Given y = a cos bx  ...(1)
Differentiating equation (1) w.r.t 'x', we get 
\(\frac{d y}{d x}=\mathrm{a}(-\sin \mathrm{b} x) \mathrm{b}=-\mathrm{ab} \sin \mathrm{b} x\)
Again differentiating, we get
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab} \cos \mathrm{b} x \cdot \mathrm{b}   \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab}^2 \cos \mathrm{b} x=-\mathrm{b}^2(\mathrm{a} \cos \mathrm{b} x)  \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{b}^2 \mathrm{y}  \)
\(\frac{d^2 y}{d x^2}+\mathrm{b}^2 \mathrm{y}  =0\)
 Therefore, y = a cos bx is a solution of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ b }^{ 2 }y=0\)

\({ { \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) } }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 2 }=xsin\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \)
The highest derivative is 2
∴ Order 2
The given differential equation is not a polynomial equation in its derivative and so its degree is not defined.

Here (x₁, y₁, z₁) = (3, -4, -3) and direction ratios of the given straight line are (a, b, c) = (-4, -7, 12). 
Direction ratios of the normal to the given plane are (A, B, C) = (5, -1, 1).
We observe that, the given point (x₁, y₁, z₁) = (3, 4, -3) satisfies the given plane 5x-y+z = 8
Next, aA+bB+cC = (-4)(5)+(-7)(-1)+(12)(1) = -1 \(\neq \) 0.
So, the normal to the plane is not perpendicular to the line.
Hence, the given line does not lie in the plane.

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

Reducing 9x²-16y² = 144 to the standard form,
we have, \(\frac { { x }^{ 2 } }{ 16 }- \frac { { y }^{ 2 } }{ 9 } =1\)
With the transverse axis is along x-axis vertices are (−4, 0) and (4, 0); and c² = a²+b² = 16 + 9 = 25, c = 5  
Hence the foci are (−5, 0) and (5, 0) 

Let sec^-1\(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\theta \).
Then, sec\(\theta\) = \(-\frac{2}{\sqrt3}\) where \(\theta\in[0,\pi]\)\{\(\frac{\pi}{2}\)}.
Thus, cos \(\theta =-\frac{\sqrt{3}}{2}\).
Now, \(cos\frac { 5\pi }{ 6 } =cos\left( \pi -\frac { \pi }{ 6 } \right) =-cos\left( \frac { \pi }{ 6 } \right) =-\frac { \sqrt { 3 } }{ 2 } .\) 
Hence, Sec-¹ \(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\frac { 5\pi }{ 6 } \)

\(Re\left( \frac { 1 }{ z } \right) \)
= \(Re\left( \frac { 1 }{ x+iy } \right) =Re\left( \frac { 1 }{ x+iy } \times \frac { x-iy }{ x-iy } \right) \)
= \(Re(\left( \frac { x-iy }{ { x }^{ 2 }({ i }^{ 2 }{ y }^{ 2 }) } \right) =Re\left( \frac { x-iy }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
∴ Real part is \(\frac { x }{ { x }^{ 2 }+{ y }^{ 2 } } \)

Δ = b²- 4ac = -20 < 0. The roots are imaginary numbers.

cos^-1(-x) = \(\pi\)-cos⁻¹(x)
Let cos^-1(-x) = \(\theta \)   ..(1)
\(\Rightarrow -x=cos\theta \)
\(\Rightarrow x=-cos\theta =cos\theta =cos\left( \pi -\theta \right) \)
\(\Rightarrow \pi -\theta ={ cos }^{ -1 }\left( x \right) \)
\(\Rightarrow \theta =\pi -{ cos }^{ -1 }x\) ...(2)
From (1) & (2) \({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \)
\({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \) is true.

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

Let x be the population at time t.Then \(\frac{dx}{dt}=kx\)
By separating the variables, we obtain \(\frac{dx}{x}=kdt\)
Integrating on both sides, we get, log|x| = kt + log |C| or x = Ce^kt, where C is an arbitrary constant.
Let x₀ be the population when t = 0 and obtain C = x₀ 
Thus, we get x = x₀e^kt 
Now x = 2x₀ , when t = 50 and thus, k = \(\frac{1}{50}log2\)
Hence, x = x₀2^{\(\frac{t}{50}\)}is the population at any time t 
Assume that the population is tripled in t₁ years 
That is, x = 3x₀ , when t = t₁ 
Thus, t₁ = 50 \(\left( \frac { log3 }{ log2 } \right) \). Therefore, the population is tripled in 50\(\left( \frac { log3 }{ log2 } \right) \)years.

