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 = 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.

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

\({ tan }^{ -1 }\frac { x }{ y } -{ tan }^{ -1 }\frac { x-y }{ x+y } \)
\(={ tan }^{ -1 }\left[ \frac { \frac { x }{ y } -\frac { x-y }{ x+y } }{ 1+\left( \frac { x }{ y } \right) \left( \frac { x-y }{ x+y } \right) } \right] \)
\(\left[ \because { tan }^{ -1 }x-{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\left[ \frac { \frac { x(x+y)-y(x-y) }{ y(x+y) } }{ \frac { y(x+y)-x(x-y) }{ y(x+y) } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { { x }^{ 2 }+xy-xy+{ y }^{ 2 } }{ { xy+y }^{ 2 }+{ x }^{ 2 }-xy } \right] \)
\(={ tan }^{ -1 }\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) ={ tan }^{ -1 }1=\frac { \pi }{ 4 } \)   Hence Proved

Let the relation \(\bot \) on the set \(\angle \) be denoted by R
For symmetry : \(l_{ 1 }Rl_{ 2 }\Rightarrow l_{ 1 }\bot l_{ 1 }\Rightarrow l_{ 2 }\bot l_{ 1 }Rl_{ 1 }\)
The relation R is symmetric
For reflective \(l_{ 1 }Rl_{ 2 }\Rightarrow l_{ 1 }\bot l\)
No line can be perpendicular to it self
For transitive : \(l_{ 1 }Rl_{ 2 }\Rightarrow l_{ 1 }\) for \(I\ \in \ \angle \)
It is not true
Relation is not transitive

A relation R on a non-empty set A is called a transitive relation if (a, b), (b, c) \(\in R\) then (a, c) \(\in R\) , i.e., aRb, bRc implies aRc.
Thus a relation R on a non empty set A is said to be transitive if there exist a, b, c \(\in A\) such that (a, b)(b, c) \(\in R\)  implies (a, c) . \(\in R\)
Example
Let A = (1, 2, 3, 6)
R = (3, 6) (6, 1) (3, 1)
3 R 6 and 6 R 1 \(\Rightarrow \)(3, 1)  \(\in R\)
\(\therefore\) A is transitive.

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, \(\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] -3\left[ \begin{matrix} x \\ 2y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ -9 \end{matrix} \right] \)
\(\Rightarrow\)  x² - 3x = - 2 and y² - 6y = - 9
\(\Rightarrow\)  x² - 3x + 2 = 0 and y² - 6y + 9 = 0
\(\Rightarrow\)  x² - 2x - x + 2 = 0 and y² - 3y - 3y + 9 = 0
\(\Rightarrow\)  x(x - 2) - 1(x - 2) = 0 and y(y - 3) - 3(y - 3) = 0
\(\Rightarrow\)  (x - 2)(x - 1) = 0 and (y - 3)(y - 3) = 0
\(\therefore\)   x = 1, 2 and y = 3, 3

Given \(A=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix};kA\begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \)
\({ A }^{ 2 }=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}=\begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix}\) [multiplying row by column]
\({ A }^{ 2 }=kA\Rightarrow \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\(\Rightarrow k=2A\)
On comparing with Eq. (ii), we got k = 2

\(A^{-1}=\frac{1}{|A|} \text { adj } A\)
\(|A|=\left[\begin{array}{lr} 2 & 3 \\ 5 & -2 \end{array}\right]=-4-15=-19 \)
\(\Rightarrow A^{-1}=-\frac{1}{19}\left[\begin{array}{rr} -2 & -3 \\ -5 & 2 \end{array}\right]=\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=\frac{1}{19} A \)

\( f(x)=|x|=\left\{\begin{array}{l} x \text { if } x>0 \\ -x \text { if } x<0 \end{array}\right.\\ f(-1)=|-1|=1, f(1)=|1|=1 \therefore f(-1)=f(1) ,\)

$f$ is not one-one
Now, consider $-1\in R$.
but it is known that f(x) = |x| is always non-negative.
Thus, there does not exist any element $x$ in  domain $R$ such that f(x) = |x| = −1 
$\therefore f$ is not onto.
Hence, the modulus function is neither one-one nor onto.

