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

As given \(\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } =0=\overrightarrow { a } \bot \)both \(\overrightarrow { b } \) and \(\overrightarrow { c } \)
⇒ \(\overrightarrow { a } \parallel \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Let \(\overrightarrow { a } =\lambda \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
then \(\left| \overrightarrow { a } \right| =\left| \lambda \right| \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| \)
⇒ \(\frac { \left| \overrightarrow { a } \right| }{ \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| } =\left| \lambda \right| \)
⇒ \(\left| \lambda \right| =\frac { 1 }{ sin\frac { \pi }{ 6 } } =2\)
λ = 士2
\(\overrightarrow { a } =\pm 2\left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Hence proved
(ii) Now \(\left[ \overrightarrow { a } +\overrightarrow { b } ,\overrightarrow { b } +\overrightarrow { c } ,\overrightarrow { c } +\overrightarrow { a } \right] \)
= \(\left[ \left( \overrightarrow { a } +\overrightarrow { b } \right) \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right] .\left( \overrightarrow { c } +\overrightarrow { a } \right) \)
= \(\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } +\left( \overrightarrow { b } \times \overrightarrow { c } \right) .\overrightarrow { a } \)
(As the scalar triple product = 0, if any two vectors are equal)
Hence, \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } \) = \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\overrightarrow { c } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= 2\(\overrightarrow { a } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= \(2\overrightarrow { a } \left( \pm \frac { 1 }{ 2 } \overrightarrow { a } \right) \)
= 士1
Hence proved

\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { dx }{ 1+\sqrt { \cot { x } } } } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { x } } }{ \sqrt { \sin { x } } +\sqrt { \cos { x } } } } dx\)
Apply property \(\int _{ a }^{ b }{ f\left( x \right) } =\int _{ a }^{ b }{ f\left( a+b-x \right) } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } }{ \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } +\sqrt { \cos { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } } } dx\)
\(\therefore I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \cos { x } } }{ \sqrt { \cos { x } } +\sqrt { \sin { x } } } } dx\)
Adding, we get \(2I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx } =\left[ x \right] _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)

\(\frac{dy}{dx}=\frac{1}{x^{ \sin x }+cot^{ 2 }x}\frac{d}{dx}(x^{ \sin x}+\cot^{ 2 }x)\)
Let \(u=x^{\sin x}and v=\cot^{2}x\)
\(\frac{dv}{dx}=2\cot x[-cosec^{2}x]\)
\(\therefore \log u=\sin x.\log x\)
\(\frac{1}{u}\frac{du}{dx}=\frac{\sin x}{x}+\log x.\cos x\)
\(\Rightarrow \frac{du}{dx}=x^{\sin x}[\frac{\sin x}{x}+\cos x\log x]\)
\(\therefore \frac{dy}{dx}=\frac{1}{x^{\sin x}+\cot^{2}x}[x^{\sin x}(\frac{\sin x}{x}+\cos x\log x)-2\cot xcosec^{2}x]\)

\(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] =8I\)
where I is the identity matrix
Let \(A=\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \)
and \(B=\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(\Rightarrow AB=8I\)
Post Multiplyingbith sides by B ^-1, we get 
\(AB{ B }^{ -1 }=8I{ B }^{ -1 }\)
\(\Rightarrow A=8{ B }^{ -1 }\)
\(\Rightarrow { B }^{ -1 }=\frac { A }{ 8 } \)
Given Equations are:
x - y + z = 4
x - 2y - 2z = 9
and 2x + y + 3z = 1
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow BX=C\)
where \(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \ and\ C=\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow X={ B }^{ -1 }C\)
Using \({ B }^{ -1 }=\frac { A }{ 8 } \)
\(\Rightarrow X=\frac { A }{ 8 } C\)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
(x, y, z)  = (3, -2, -1)