Given that the equation is \(\frac{dv}{dx}+2y\ cot\ x=3x^2 cosec^2x\)
This is a linear differential equation. Here, P = 2 cot x ; Q = 3x²cosec²x.
\(\int { Pdx=\int { 2cot\quad xdx=2log|sin\quad x|=log|sin\quad x{ | }^{ 2 } } =log{ sin }^{ 2 }x } \)
Thus, \(I.F={ e }^{ \int { Pdx } }={ e }^{ log{ sin }^{ 2 } }x={ sin }^{ 2 }x\)
Hence, the solution is \({ ye }^{ \int { Pdx } }={ \int { Qe } }^{ \int { Pdx } }dx+C\)
That is, \(y{ sin }^{ 2 }x=\int { { 3x }^{ 2 }cose{ c }^{ 2 }x.{ sin }^{ 2 }xdx+C=\int { { 3x }^{ 2 }dx+C={ x }^{ 3 }+C } } \)
Hence, \(y{ sin }^{ 2 }x={ x }^{ 3 }+C\) is the required solution

Given equation is m \(\frac{dV}{dt}\) = F- kv  
The given equation can be written as
\(\frac { dv }{ F-kv } =\frac { dt }{ m } \)
Now Integrating, we get
\(\int { \frac { dv }{ F-kv } } =\int { \frac { dt }{ m } } \)
\(\int \frac{d V}{F-k V}=\int \frac{d t}{M}   \frac{\log (F-k V)}{-k}=\frac{t}{M}+C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}-k C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}+\log \mathrm{C} \\ \log [\mathrm{F}-\mathrm{kV}]-\log \mathrm{C}=-\frac{k t}{M} \\ \log \left(\frac{F-k V}{C}\right)=-\frac{k t}{M} \\ \frac{F-k V}{C}=e^{\frac{-k t}{M}} \\ \frac{F-k V}{e^{\frac{-t}{M}}}=\mathrm{C} \Rightarrow \mathrm{C}=e^{\frac{k t}{M}}(F-k V) \\\)
Initial condition:
Given V = 0 when t = 0
\(\mathrm{C}=e^{\frac{k(0)}{M}}[\mathrm{~F}-\mathrm{k}(0)] \\ =\mathrm{e}^0[\mathrm{~F}-0] \\ \mathrm{C}=\mathrm{F} \\ \therefore \mathrm{F}=(F-k V) e^{\frac{k t}{M}}\)

The required plane is parallel to the given line and so it is parallel to the vector \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and the plane passes through the points \(\vec { a } =-\hat { i } +2\hat { j } ,\vec { b } =2\hat { i } +2\hat { j } -\hat { k } \)
(i) vector equation of the plane in parametric form is \(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } \), where s, t ∈ R
which implies that \(\vec { r } =(-\hat { i } +2\hat { j } )+s(3\hat { i } -\hat { k } )+t(\hat { i } +\hat { j } -\hat { k } )\), where s, t ∈ R
(ii) vector equation of the plane in non-parametric form is \((\vec { r } -\vec { a } ).(\vec { b } -\vec { a } )\times \vec { c } )\) = 0
Now, \((\vec { b } -\vec { a } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 0 & -1 \\ 1 & -1 & -1 \end{matrix} \right| =\hat { i } +2\hat { j } +3\hat { k } \)
we have \((\vec { r } -(-\hat { i } +2\hat { j } ).(\hat { i } +2\hat { j } +3\hat { k } )\) = 0 ⇒ \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) = 3
If  \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) is the position vector of an arbitrary point on the plane, then from the above equation, we get the Cartesian equation of the plane as x + 2y + 3z = 3