\(\left\{ \frac { { 2xy }^{ 2 }-y-{ y }^{ 3 }cos(xy) }{ { y }^{ 2 }ccos(xy)-x+{ y }^{ 2 } } \right\} \)

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

\(\Delta =\left| \begin{matrix} a & { a }^{ 2 } & bc \\ b & { b }^{ 2 } & ca \\ c & { c }^{ 2 } & ab \end{matrix} \right| \)
Using R₁-->aR₁,R₂ --> bR₂,R₃ -->c R₃
\(\Delta =\frac { 1 }{ abc } \left| \begin{matrix} { a }^{ 2 } & { a }^{ 2 } & abc \\ { b }^{ 2 } & { b }^{ 3 } & 1 \\ { c }^{ 2 } & { c }^{ 3 } & abc \end{matrix} \right| \)
Taking abc common from C₃
Using R₂-->R₂-R₁,R₃-->R₃-R₁
\(\Delta =\left| \begin{matrix} { a }^{ 2 } & { a }^{ 3 } & 1 \\ { b }^{ 2 }-{ a }^{ 2 } & { b }^{ 3 }-{ a }^{ 3 } & 0 \\ { c }^{ 2 }-{ a }^{ 2 } & { c }^{ 3 }-{ a }^{ 3 } & 0 \end{matrix} \right| \)
Taking common (b-a) from R₂, (c-a) from R₃
\(\Delta =(b-a)(c-a)\left| \begin{matrix} { a }^{ 2 } & { a }^{ 3 } & 1 \\ b+a & { b }^{ 2 }+ab+{ a }^{ 2 } & 0 \\ c+a & { c }^{ 2 }+ac+{ a }^{ 2 } & 0 \end{matrix} \right| \)
Expanding along C₃, we get
\(\Delta =(b-a)(c-a)\left| \begin{matrix} b+a & { b }^{ 2 }+ab+{ a }^{ 2 } \\ c+a & { a }^{ 2 }+ac+{ a }^{ 2 } \end{matrix} \right| \)
R₂ --> R₂ - R₁
=\((b-a)(c-a)(b-c)\left| \begin{matrix} b+a & { b }^{ 2 }+ab+{ a }^{ 2 } \\ c-b & { c }^{ 2 }-{ b }^{ 2 }+ac-ab \end{matrix} \right| \)
= \((b-a)(c-a)(b-c)\left| \begin{matrix} a+b & { a }^{ 2 }+{ b }^{ 2 }+ab \\ 1 & a+b+c \end{matrix} \right| \)
= \((a-b)(b-c)(c-a)\{ { a }^{ 2 }+ab+ac+a\underline { b } +\underline { { b }^{ 2 } } +bc-{ a }^{ 2 }-{ b }^{ 2 }-ab^{ 2 }\} \)
= (a-b)(b-c)(c-a)(ab+bc+ac)