Let ABCD be a rectangle having area A inscribed in an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Let the co-ordinate of A be \((\alpha, \beta)\)
ஃ Coordinate of \(B\equiv(\alpha,- \beta)\)
\(C\equiv(-\alpha,- \beta)\)
\(D\equiv(-\alpha, \beta)\)
Now   A = Length x Breadth
\(=2\alpha\times 2\beta\)
\(\Rightarrow A=4\alpha\beta\)
\(\Rightarrow A=4\alpha\sqrt{b^2(1-\frac{\alpha^2}{a^2})}\)
\(\Rightarrow A^2=16a^2\{{b^2(1-\frac{\alpha^2}{a^2})}\}\)
\(\Rightarrow A^2=\frac{16b^2}{a^2}(a^2\alpha^2-\alpha^4)\)
\(\Rightarrow \frac{d(A^2)}{d\alpha}=\) \(\frac{16b^2}{a^2}(2a^2\alpha^2-4\alpha^3)\)

For  maximum or minimum value
\(\frac{d(A^2)}{d\alpha}=0\)
\(\Rightarrow 2a^2\alpha-4\alpha^3=0\)
\(\Rightarrow 2\alpha(a^2-2\alpha^3)=0\)
\(\Rightarrow \alpha=0, \ \ a=\frac{a}{\sqrt2}\)
Again \(\Rightarrow\frac{d(A^2)}{d\alpha}=\)\(\frac{16b^2}{a^2}(2a^2-12\alpha^2)\)
⇒  For \(\alpha=\frac{a}{\sqrt2}\) A is maximum.
i.e., for greatest area A   \(\alpha=\frac{a}{\sqrt2}\) and \(\beta=\frac{b}{\sqrt2}\)
ஃ Greatest are a = \(4\alpha\cdot\beta=4\frac{a}{\sqrt2}\times\frac{b}{\sqrt2}=2ab\)

Let the events be:
E₁ = Choosing 1^st coin
E₂ = Choosing 2^nd coin
E₃ = Choosing 3^rd coin
A: Getting Heads
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P(E_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=\frac { 40 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 1 }{ 2 } \)
\(P({ E }_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .\frac { 40 }{ 100 } }{ \frac { 1 }{ 3 } .\frac { 40 }{ 100 } +\frac { 1 }{ 3 } .\frac { 75 }{ 100 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } } =\frac { 8 }{ 33 } \)

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 \(a\neq 0\quad \left( \frac { 1 }{ a } ,\frac { -b }{ a } \right) \)

\(LHS=2{ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(={ tan }^{ -1 }\frac { 2\times \frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 4 } } +tan^{ -1 }\frac { 1 }{ 7 } \)
\(={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(={ tan }^{ -1 }\frac { \frac { 4 }{ 3 } +\frac { 1 }{ 7 } }{ 1-\frac { 4 }{ 3 } .\frac { 1 }{ 7 } } \)
\(={ tan }^{ -1 }\frac { 31 }{ 17 } \)
Using the above triangle, we get
\({ tan }^{ -1 }\frac { 31 }{ 17 } ={ sin }^{ -1 }\left( \frac { 31 }{ 25\sqrt { 2 } } \right) \)
=RHS
Hence   LHS = RHS

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 \(n\epsilon N\)

We have: \((\mathrm{X}+\mathrm{Y})+(\mathrm{X}-\mathrm{Y}) =\left[\begin{array}{ll} 5 & 2 \\ 0 & 9 \end{array}\right]+\left[\begin{array}{cc} 3 & 6 \\ 0 & -1 \end{array}\right] \)
or \((\mathrm{X}+\mathrm{X})+(\mathrm{Y}-\mathrm{Y}) =\left[\begin{array}{ll} 8 & 8 \\ 0 & 8 \end{array}\right] \Rightarrow 2 \mathrm{X}=\left[\begin{array}{ll} 8 & 8 \\ 0 & 8 \end{array}\right] \)
or \(\mathrm{X} =\frac{1}{2}\left[\begin{array}{ll} 8 & 8 \\ 0 & 8 \end{array}\right]=\left[\begin{array}{ll} 4 & 4 \\ 0 & 4 \end{array}\right] \)
Also \((\mathrm{X}+\mathrm{Y})-(\mathrm{X}-\mathrm{Y}) =\left[\begin{array}{ll} 5 & 2 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{rr} 3 & 6 \\ 0 & -1 \end{array}\right] \)
or \((\mathrm{X}-\mathrm{X})+(\mathrm{Y}+\mathrm{Y}) =\left[\begin{array}{cc} 5-3 & 2-6 \\ 0 & 9+1 \end{array}\right] \Rightarrow 2 \mathrm{Y}=\left[\begin{array}{ll} 2 & -4 \\ 0 & 10 \end{array}\right] \)
or \(\mathrm{Y} =\frac{1}{2}\left[\begin{array}{rr} 2 & -4 \\ 0 & 10 \end{array}\right]=\left[\begin{array}{rr} 1 & -2 \\ 0 & 5 \end{array}\right] \)