Let \({ z }^{ 3 }+8i=0\)
\(\Rightarrow\) z³ = -8i
= \(8(-i)=8\left( cos\left( -\frac { \pi }{ 2 } +2k\pi \right) isin\left( -\frac { \pi }{ 2 } +2k\pi \right) \right) \),k\(\in Z\)
\(z=\sqrt [ 3 ]{ 8 } \left( cos\left( \frac { -\pi +4k\pi }{ 6 } \right) +isin\left( \frac { -\pi +4k\pi }{ 6 } \right) \right) \)
Taking k = 0, 1, 2 we get,
k = 0, \(z=2\left( cos\left( -\frac { \pi }{ 6 } \right) +isin\left( -\frac { \pi }{ 6 } \right) \right) =2\left( -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) =2\left( \frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) \)
k = 1, \(z=2\left( cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) \right) =2=\left( 0+i \right) =0+2i=2i\)
k = 2,\(z=2\left( xcos\left( \frac { 7\pi }{ 6 } \right) +isim\left( \frac { 7\pi }{ 6 } \right) \right) =2\left( cos\left( \pi +\frac { \pi }{ 6 } \right) \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \)
= \(2\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) =2\left( -\frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) =-\sqrt { 3 } -i\)
The values of z are \(\sqrt { 3 } -i,2i\) and \(-\sqrt { 3 } -i\)

Let the of the parbola be x² = 4ay (1)

Since (30, 16) is a point on (1),
we get 30² = 4 \(\times\) a \(\times\) 16
⇒ a = \(\frac { 30\times 30 }{ 4\times 16 } =\frac { 225 }{ 16 } \)
∴ becomes, x² = \({ x }^{ 2 }=\frac { 4\times 225 }{ 16 } y=\frac { 225 }{ 4 } y\)
Let AC = h m and BD = lm
∴ A(6, h) is a point on the parabola [∵ OD = 6]
∴ \({ 6 }^{ 2 }=\frac { 225 }{ 4 } \times h\)
⇒ \(h=\frac { 36\times 4 }{ 225 } \Rightarrow h=0.52\)
∴ AD = 3 + h = 3 + 0.52 = 3.52 m
Also (12, 1) is a point on the parabola
[∵ ON = 6 + 6 = 12]
∴ \({ 12 }^{ 2 }=\frac { 225 }{ 4 } \times l\)
⇒ l = \(\frac { 12\times 12\times 4 }{ 225 } =\frac { 576 }{ 225 } =2.08\) = 5.08 m
Hence the length of first two vertical cables are 3.52 m and 5.08 m.

Consider \({ tan }^{ -1 }\left( x-1 \right) +{ tan }^{ -1 }\left( x+1 \right) \)
= \({ tan }^{ -1 }\left( \frac { x-1+x+1 }{ 1-(x-1)(x+1) } \right) ={ tan }^{ -1 }\left( \frac { 2x }{ 1-\left( { x }^{ 2 }-1 \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { x-1+x+1 }{ 1-(x-1)(x+1) } \right) ={ tan }^{ -1 }\left( \frac { 2x }{ 1-\left( { x }^{ 2 }-1 \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 }+1 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) \)
\(\therefore { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) +{ tan }^{ -1 }\left( x \right) ={ tan }^{ -1 }\left( 3x \right) \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) ={ tan }^{ -1 }\left( 3x \right) -{ tan }^{ -1 }\left( x \right) \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) ={ tan }^{ -1 }\left( \frac { 3xx }{ 1+3x^{ 2 } } \right) \)
\(\Rightarrow \frac { 2x }{ 2-{ x }^{ 2 } } =\frac { 2x }{ 1+{ 3x }^{ 2 } } \)
x(1+ 3x²) = x(2-x²)
x[1+ 3x²-2 + x²] = 0
x(4x²-1) = 0
x(2x+ 1) (2x-1) = 0
x = 0, x = \(\frac{1}{2}\), x = -\(\frac{1}{2}\) are the roots 
Hence, there are 3 solutions

Now, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } =tan^{ -1 }{ a }_{ 2 }-tan^{ -1 }{ a }_{ 1 }\)
Similarly, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) =tan^{ -1 }{ a }_{ 3 }-tan^{ -1 }{ a }_{ 2 }\)
Continuing inductively, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ n-1 } }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }{ a }_{ n }-tan^{ -1 }{ a }_{ n-1 }\)
Adding vertically, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)

