tan\(\left( { sin }^{ -1 }\left( x+\frac { 1 }{ 2 } \right) \right) \)

We know that \(sin^{ -1 }\left( x \right) ={ tan }^{ -1 }\left( \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) \) if-1 \(\therefore { sin }^{ -1 }\left( x+\frac { 1 }{ 2 } \right) ={ tan }^{ -1 }\left( \frac { x\frac { 1 }{ 2 } }{ 1-\left( x+\frac { 1 }{ 2 } \right) ^{ 2 } } \right) \)
= \({ tam }^{ -1 }\left( \frac { x+\frac { 1 }{ 2 } }{ \sqrt { 1-{ x }^{ 2 }-\frac { 1 }{ 4 } -x } } \right) \)
= \({ tan }^{ -1 }\left( \frac { x+\frac { 1 }{ 2 } }{ \sqrt { { x }^{ 2 }-x+\frac { 3 }{ 4 } } } \right) \)
\(\therefore tan\left( { sin }^{ -r }\left( x+\frac { 1 }{ 2 } \right) \right) =tan\left( { tan }^{ -1 }\left( \frac { x+\frac { 1 }{ 2 } }{ \sqrt { -{ x }^{ 2 }-x+\frac { 3 }{ 4 } } } \right) \right) \)
= \(\frac { x+\frac { 1 }{ 2 } }{ \sqrt { -{ x }^{ 2 }-x+\frac { 3 }{ 4 } } } =\frac { 2 }{ \frac { \sqrt { { -4x }^{ 2 }-4x+3 } }{ 2 } } \)
\(tan\left( { sin }^{ -1 }\left( x+\frac { 1 }{ 2 } \right) \right) =\frac { 2x+1 }{ \sqrt { 3+4x-{ 4x }^{ 2 } } } \)

|z-4|²-|z-1|² = 16
|x+iy-4|² - |x+iy-1|² = 16
⇒ |(x-4)+iy|² - |(x-1)+iy2|² = 16
⇒ [(x-4)²+y²] - [(x-1)²+y²] = 16
⇒ x²-8x+16+y²-[x²-2x+1+y²] = 16
\(\Rightarrow \not x^{2}-8 x+16+\not y^{2}-\not x^{2}+2 x-1-\not y^{2}=16\)
⇒ -6x+15-16 = 0
⇒ -6x-1 = 0
⇒ 6x+1 = 0 Which is the required Cartesian equation.
The locus of the point is a straight line.

Im[(1−i)z + 1] = 0
(1-i)z + 1 = (1-i)( x + iy) +1
= x + iy-ix-i²y+1
= x+iy-ix+y+1
= (x + y + 1) + i(y - x)
∴ Im[(1-i)z + 1] = y-x = 0
⇒ x - y = 0
Hence, the Cartesian equation is x - y = 0

Consider \(\frac { 19-7i }{ 9+i } =\frac { 19-7i }{ 9+i } \times \frac { 9-i }{ 9-i } \)
= \(\frac { 171-19i-63i+7i^{ 2 } }{ (9)^{ 2 }-{ i }^{ 2 } } =\frac { 171-82i-7 }{ 81+1 } \)
= \(\frac { 164-82i }{ 82 } =\frac { 82(2-i) }{ 82 } \) = 2- i
Also \(\frac { 20-5i }{ 7-6i } =\frac { 20-5i }{ 7-6i } \times \frac { 7+6i }{ 7+6i } \)
= \(\frac { 140+120i-35i-30^{ 2 } }{ { 7 }^{ 2 }-(6i)^{ 2 } } \)
\(=\frac{140+85 i+30}{49+36}=\frac{170+85 i}{85}=\frac{\not 85(2+i)}{\not 85}=2+i\)
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) = (2-i)12+(2+i)12
Let z = (2-i)¹²+(2+i)¹²
∴ \(\overline { z } \) = \(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
[∵ \(\overline { { z }_{ 1 }+{ z }_{ 2 } } =\overline { { z }_{ 1 } } +\overline { { z }_{ 2 } } \)]
= (2 + i)¹²+(2 - i)¹²= z
∴ \(\overline { z } \) = z ⇒ z is purely real
∴ \(\left( \frac { 19-7i }{ 9+i } \right) ^{ 12 }+\left( \frac { 20-5i }{ 7-6i } \right) ^{ 12 }\) is real.

