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

Given, y = tan^-1\(\sqrt { \frac { sinx }{ 1+cosx } } \)
y = tan^-1\(\sqrt { \frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ 2{ cos }^{ 2 }\frac { x }{ 2 } } } \)
y = tan^-1\(\left( \sqrt { tan\frac { x }{ 2 } } \right) \)
y = tan^-1

(i) f(g) = f(x)g(x)
\(fg=(\sqrt { x+4 } )(\sqrt { x-4 } )=\sqrt { x^{ 2 }-4 } \)
(ii) \(\frac { f }{ g } =\frac { f(x) }{ g(x) } =\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \times \frac { \sqrt { x-4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x^{ 2 }-16 } }{ x-4 } \)

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

Let A be a skew-symmetric matrix. Then by definition \({ A }^{ \prime }=-A\)
\(\Rightarrow\) the (i, j)^th element of \({ A }^{ \prime }\) = the (i, j)^th element of (- A)
\(\Rightarrow\) the (j, i)^th element of A = - the (i, j)^th element of A
For the diagonal elements i = j \(\Rightarrow\) the (i, j)the element of A = - the (i, j)^th element of A.
\(\Rightarrow\)  the (i, j)^th element of A = 0
Hence the diagonal elements are all zero.

\({ tan }^{ -1 }\frac { 3 }{ 4 } +{ tan }^{ -1 }\frac { 3 }{ 5 } -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(={ tan }^{ -1 }\left( \frac { \frac { 15+12 }{ 20 } }{ \frac { 20-9 }{ 20 } } \right) -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(\left[ \because { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\frac { 27 }{ 11 } -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(\left[ \because { tan }^{ -1 }x-{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\left( \frac { \frac { 27 }{ 11 } -\frac { 8 }{ 19 } }{ 1+\frac { 27\times 8 }{ 11\times 19 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 513-88 }{ 209 } }{ \frac { 209+216 }{ 209 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 425 }{ 425 } \right) \)
\(={ tan }^{ -1 }1=\frac { \pi }{ 4 } \)

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, \(A=\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \) and \(B=\left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} -2+0 & 8+2 \\ -3+0 & 12+8 \end{matrix} \right] \)
\(\left[ \begin{matrix} -2 & 10 \\ -3 & 20 \end{matrix} \right] \)
and \(BA=\left[ \begin{matrix} -1 & 4 \\ 0 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 10 & 15 \\ 6 & 8 \end{matrix} \right] \)
\(\therefore\)   \(AB\neq BA\)

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

Function f (x) = x² is continuous in [-1, 1], differentiable in ]-1, 1[and f (-1) = f (1)]. Hence, Rolle's Theorem verified.
⇒ f'(c) = 0  ⇒ 2c = 0 ⇒ c = 0. 

\(X=\begin{bmatrix} 2 & 4 \\ 5 & 0 \end{bmatrix}-\begin{bmatrix} 2 & -1 \\ 3 & -1 \end{bmatrix} \)
\(=\begin{bmatrix} 2-2 & 4+1 \\ 5-3 & 0+1 \end{bmatrix}=\begin{bmatrix} 0 & 5 \\ 2 & 1 \end{bmatrix}. \)

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

$f$ 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.

\(Let\quad y={ x }^{ x }+{ x }^{ a }+{ a }^{ x }+{ a }^{ a }\)
\(\frac { dy }{ dx } =\frac { d }{ dx } \left( { x }^{ x } \right) +\frac { d }{ dx } \left( { x }^{ a } \right) +\frac { d }{ dx } \left( { a }^{ x } \right) +\frac { d }{ dx } \left( { a }^{ a } \right) \)
\( { x }^{ x }(1+log\quad x)+a\quad { x }^{ a-1 }+{ a }^{ x }\quad log\quad a+0\)
\( { x }^{ x }(1+log\quad x)+a\quad { x }^{ a-1 }+{ a }^{ x }\quad log\quad a\)

Let f(x) = cos ^–1 (sin x). Observe that this function is defined for all real numbers. We may rewrite this function as
f(x) = cos ^–1 (sin x)
\(={ cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 2 } -x \right) \right] \)
\(=\frac { \pi }{ 2 } -x\)
\(Hence,\ f(x)=0-1=-1\)
\((ii)\ Let\ f\left( x \right) ={ tan }^{ -1 }\left( \frac { sin\quad x }{ 1+cosx } \right) \)
Observe that this function is defined for all real numbers, where cos x ≠ – 1; i.e., at all odd multiplies of π. We may rewrite this function as
\(={ tan }^{ -1 }\left( \frac { 2sin\quad x/2\quad cos\quad x/2 }{ 2{ cos }^{ 2 }x/2 } \right) \)
\(=\tan ^{-1}\left[\tan \left(\frac{x}{2}\right)\right]=\frac{x}{2}\)
Observe that we could cancel \(\cos \left(\frac{x}{2}\right)\) in both numerator and denominator as it is not equal to zero. Thus \(f^{\prime}(x)=\frac{1}{2}\)
\((iii)Let\ f\left( x \right) ={ sin }^{ -1 }\left( \frac { { 2 }^{ x+1 } }{ 1+{ 4 }^{ x } } \right) \)
To find the domain of this function we need to find all x such that \(-1 \leq \frac{2^{x+1}}{1+4^x} \leq 1\). Since the quantity in the middle is always positive, we need to find all x such that \(\frac{2^{x+1}}{1+4^x} \leq 1, \text { i.e.}\) all x such that \(2^{x+1} \leq 1+4^x\). We may rewrite this as \(2 \leq \frac{1}{2^x}+2^x\) which is true for all x. Hence the function is defined at every real number. By putting 2^x = tan θ, this function may be rewritten as
\( f(x)= \sin ^{-1}\left[\frac{2^{x+1}}{1+4^x}\right] \)
\(= \sin ^{-1}\left[\frac{2^x \cdot 2}{1+\left(2^x\right)^2}\right] \)
\( =\sin ^{-1}\left[\frac{2 \tan \theta}{1+\tan ^2 \theta}\right] \)
\(= \sin ^{-1}[\sin 2 \theta] \\ = 2 \theta=2 \tan { }^{-1}\left(2^x\right) \)
\(f^{\prime}(x)= 2 \cdot \frac{1}{1+\left(2^x\right)^2} \cdot \frac{d}{d x}\left(2^x\right) \)
\(= \frac{2}{1+4^x} \cdot\left(2^x\right) \log 2 \)
\(= \frac{2^{x+1} \log 2}{1+4^x}\)