Given z = x + iy and arg\(\left( \frac { z-i }{ z+2 } \right) =\frac { \pi }{ 4 } \)
⇒ arg(z-i) - arg(z+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x + iy-i) - arg(x+iy+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x+i(y-1)-arg((x+2)+iy) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { y-1 }{ x } \right) -tan^{ -1 }\left( \frac { y }{ x+2 } \right) \) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { \frac { y-1 }{ x } -\frac { y }{ x+2 } }{ 1+\frac { y-1 }{ x } .\frac { y }{ x+2 } } \right) \)
= \(\frac { \pi }{ 4 } \)\(\left[ \because tan^{ -1 }x-tan^{ -1 }y=tan^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(\Rightarrow \frac{\left(\frac{(x+2)(y-1)- x y}{\not {x (\not x+\not2)}}\right)}{\left(\frac{x(x+2)+y(y-1)}{\not x(\not x+\not 2)}\right)}=\tan \frac{\pi}{4}=1\)
⇒ \(\frac { (x+2)(y-1)-xy }{ x(x+2)+y(y-1) } \) = 1
⇒ -x + 2y-2 = x²+ 2x + y²-y
⇒ x² + 2x + y²-y + x-2y + 2 = 0
⇒ x² + y²+3x-3y + 2 = 0
Hence proved.

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as
x²\(\left[ \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -10 \left( x+\frac { 1 }{ x } \right) +26 \right] \)= 0 Since x \(\neq\) 0, we get \(\left(x^{2}+\frac{1}{x^{2}}\right)-10\left(x+\frac{1}{x}\right)+26=0\)
Let y = \(\left( x+\frac { 1 }{ x } \right) \)
[(y²-2)-10y+26] = 0 ⇒ (y²-10y+24) = 0 ⇒ (y-6)(y-4) = 0 ⇒ y = 6 or y = 4
Case (i)
y = 6 ⇒ x +\(\frac{1}{x}\) = 6 ⇒ x = 3+2\(\sqrt{2}\), x = 3 - 2\(\sqrt{2}\)
Case (ii)
y = 4 ⇒ x = 2+\(\sqrt{3}\), x = 2-\(\sqrt{3}\).
Hence, the roots are \(3 \pm 2 \sqrt{2}, 2 \pm \sqrt{3}\)

Rearranging terms in the equation of hyperbola to bring it to standard form,
we have, 11(x²-4x)-25(y²-2y)-256 = 0
11(x− 2)²−25(y−1)² = 256−44+25
11(x−2 )²− 25 (y−1)² = 275
\(\frac { { \left( x-2 \right) }^{ 2 } }{ 25 } -\frac { { \left( y-1 \right) }^{ 2 } }{ 11 } =1\)
Centre (2, 1) a² = 25, b² = 11
c² = a² +b²
= 25 +11 = 36
Therefore, c = ±6
and e = \(\frac { c }{ a } =\frac { 6 }{ 5 } \)and the coordinates of foci are(8, 1) and(-4, 1) from figure.

Consider a triangle ABC in which the two altitudes AD and BE intersect at O. Let CO be produced to meet AB at F. We take O as the origin and let \(\vec { OA } =\vec { a } \), \(\vec { OB } =\vec { b} \) and  \(\vec { OC } =\vec { c } \)

Since \(\vec { AD } \) is perpendicular to \(\vec { BC } \), we have \(\vec { OA } \) is perpendicular to \(\vec { BC } \), and
hence we get \(\vec { OA } \) . \(\vec { BC } \) = 0. That is, \(\vec { a } .(\vec { c } -\vec { b } )=0\), which means
\(\vec { a } .\hat{c}-\hat{a}.\hat{b}=0\)....(1)
Similarly, since \(\vec { BE } \) is perpendicular to \(\vec { CA } \), we have \(\vec { OB } \) is perpendicular to \(\vec { CA } \), and hence we get \(\vec { OB } .\vec { CA } \) = 0.
That is, \(\vec {b } .(\vec {a } -\vec { c } )=0\)
\(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\).......(2)
Adding equations (1) and (2), gives  \(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\). That is, \(\hat{c}(\hat{a}-\hat{b})=0\)
That is \(\vec { OC } \) . \(\vec { BA } \) = 0.
Therefore, \(\vec { BA } \) is perpendicular to \(\vec { OC} \). 
Which implies that \(\vec { CF} \) is perpendicular to \(\vec { AB } \). 
Hence, the perpendicular drawn from C to the side AB passes through O. Therefore, the altitudes are concurrent.