Given, differential equation is (x(1 + y²) dx - y(1 + x²) dy = 0
\(\Rightarrow\)    x(1 + y²)dx = y(1 + x²) dy
On separating the variables, we get
\(\frac{y}{\left(1+y^2\right)} d y=\frac{x}{\left(1+x^2\right)} d x\)
On integrating both sides, we get
\(\begin{aligned} \int \frac{y}{1+y^2} d y & =\int \frac{x}{\left(1+x^2\right)} d x \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{1}{2} \log \left|1+y^2\right| & =\frac{1}{2} \log \left|1+x^2\right|+C \end{aligned}\)        ...(i)
\(\left[\begin{array}{l} \text { put } 1+y^2=u \Rightarrow 2 y d y=d u \\ \text { then } \int \frac{y}{1+y^2} d y=\int \frac{1}{2 u} d u=\frac{1}{2} \log |u| \\ \text { and put } 1+x^2=v \Rightarrow 2 x d x=d v \\ \text { then } \int \frac{x}{1+x^2} d x=\frac{1}{2} \int \frac{1}{v} d v=\frac{1}{2} \log |v| \end{array}\right]\)
Also, given that y = 1, when x = 0.
On substituting the values of x and y in Eq. (i), we get
\(\begin{aligned} \quad \frac{1}{2} \log \left|1+(1)^2\right| & =\frac{1}{2} \log \left|1+(0)^2\right|+C \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{1}{2} \log 2 & =C \end{aligned}\)                [\(\because\) log 1 = 0]
On putting C = \(\frac{1}{2}\) log 2 in Eq. (i), we get
\(\begin{array}{cc} & \frac{1}{2} \log \left|1+y^2\right|=\frac{1}{2} \log \left|1+x^2\right|+\frac{1}{2} \log 2 \end{array}\)
\(\begin{array}{cc} \Rightarrow & \log \left|1+y^2\right|=\log \left|1+x^2\right|+\log 2 \\ \end{array}\)
\(\begin{array}{cc} \Rightarrow & \log \left|1+y^2\right|-\log \left|1+x^2\right|=\log 2 \\ \end{array}\)
\(\begin{array}{cc} \Rightarrow & \log \left|\frac{1+y^2}{1+x^2}\right|=\log 2 \end{array}\)
                  \(\begin{aligned} {\left[\because \log m-\log n=\log \frac{m}{n}\right] } \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & \frac{1+y^2}{1+x^2}=2 \Rightarrow 1+y^2=2+2 x^2 \end{aligned}\)
\(\Rightarrow\)    y² - 2x² -1 = 0
which is the required particular solution of given differential equation.

200 at (50,0)

The given equation is:
\(\left( x+y \right) \frac { dy }{ dx } =2{ e }^{ -y }-1\)
\(\Rightarrow\)  \(\frac { dy }{ 2{ e }^{ -y }-1 } =\frac { dx }{ x+1 } \)
Variables Separable 
Integrating, \(\int { \frac { dy }{ 2{ e }^{ -y }-1 } } =\int { \frac { dx }{ x+1 } +C } \)
\(\Rightarrow\) \(\int { \frac { dy }{ 2{ e }^{ -y }-1 } =log|x+1|+C } \)  .(1)
Now  \(I=\int { \frac { dy }{ 2{ e }^{ -y }-1 } } =\int { \frac { { e }^{ y } }{ 2-{ e }^{ y } } } dy\)
Put \({ e }^{ y }=t\) so that \({ e }^{ y }\quad dy=dt.\)
\(\therefore\) \(I=\int { \frac { dt }{ 2-t } =- } log|2-t|=-log|2-{ e }^{ y }|.\)
From(1),  \(-log2-{ e }^{ y }|=log|x+1|+C\)...(2)
When \(x=0,y=0,\)                   
\(\therefore\)  \(log|2-1|=log|0+1|+C\)
\(\Rightarrow\) \(log|1|=log|1|+C\)
\(\Rightarrow\) \(0=0+C\Rightarrow C=0.\)
Putting in (2), \(-log|2-{ e }^{ y }|=log|x+1|\) 
\(\Rightarrow\)  \(log|2-{ e }^{ y }|=log|\frac { 1 }{ x+1 } |\)
\(\Rightarrow\) \(2-{ e }^{ y }=\frac { 1 }{ x+1 } \)
\(\Rightarrow\)  \({ e }^{ y }=2-\frac { 1 }{ x+1 } =\frac { 2x+1 }{ x+1 } \)
\(\Rightarrow\) \(y=log|\frac { 2x+1 }{ x+1 } |,x\neq -1.\)
Which is the requored solution.