\(\tan ^{ -1 }{ (x-1) } +\tan ^{ -1 }{ (x+1) } =\tan ^{ -1 }{ 3x } -\tan ^{ -1 }{ x } \)
\(\Rightarrow \tan ^{ -1 }{ \left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) } =\tan ^{ -1 }{ \left( \frac { 2x }{ 2-3{ x }^{ 2 } } \right) } \)
\(\Rightarrow \frac { 2x }{ 2-{ x }^{ 2 } } =\frac { 2x }{ 2-3{ x }^{ 2 } } \)
\(\Rightarrow 2x(1+3{ x }^{ 2 }-2+{ x }^{ 2 })=0\)
\(\Rightarrow x=0,\frac { 1 }{ 2 } ,-\frac { 1 }{ 2 } \)
Alternative method:
We have, 
\(\tan ^{ -1 }{ (x-1) } +\tan ^{ -1 }{ x } +\tan ^{ -1 }{ (x+1) } =\tan ^{ -1 }{ 3x } \)
\(\tan ^{ -1 }{ (x-1) } +\tan ^{ -1 }{ (x+1) } +\tan ^{ -1 }{ x } =\tan ^{ -1 }{ 3x } \)
Taking LHS
\(\tan ^{ -1 }{ \left( \frac { (x-1)+(x+1) }{ 1-(x-1)(x+1) } \right) } +\tan ^{ -1 }{ x } \)
\(= \tan ^{ -1 }{ \left( \frac { 2x }{ 1-{ x }^{ 2 }+1 } \right) } +\tan ^{ -1 }{ x } \)
\(=\tan ^{ -1 }{ \left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) +\tan ^{ -1 }{ x } } \)
\(=\tan ^{ -1 }{ \left[ \frac { \frac { 2x }{ 2-{ x }^{ 2 } } +x }{ 1-\frac { 2x }{ 2-{ x }^{ 2 } } \times x } \right] } \)
\(= \tan ^{ -1 }{ \left( \frac { 2x+2x-{ x }^{ 3 } }{ 2-{ x }^{ 2 }-2{ x }^{ 2 } } \right) } \)
\(\therefore \tan ^{ -1 }{ \left( \frac { 4x-{ x }^{ 3 } }{ 2-3{ x }^{ 2 } } \right) } =\tan ^{ -1 }{ 3x } \)
\(\Rightarrow \frac { 4x-{ x }^{ 3 } }{ 2-3{ x }^{ 2 } } =3x\)
\(\Rightarrow 6x-9{ x }^{ 3 }=4x-{ x }^{ 3 }\)
\(\Rightarrow 8{ x }^{ 2 }-2x=0\)
\(\Rightarrow x=0\quad or\quad 4{ x }^{ 2 }-1=0\)
\(\Rightarrow x=0\quad or\quad 4{ x }^{ 2 }=1\)
\(\Rightarrow x=0\quad or\quad x=\pm \frac { 1 }{ 2 } \)
\(\Rightarrow x=0,\quad x=\frac { 1 }{ 2 } \quad x=-\frac { 1 }{ 2 } \)
\(\Rightarrow x=0\quad or\quad { x }^{ 2 }=\frac { 1 }{ 4 } \)

For getting A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \)
For getting A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
Simplifying A³ - 6A² + 7A + kI³ as
\(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
Equating  \(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \)
\(\Rightarrow\)  k - 2 = 0
\(\Rightarrow\)  k = 2
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)
A³ = A² . A = \(\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] \)
Now A³ - 6A² + 7A + kP₃ = 0
\(\Rightarrow \left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] -\left[ \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right] +\left[ \begin{matrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow\)   21 - 30 + 7 + k = 0
\(\Rightarrow\)  k = 2

\(\text { We have } A=I A \Rightarrow\left[\begin{array}{ll} 6 & 5 \\ 5 & 4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A\)
\(\Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 5 & 4 \end{array}\right]=\left[\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right] A\)
\(\text { [by performing } \left.R_{1} \rightarrow R_{1}-R_{2}\right]\)
\(\Rightarrow\left[\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{rr} 1 & -1 \\ -5 & 6 \end{array}\right] A\)
\(\text { [by performing } \left.R_{2} \rightarrow R_{2}-5 R_{1}\right]\)
\(\Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 5 & -6 \end{array}\right] A\)
\(\text { [by performing } \left.R_{2} \rightarrow(-1) R_{2}\right]\)
\(\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} -4 & 5 \\ 5 & -6 \end{array}\right] A\)
\(\text { [by performing } \left.R_{1} \rightarrow R_{1}-R_{2}\right]\)
\({ A }^{ -1 }=\begin{bmatrix} -4 & 5 \\ 5 & -6 \end{bmatrix}\)