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors making angles α and β respectively, with positive x -axis, where A and B are as shown in the diagram.
Then, we get
\(\hat { a } \) = cosα\(​​\hat{i}\)+ sinα\(\hat {j } \) and \(\hat{b}\)ˆ = cosβ\(​​\hat{i}\)+ sin β \(\hat {j } \),

The angle between \(\hat { a } \) and \(\hat{b}\) is α −β and, the vectors \(\hat{b}\), \(\hat { a } \), \(\hat{k}\) form a right-handed system.
Hence, we get \(\hat { b } \times \hat { a } =\left| \hat { b } \right| \left| \hat { a } \right| \)sin(α −β )\(\hat {k}\) = sin(α −β)\(\hat {k}\)
On the other hand, \(\hat { b } \times \hat { a } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\beta & sin\beta & 0 \\ sin\alpha & sin\alpha & 0 \end{matrix} \right| \) = (sin α cos β -cos α sin β)\(\hat { k } \)
Hence, by equations (1) and (2), we get
sin(α −β ) = sin α cos β − cos α sin β

\( \frac{1}{1-\cos \theta+i \sin \theta} \\ =\frac{1}{1-\cos \theta+i \sin \theta} \times \frac{(1-\cos \theta)-i \sin \theta}{(1-\cos \theta)-i \sin \theta} \\ =\frac{(1-\cos \theta)-i \sin \theta}{(1-\cos \theta)^2+\sin ^2 \theta} \\ =\frac{(1-\cos \theta)-i \sin \theta}{1-2 \cos \theta+\cos ^2 \theta+\sin ^2 \theta} \\ =\frac{(1-\cos \theta)-i \sin \theta}{1-2 \cos \theta+1} \text { where } \cos ^2 \theta+\sin ^2 \theta=1 \\ =\frac{(1-\cos \theta)-i \sin \theta}{2+2 \cos \theta} \)
\(=\frac{1}{2}-i \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{4 \sin ^2 \frac{\theta}{2}} \\ =\frac{1}{2}-i \frac{\cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}} \\ =\frac{1}{2}-\frac{i}{2} \cot \frac{\theta}{2} \)

\(\frac { dy }{ dx } ={ e }^{ x }{ e }^{ y }+{ x }^{ 3 }(e^y)={ e }^{ y }({ e }^{ x }+{ x }^{ 3 })\)
\(\Rightarrow \frac { dv }{ { e }^{ y } } =({ e }^{ x }+{ x }^{ 3 })dx\)
The equation can be written as
\(\Rightarrow \frac { dv }{ { e }^{ y } } =({ e }^{ x }+{ x }^{ 3 })dx\)
Taking integration on both sides, we get
\(\therefore \int { { e }^{ -y } } dy=\int { ({ e }^{ x }+{ x }^{ 3 })dx } \)
\(\frac{e^{-y}}{-1}=e^x+\frac{x^4}{4}+C\)
\(\Rightarrow { e }^{ x }+{ e }^{ -y }+\frac { { x }^{ 4 } }{ 4 } =-C=C\)
[which is also a constant]
\(\Rightarrow { e }^{ x }+{ e }^{ -y }+\frac { { x }^{ 4 } }{ 4 } = C\)

Given that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

If the position vector of the given point is \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \), then the equation of the plane passing through a point and normal to a vector is given by \((\vec { r } -\vec { a } ).\vec { n } =0\) or \(\vec { r } .\vec { n } =\vec { a } .\vec { n } \)
Substituting \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \) in the above equation, we get
\(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )=(4\hat { i } +2\hat { j } -3\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )\)
Thus, the required vector equation of the plane is \(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )\)= 3. If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) then
we get the Cartesian equation of the plane 2x − y + z = 3.