We have:  \(A=\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}.\)
\(\therefore\ A\prime =\begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix}\)
\( \therefore \ A-A\prime =\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}+\begin{bmatrix} 1 & 3 \\ 4 & 7 \end{bmatrix}\)
\(=\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}+\begin{bmatrix} -1 & -3 \\ -4 & -7 \end{bmatrix}\)
\(=\begin{bmatrix} 1-1 & 4-3 \\ 3-4 & 7-7 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} ...(1)\)
Now    \((A-A\prime )\prime =\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}=-\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\)
Hence,  \(A-A\prime \)  is skew-symmetric matrix.

Part (B) is the correct answer.
Reason: The parabola y² = 4x
intersects the line y = 2x
at. O(0,0) and A (1,2)

\(\therefore \ Reqd.area=\overset { 1 }{ \underset { 0 }{ \int { } } } \sqrt { 4x } dx-\overset { 1 }{ \underset { 0 }{ \int { } } } (2x)dx\)
\(=2\left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] _{ 0 }^{ 1 }-2\left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 1 }\)
\(=\frac { 4 }{ 3 } [1-0]-\frac { 2 }{ 2 } [1-0]=\frac { 4 }{ 3 } -1\)
\(=\frac { 1 }{ 3 } sq.unit.\)

the given system of equations is:
4x-3y = 3
3x-5y = 7
These can be written as AX = B
\(\Rightarrow\) X = A^-1B  ...(1)
where \(A\begin{bmatrix} 4&-3\\3&-5\end{bmatrix}, X=\begin{bmatrix} x\\y\end{bmatrix}\ and\ B=\begin{bmatrix} 3\\7\end{bmatrix}\)
\(\therefore\ |A|=\begin{bmatrix} 4&-3\\3&-5\end{bmatrix}=-20+9=-11\neq0\)
\(\Rightarrow A^{-1}\) exists
Now \(adj\ A=\begin{bmatrix}-5&-3\\3&4 \end{bmatrix}'=\begin{bmatrix}-5&3\\-3&4 \end{bmatrix}\)
\(\therefore\ A^{-1}={1\over|A|}(adj\ A)={1\over-11}\begin{bmatrix}-5&3\\-3&4 \end{bmatrix}\)
From (1), \(X={1\over-11}\begin{bmatrix} -5&3\\-3&4\end{bmatrix}\begin{bmatrix} 3\\7\end{bmatrix}\)
\(={1\over-11}\begin{bmatrix}-15+21\\-9+28 \end{bmatrix}={1\over-11}\begin{bmatrix}6\\19 \end{bmatrix}\)
\(\Rightarrow\begin{bmatrix}x\\y \end{bmatrix}=\begin{bmatrix}-6/11\\-19/11 \end{bmatrix}\)
Hence, \(x={-6\over11}, y={-19\over11}\)

Given that
\(x=a\ cos\theta ,\ y=a\ sin\theta \)
\(\frac { dx }{ d\theta } =-a\ sin\theta \ and\ \frac { dy }{ d\theta } =a\ cos\theta \)
\(Hence,\ \frac { dy }{ dx } =\frac { { dy }/{ d\theta } }{ { dx }/{ d\theta } } =\frac { a\ cos\theta }{ -a\ sin\theta } =-cot\theta \)