(a) (i) By Definition \(a*b=b*a=1,\forall a,b\in R\)
Hence, R is commutative
(ii) \((a*b)*c=1*c=1\)
and \(a*(b*c)=a*1=1\forall a,b,c\in R\)
Hence, R is associative.
(b) We have: \(a*b=\frac { a+b }{ 2 } \forall a,b,\in R.\)
(i) \(a*b=\frac { a+b }{ 2 } =\frac { b+a }{ 2 } \)   [∵ Real numbers are commutative under addition]
\(=b*a.\)
Thus \(a*b=b*a.\)
Hence, ' * ' is commutative in R.
(ii) For \(a,b,c\in R\).
\((a*b)*c=\frac { a+b }{ 2 } *c\)
\(=\frac { \frac { a+b }{ 2 } +c }{ 2 } =\frac { a+b+2c }{ 4 } \)
And \(a*(b*c)=a*\frac { b+c }{ 2 } \)
\(=\frac { a+\frac { b+c }{ 2 } }{ 2 } =\frac { 2a+b+c }{ 4 } \)
Thus \((a*b)*c\neq a*(b*c).\)
Hence, ' * ' is not associative in R.

\(We\quad have:\quad ax+{ by }^{ 2 }=cos\quad y\)
\( Diff\quad w.r.t.\quad x,\quad a+2b\quad y\frac { dy }{ dx } =-sin\quad y\quad \frac { dy }{ dx } \)
\((2by+siny)\frac { dy }{ dx } \)
\(Hence, \frac { dy }{ dx } =-\frac { a }{ 2by+sin\quad y }\)

\({ \left( I+A \right) }^{ 2 }=(I+A)(I+A)\)
\(=II+IA+AI+AA\)
\(=I+A+A+{ A }^{ 2 }\)
\( =I+2A+A \quad \left[ \because \quad { A }^{ 2 }=A \right] \)
\(=I+3A ....(1)\)
\(\therefore \ { \left( I+A \right) }^{ 3 }={ (I+A) }^{ 2 }(I+A)\)
\(=(I+3A)(I+A)\ \left[ Using\quad (1) \right] \)
\( =II+IA+3AI+3AA\)
\(=I+A+3A+3{ A }^{ 2 }\)
\(=I+A+3A+3A\ \left[ \because \quad { A }^{ 2 }=A \right] \)
\(=I+7A...(2)\)
\(Hence,\ 7A-{ \left( I+A \right) }^{ 3 }=7A-(I=7A) = -I\) Using (2)

\(Let\quad y=sin({ x }^{ 2 }+5)\)
\(Put\quad { x }^{ 2 }+5=t\)
\(y=sin\quad t,\quad where\quad t={ x }^{ 2 }+5\)
\( \frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(=cos\quad t.\quad (2x+0)=2xcos({ x }^{ 2 }+5)\)

We have   \(=\left[\begin{array}{c} 4000+50000+250000 \\ 120000+100000+500000 \end{array}\right]=\left[\begin{array}{l} 304,000 \\ 720,000 \end{array}\right] \begin{aligned} &\rightarrow X \\ &\rightarrow Y \end{aligned}\)
\(=\left[\begin{array}{ll} 340,000 \\ 720,000 \end{array}\right] \begin{aligned} &\rightarrow \mathrm{X} \\ &\rightarrow \mathrm{Y} \end{aligned}\)
So the total amount spent by the group in the two cities is Rs. 340,000 paise and Rs. 720,000 paise, i.e., Rs. 3400 and Rs. 7200, respectively.

(i) Here \(f(x)=17\)
\(\Rightarrow \)  \(x^{ 2 }+1=17\Rightarrow x^{ 2 }=16\Rightarrow x=\pm 4\).
\(\therefore \) Pre-image of 17 = {-4, 4}.
(ii)   \(f(x)=-3\)
\(\Rightarrow \) ​​​​​​​​​​​​​​ \(x^{ 2 }+1=-3\Rightarrow x^{ 2 }=-4\Rightarrow x\) is not real.
\(\therefore \) Pre-image of -3 = \(\phi \)

