-z-x&2y\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}\)
= \(\begin{vmatrix}x+y+z&x+y+z&x+y+z\\2z&2z&z-x-y\\x-y-z&2x&2x \end{vmatrix}\)
= \((x+y+z)\begin{vmatrix}1&1&1\\2z&2z&z-x-y\\x-y-z&2x&2x \end{vmatrix}\)
= \((x+y+z)\begin{vmatrix} 1&0&0\\2z&0&-(x+y+z)\\x-y-z&x+y+z&0\end{vmatrix}\)
= \((x+y+z)(1)\begin{vmatrix}0&-(x+y+z)\\x+y+z&0 \end{vmatrix}\)
= \((x+y+z)[0+(x+y+z)^2]\)
= \((x+y+z)^3\)= RHS

Since a * b = L.C.M of a,b,
\(\therefore \) 2*3 = L.C.M of 2 and 3 = 6 \(\notin \){1, 2, 3, 4, 5}.
Hence ' * ' is not a binary operation.

a\(\wedge \)b=min {a,b}.
The table is as below:

Let A=\(\left|\begin{matrix}1&2\\ 3&4\end{matrix}\right|\)
Then =\(A_11=(-1)^{1+1}M_{11}=(+1)(4)\)
\(=4\)
\(=A_{12}=(-1)^{1+2}M_{12}=(-1)(3)\)
\(=-3\)
\(A_{21}=(-1)^{2+1}M_{21}=(-1)(2)\)
\(=-2\)
\(A_{22}=(-1)^{2+2}M_{22}=(+1)(1)\)
\( =1\)
\(\therefore\ adj\ A=\begin{vmatrix} A_{11}&A_{12}\\A_{21}&A_{22}\end{vmatrix}'\)
\(\left[\begin{matrix}4&-3\\ -2&1\end{matrix}\right]'=\left[\begin{matrix}4&-2\\ -3&1\end{matrix}\right]\)

\(We\quad have:\quad 3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix},\)
\(\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix}.\)
Equating corresponding elements:
3x = x + 4 \(\Rightarrow \) 2x = 4 \(\Rightarrow \) x = 2
3y = x + y + 6 \(\Rightarrow \) 2y = 2 + 6 \(\Rightarrow \) 2y = 8  \(\Rightarrow \) y = 4
3w = 2w + 3  \(\Rightarrow \) w = 3 
and 3z = z + w-1 \(\Rightarrow \) 2z = 3-1 = 2  \(\Rightarrow \) z = 1.
Hence, x = 2, y = 4, z = 1 and w = 3.

We have 
\(\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]=\tan ^{-1}\left[\frac{\frac{a \cos x-b \sin x}{b \cos x}}{\frac{b \cos x+a \sin x}{b \cos x}}\right]=\tan ^{-1}\left[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right]\)
\(=\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x)=\tan ^{-1} \frac{a}{b}-x\)

Let x = sec θ, then \(\sqrt{x^2-1}=\sqrt{\sec ^2 \theta-1}=\tan \theta\)
Therefore, \(\cot ^{-1} \frac{1}{\sqrt{x^2-1}}=\cot ^{-1}(\cot \theta)=\theta=\sec ^{-1} x\) which is the simplest form

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

Performing \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ cosyR }_{ 1 }+{ siny }{ R }_{ 2 }\)
we get on simplification
\(\triangle =sin \ y-cos \ y=\sqrt { 2 } .sin\left( y-\frac { \pi }{ 4 } \right) \le \sqrt { 2 } \)
As -1 \(\le sin\left( y-\frac { \pi }{ 4 } \right) \le 1\)
⇒ \(-\sqrt { 2 } \le \sqrt { 2 } \) \(sin\left( y-\frac { \pi }{ 4 } \right) \le \sqrt { 2 } \)
∴ \([-\sqrt { 2 } ,\sqrt { 2 } ]\)