Given lines are
\(\frac { x-1 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-1 }{ 4 } \)
\(\vec { a } =\hat { i } -\hat { j } +\hat { k } \vec { b } \quad and\quad \vec { b } =2\hat { i } +3\hat { j } +4\hat { k } \)
\(\frac { x-3 }{ 1 } =\frac { y-m }{ 2 } =\frac{z-0}{1}\)\(\Rightarrow \vec { c } =3\hat { i } +m\hat { j } \)
\(\vec { d } =\hat { i } +2\hat { j } +\hat { k } \quad \)
\(\vec { c } -\vec { a } =(3\hat { i } +m\hat { j } )-(\hat { i } -\hat { j } +\hat { k } )\)
\(=2\hat { i } +(m,+1)\hat { j } -\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =\hat { i } (3-8)-\hat { j } (2-4)+\hat { k } (4-3)\)
\(=-5\hat { i } +2\hat { j } +\hat { k } \)
Since the lines intersect at a point, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=0,\)
\(\Rightarrow (2\hat { i } +(m+1)\hat { j } -\hat { k } ).(-5\hat { i } +2\hat { j } +\hat { k } )=0\)
\(\Rightarrow -10+2(m+1)-1=0\Rightarrow -10+2m+2-1=0\)
\(\Rightarrow 2m-9=0\Rightarrow 2m=9\Rightarrow m=\frac { 9 }{ 2 } \)
\(\therefore m=\frac { 9 }{ 2 } \)

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

Using the definition of the scalar triple product, we get 
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).[(\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )]\) ....(1)
By treating \((\vec { b } \times \vec { c } )\) as the first vector in the vector triple product, we find
\((\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )\) = \(((\vec { b } \times \vec { c } ).\vec { a } )\vec { c } \) -  \(((\vec { b } \times \vec { c } ).\vec { c } )\vec { a } )\) = \([{ \vec { a } ,\vec { b } ,\vec { c } }]\vec { c } \)
Using this value in (1), we get
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).([\vec { a } ,\vec { b } ,\vec { c } ]\vec { c } )=[\vec { a } ,\vec { b } ,\vec { c } ](\vec { a } \times \vec { b } ).\vec { c } ={ [\vec { a } ,\vec { b } ,\vec { c } ] }^{ 2 }\)

End points of latus rectum are (4, -8) and (4, 8)
Focus is the mid-point of (4, -8) and (4, 8)

∴ Focus = \(\left( \frac { 4+4 }{ 2 } ,\frac { -8+8 }{ 2 } \right) \) = (4, 0)
∴ a = 4 and vertex is (0, 0)
Equation of the parabola is y² = 4ax
⇒ y² = 4(4)x
⇒ y² = 16x

Let z = x + iy
Then \(\frac { 1+z }{ 1-z } \) = cos2θ + i sin 2θ
⇒ \(\frac { 1+x+iy }{ 1-x-iy } \) = cos 2θ + i sin 2θ ....(1)
Taking modulus,
\(\left| \frac { 1+x+iy }{ 1-x-iy } \right| \) = |cos 2θ+i sin 2θ| ⇒ \(\frac { |1+x+iy| }{ |1-x-iy| } \)
=\(\sqrt { cos^{ 2 }2\theta +sin^{ 2 }2\theta } \) = 1
⇒ |1 + x + iy| = |1 - x - iy|
⇒ \(\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } =\sqrt { (1+x)^{ 2 }+{ y }^{ 2 } } \)
⇒ (1 + x)²+ y² = (1-x)²+ y²

⇒ 4x = 0 ⇒ x = 0
From (1) \(\frac { (1+x)+iy }{ (1-x)-iy } \times \frac { (1-x)+iy }{ (1-x)+iy } \)
= cos2θ + isin 2θ
Choosing the imaginary part alone we get,
\(\frac { y(1+x)+y(1-x) }{ (1-x)^{ 2 }+{ y }^{ 2 } } \)= sin 2θ

\(\frac { 2y }{ 1+y^{ 2 } } \) = sin 2θ
⇒ \(\frac { 2tan\theta }{ 1+tan^{ 2 }\theta } \) = sin 2θ
∴ y must be equal to tan θ
⇒ y = tan θ
z = x+ iy
∴ z = 0 + i tan θ
⇒ z = tan θ

sin²x - 5 sinx + 4 = 0
put y = sin x
⇒ y²-5y+4 = 0
⇒ (y-4)(y-1) = 0
⇒ y = 4, 1
Case(i) 
When y = 4, sin x = 4 and no solution for sin x = 4 since the range the sin function is [-1, 1]
Case (ii)
When y = 1, sin x = 1
⇒ sin x = sin \(\frac{\pi}{2}\) [\(\because sin \frac {\pi}{2}=1\)]
\(x=n \pi+(-1)^{n} \frac{\pi}{2} \forall n \in z\).