We have:   \(A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] \) and\(B=\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] .\)
\(\therefore \ 2A+3B=2\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] +3\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 14 \end{matrix} \right] +\left[ \begin{matrix} -3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6-3 & 0+0 & 0+0 \\ 0+0 & -10+6 & 0+0 \\ 0+0 & 0+0 & 14+12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 26 \end{matrix} \right] \) =diag.\(\left[ \begin{matrix} 3, & -4, & 26 \end{matrix} \right] .\)

\(\Delta=\begin{vmatrix}x+4&x+4&x+4\\1&x&1\\3&2&x \end{vmatrix}\)
\((x+4)\begin{vmatrix} 1&1&1\\1&x&1\\3&2&x\end{vmatrix}\)
\(=(x+4)\begin{vmatrix}1&0&0\\1&x-1&0\\3&-1&x-3 \end{vmatrix}\)
= (x+4) (1)[(x-1) (x-3)+0]
= (x+4) (x-1) (x-3)
Thus \(\Delta=0\Rightarrow x=-4,1,3.\)
Hence the other two roots are 1 and 3.

A binary operation * on {1, 2} is a function from {1, 2} × {1, 2} to {1, 2}, i.e., a function from {(1, 1), (1, 2), (2, 1), (2, 2)} \(\rightarrow \) {1, 2}. Since 1 is the identity for the desired binary operation *, * (1, 1) = 1, * (1, 2) = 2, * (2, 1) = 2 and the only choice left is for the pair (2, 2). Since 2 is the inverse of 2, i.e., * (2, 2) must be equal to 1. Thus, the number of desired binary operation is only one.

\(\left| \begin{matrix} x & { x }^{ 2 } & yz \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
= \(\left| \begin{matrix} x-y & { x }^{ 2 }-y^2 & yz-zx \\ y-z & { y }^{ 2 }-z^2 & zx-xy \\ z & { z }^{ 2 } & xy \end{matrix} \right| \)
[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)

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 \({ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) =y\) 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 = \(-\frac { \pi }{ 6 } \)

3*4 = 2(3) + 4 - 3 = 6 + 4 - 3 = 7

x = \(-\frac { a }{ 3 } \) or x = \(-\frac { a }{ 2 } \)

x =8 y = 8

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

\((x-y)^{ \frac { x }{ x-y } }\)
\(\Rightarrow log(x-y)+\frac { x }{ x-y } =log\quad a\)
Differentiate w.r.t Ix|,
\(\frac { 1+y^{ ' } }{ x-y } +\frac { 1(x-y)-x(1-y) }{ (x-y)^{ 2 } } =0,\left( y=\frac { dy }{ dx } \right) 2\)
\(\Rightarrow (x-y)(1-y)+(x-y)-x(1-y)=0\)
\(\Rightarrow yy+x-2y=0\)
\(y\frac { dy }{ dx } +x=2y\)

\(A=\left[ \begin{matrix} \overset { F }{ 30 } & \overset { M }{ 12 } & \overset { T }{ 70 } \\ 40 & 15 & 55 \\ 35 & 20 & 75 \end{matrix} \right] \begin{matrix} X \\ Y \\ Z \end{matrix}\)
\(B=\left[ \begin{matrix} 25 \\ 100 \\ 50 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} 30 & 12 & 70 \\ 40 & 15 & 55 \\ 35 & 20 & 75 \end{matrix} \right] \left[ \begin{matrix} 25 \\ 100 \\ 50 \end{matrix} \right] \)
\(X=AB=\left[ \begin{matrix} 750+1200+3500 \\ 1000+1500+2750 \\ 875+2000+3750 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} 5450 \\ 5250 \\ 6625 \end{matrix} \right] \begin{matrix} x \\ y \\ z \end{matrix}\)
fund by X = Rs. 5450
fund by Y = Rs. 5250
fund by Z = Rs. 6625
Total fund = Rs. 17325
They are helping victims and hence the value of helping other is generator.