\(\text {Let } \sin ^{-1} \frac{3}{5}=x \text { and } \sin ^{-1} \frac{8}{17}=y\)
\( \sin x=\frac{3}{5} \text { and } \sin y=\frac{8}{17}\)
\(\text {Now }\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
\(\text {and }\cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\frac{64}{289}}=\frac{15}{17}\)
We have cos(x−y) = cos x cos y + sin x siny
\(=\frac{4}{5} \times \frac{15}{17}+\frac{3}{5} \times \frac{8}{17}=\frac{84}{85}\)
\( x-y=\cos ^{-1} \frac{84}{85}\)
\(\text {Hence } \ \sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85}\)

 line is  \(\vec { r } .(\hat { i } -2\hat { j } +3\hat { k } )+\lambda (-3\hat { i } +5\hat { j } +4\hat { k } )\)
In Cartesian from : \(\frac { x-1 }{ -3 } =\frac { y-2 }{ 5 } =\frac { z-3 }{ 4 } \)

Let E_I = Letter has come from CALCUITA
E₂ = Letter has come from TATANAGAR
and E = Two consecutive letters (i.e. alphabets) TA are visible on envelope
\(\therefore P\left(E_{1}\right)=\frac{1}{2}, P\left(E_{2}\right)=\frac{1}{2}, P\left(\frac{E}{E_{1}}\right)=\frac{n\left(E \cap E_{1}\right)}{n\left(E_{1}\right)}=\frac{1}{7}\) 
[\(\therefore\) pairs of consecutive letters are CA, AL, LC, CU, UT, TT, TA]
and \(P\left(\frac{E}{E_{2}}\right)=\frac{n\left(E \cap E_{2}\right)}{n\left(E_{2}\right)}=\frac{2}{8}\) 
[\(\therefore\)8 pairs of consecutive letters are TA, AT, TA;AN, NA, AG, GA, AR]
\(\therefore P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}\) 
[using Baye's theorem]
\(=\frac{\frac{1}{2} \times \frac{2}{8}}{\frac{1}{2} \times \frac{1}{7}+\frac{1}{2} \times \frac{2}{8}}=\frac{\frac{2}{16}}{\frac{8+14}{2 \times 7 \times 8}}=\frac{2 \times 7}{22}=\frac{7}{11}\) 
Hence, the probability that the letter TA came from TATANAGAR is \(\frac{7}{11}\).

Given equation of ellipse is \(9 x^{2}+y^{2}=36\)  which can be expressed as \(\frac{x^{2}}{4}+\frac{y^{2}}{36}=1 \text { or } \frac{x^{2}}{2^{2}}+\frac{y^{2}}{6^{2}}=1\) and hence, its shape 
\(y-0=\frac{6-0}{0-2}(x-2)\)
or y = – 3(x – 2)
or y = – 3x + 6
Area of the shaded region 
\(=3 \int_{0}^{2} \sqrt{4-x^{2}} d x-\int_{0}^{2}(6-3 x) d x\)
\(=3\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}-\left[6 x-\frac{3 x^{2}}{2}\right]_{0}^{2}\)
\(=3\left[\frac{2}{2} \times 0+2 \sin ^{-1}(1)\right]-\left[12-\frac{12}{2}\right]=3 \times 2 \times \frac{\pi}{2}-6=3 \pi-6\)

\(=\sqrt{(1+x^2)}+{1\over2}log|{\sqrt{1+x^2}-1\over\sqrt{1+x^2}+1}|+c\)

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

Let A be the point (5, 4, - 6)
Image A' be the point (5, 4, - 6)
\(\therefore\) A'(5, 4, - 6)
Distance between AA'
\(=\sqrt { (5-5)^{ 2 }+(4-4)^{ 2 }+(6+6)^{ 2 } } \)
\(=\sqrt { 0+0+12^{ 2 } } \)
=123 units

\(\overset\rightarrow c=\overset\rightarrow a+\overset\rightarrow b\)
\(=\overset\wedge i+2\overset\wedge j+\overset\wedge k+3\overset\wedge i+\overset\wedge j-\overset\wedge k\)
\(\Rightarrow \overset\rightarrow c=4\overset\wedge i+3\overset\wedge j\)
\(\Rightarrow \overset\rightarrow d=\overset\rightarrow a-\overset\rightarrow b\)
=(i+2j+k)-(3i+j-k)
\(\Rightarrow\overset\rightarrow d =-2\overset\wedge i+\overset\wedge j+2\overset\wedge k\)
Projection \(\overset\rightarrow c \) on \(\overset\rightarrow d\)\(=\frac{\overset\rightarrow c.\overset\rightarrow d}{\left| \overset\rightarrow d \right| }\)
\(=\frac{(4\overset\wedge i+3\overset\wedge j).(-2\overset\wedge i+\overset\wedge j+2\overset\wedge k)}{\left|-2\overset\wedge i+\overset\wedge j+2\overset\wedge k \right| }\)
\(=\frac{-8+3}{\sqrt{4+1+4}}=-\frac{5}{3}\)