The given equations are:
x+3y=5
2x+6y=8
Here \(A=\begin{bmatrix}1&3\\2&6 \end{bmatrix}and\ B=\begin{bmatrix}5\\8 \end{bmatrix}\)
Now |A|=\(\begin{vmatrix}1&3\\2&6 \end{vmatrix}=6-6=0\)
Now  \(adj\ A=\begin{bmatrix} 6&-2\\-3&1\end{bmatrix}'=\begin{bmatrix}6&-3\\-2&1 \end{bmatrix}\) and \(\begin{bmatrix} 5\\8\end{bmatrix}\)
\(\therefore(adj\ A)B=​​​​\begin{bmatrix} 6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8 \end{bmatrix}\)
\(=\begin{bmatrix}30-24\\-10+8 \end{bmatrix}=\begin{bmatrix}6\\-2 \end{bmatrix}\neq0\)
Hence the given system of equation is inconsistent

Let \(A=\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1 \end{bmatrix}\)
\(\therefore |A|=\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1 \end{bmatrix}\)
\(=2(-1-0)-(1)(4+0)+3(8-7)\)
\(=-2-4+3=-3\neq0\)
A is non-singular = A^-1 exists.
Now \(A_{11}=(-1)^{1+1}\begin{vmatrix} -1&0\\2&1\end{vmatrix}=(+1)(-1-0)=1;\)
\(A_{12}=(-1)^{1+2}\begin{vmatrix}4&0\\-7&1 \end{vmatrix}=(-1)(4)=-4;\)
\(A_{13}=(-1)^{1+3}\begin{vmatrix} 4&-1\\-7&2\end{vmatrix}=(+1)(1)=1;\)
\(A_{21}=(-1)^{2+1}\begin{vmatrix} 1&3\\2&1\end{vmatrix}=(-1)(-5)=5;\)
\(A_{22}=(-1)^{2+2}\begin{vmatrix} 2&3\\-7&1\end{vmatrix}=(1)(23)=23;\)
\(A_{23}=(-1)^{2+3}\begin{vmatrix}2&1\\-7&2 \end{vmatrix}=(-1)(4+7)=-11;\)
\(A_{31}=(-1)^{3+1}\begin{vmatrix} 1&3\\-1&0\end{vmatrix}=(+1)(3)=3;\)
\(A_{32}=(-1)^{3+2}\begin{vmatrix} 2&3\\4&0\end{vmatrix}=(-1)(-12)=12;\)
\(A_{33}=(-1)^{3+3}\begin{vmatrix} 2&1\\4&-1\end{vmatrix}=(+1)(-2-4)=-6\)
\(\therefore\ adj\ A\begin{bmatrix}-1&5&3\\-4&23&12\\3&12&-6 \end{bmatrix}=\begin{bmatrix} -1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}\)
\(\therefore\ adj\ A^{-1}={1\over|A|}(adj\ A)\)
\(={1\over-3}\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6 \end{bmatrix}\)
\(=\left[ \begin{matrix} \frac { 1 }{ 3 } & \frac { -5 }{ 3 } & -1 \\ \frac { 4 }{ 3 } & \frac { -23 }{ 3 } & -4 \\ -\frac { 1 }{ 3 } & \frac { 11 }{ 3 } & 2 \end{matrix} \right] \)

Let tan^-1(-1) = y, where \(y\in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(\Rightarrow tany=-1=-tan\frac { \pi }{ 4 } =tan\left( -\frac { \pi }{ 4 } \right) \)
\( \Rightarrow y=-\frac { \pi }{ 4 } \)
Hence, the required principal value = \(-\frac { \pi }{ 4 } \)

Let \({ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) =y\) where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\(\Rightarrow siny=-\frac { \pi }{ 2 } \)
\(\Rightarrow siny=-sin\frac { \pi }{ 6 } =sin\left( -\frac { \pi }{ 6 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 6 } \)
Hence, the required principal value = \(-\frac { \pi }{ 6 } \)

As y' = 5Ae^5x - 5Be^-5x
and y'' = 25Ae^5x + 25Be^-5x
= 25y

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 = \(\frac{\pi}{2}\)