Sample space ={B₁B₂, B₁G₂, G₁B₂, G₁G₂}, B₁ and G₁ are the older boy and girl respectively.
Let   E₁ = both the children are boys;
E₂ = one of the children are boys;
E₃ = the older child is a boy
Then, (a) P(E₁/E₂) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 2 } }{ { E }_{ 2 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 3 } \)
(b) P(E₁/E₃) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 3 } }{ { E }_{ 3 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 2 }{ 4 } } =\frac { 1 }{ 2 } \)

\(\log { \left( \frac { dy }{ dx } \right) } =x+1\)
\(\Rightarrow \frac { dy }{ dx } ={ e }^{ x+1 }\)
\(\Rightarrow dy={ e }^{ x+1 }dx\)
Integrating both the sides,
\(\int { dy } =\int { { e }^{ x+1 }dx } \)
\(\Rightarrow y={ e }^{ x+1 }+C\)

\(f(x)=x^{ 2 }+1,g(x)=2x+1\)
\(fog=f[(g(x)\} =f(2x+1)\)
\(=4x^{ 2 }+4x+1+1\)
\(=4x^{ 2 }+4x+2\)
\(gof=g(f(x)=g(x^{ 2 }+1)\)
\(=2(x^{ 2 }+1)+1\)
\(=2x^{ 2 }+2+1\)
\(=2x^{ 2 }+3\)

\(\int { \frac { dx }{ 1+sinx } } \times \frac { (1-sinx }{ (1-sinx) } \)
\(=\int { \frac { (1-sinx) }{ 1-sin^{ 2 }x } } dx\)
\(=\int { \frac { dx }{ { cos }^{ 2 }x } -\int { \frac { sinx }{ { cos }^{ 2 }x } dx } } \)
\(=\int { { sec }^{ 2 }xdx } -\int { tanx\quad secx\quad dx } \)
\(=tanx-secx+C\)

\(L.H.S.={ tan }^{ -1 }\left( \frac { 3{ a }^{ 2 }x-{ x }^{ 3 } }{ { a }^{ 3 }-3a{ x }^{ 2 } } \right) \)
\(Divide\quad by\quad a^{ 3 },\)
\(={ tan }^{ -1 }\left( \frac { 3\frac { x }{ a } -\frac { { x }^{ 3 } }{ { a }^{ 3 } } }{ 1-3\left( \frac { { x }^{ 2 } }{ { a }^{ 2 } } \right) } \right) \)
\(={ tan }^{ -1 }\left[ \frac { 3\left( \frac { x }{ a } \right) -{ \left( \frac { x }{ a } \right) }^{ 3 } }{ 1-3{ \left( \frac { x }{ a } \right) }^{ 2 } } \right] \)
\(Put,\frac { x }{ a } =tan\theta ,\quad \theta ={ tan }^{ -1 }\frac { x }{ a } \)
\(={ tan }^{ -1 }\left[ \frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \right] \)
\(={ tan }^{ -1 }(tan\quad 3\theta )\)
\(3\theta =3{ tan }^{ -1 }\left( \frac { x }{ a } \right) =R.H.S.\)

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|

As for curve ay + x² = 7
⇒ a\(\frac { dy }{ dx } \) + 2x =0
⇒ \(\frac { dy }{ dx } \) = \(-\frac { 2x }{ a } \Rightarrow \frac { dy }{ dx } \)]_(1,1) = -\(\frac2a\)
and for curve x³ = y, 3x² = \(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } \) ]_(1,1) = 3
If curves cut orthogonally then
\(\left( -\frac { 2 }{ a } \right) \) x 3 = -1 ⇒ a = 6

Here (1,2) e R, (2,1) € R, if transitive (1,1) should belong to R.