\(\frac { 1 }{ 3 } \left[ \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{matrix} \right] \) and λ = \(\frac { 1 }{ 3 } \)
Since adj (λA) = λ^n-1(adj A)
we get adj \(\left( \frac { 1 }{ 3 } \left[ \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{matrix} \right] \right) =\left( \frac { 1 }{ 3 } \right) ^{ 2 }\)
adj\(\left( \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 1 \end{matrix} \right) \)
∴ Required adjoint matrix
\(\frac { 1 }{ 9 } \left[ \begin{matrix} +\left| \begin{matrix} 1 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} -2 & 2 \\ 1 & 2 \end{matrix} \right| & +\left| \begin{matrix} -2 & 1 \\ 1 & -2 \end{matrix} \right| \\ -\left| \begin{matrix} 2 & 1 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| \\ +\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
= \(\frac { 1 }{ 9 } \left[ \begin{matrix} (2+4)-(-4-2)+(4-1) \\ -(4+2)+(4-1)+(-4-2) \\ +(4-1)-(4+2)+(2+4) \end{matrix} \right] ^{ T }\)
= \(\frac { 1 }{ 9 } \left[ \begin{matrix} 6 & 6 & 3 \\ -6 & 3 & 6 \\ 3 & -6 & 6 \end{matrix} \right] ^{ T }=\frac { 1 }{ 9 } \left[ \begin{matrix} 6 & -6 & 3 \\ 6 & 3 & -6 \\ 3 & 6 & 6 \end{matrix} \right] \)
= \(\frac { 3 }{ 9 } \left[ \begin{matrix} 2 & -2 & 1 \\ 2 & -1 & -2 \\ 1 & 2 & 2 \end{matrix} \right] \)
[Taking 3 common from each entry]
=\(\frac { 1 }{ 3 } \left[ \begin{matrix} 2 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 2 & 2 \end{matrix} \right] \).

 Passing through (5, -2) length of the transverse axis is a long x-axis and of length 8 units.
2a = 8 ⇒ a = 4
Since the transverse axis is along x-axis, centre is (0, 0)
Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ b^{ 2 } } =1\)
Since (5, -2)passes through the parabola,
\(\frac { 25 }{ 16 } -\frac { 4 }{ { b }^{ 2 } } \Rightarrow \frac { 4 }{ { b }^{ 2 } } =\frac { 25 }{ 16 } -1=\frac { 25-16 }{ 16 } =\frac { 9 }{ 16 } \)
∴ \({ b }^{ 2 }=\frac { 16\times 4 }{ 9 } =\frac { 64 }{ 9 } \)
∴ Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ \frac { 64 }{ 9 } } =1\Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 64 } =1\)

Given z³+2\(\overline { z } \) = 0
⇒ z³ = -2\(\overline { z } \)
Taking modulus, |z³| = |-2\(\overline { z } \)|
⇒ |z|³ = 2|z|
⇒ |z| [|z|2-2] = 0
|z| = 0 or |z|²-2 = 0
⇒ |z| = 0
⇒ z = 0 is a solution ....(1)
|z²| = 2
⇒ |z²| = 2 ⇒ (z\(\overline { z } \))² = 2
⇒ z\(\overline { z } \) = \(\sqrt { 2 } \Rightarrow \overline { z } \frac { \sqrt { 2 } }{ z } \)
Given z + 2\(\overline { z } \) = 0
⇒ z³+2\(\frac { \sqrt { 2 } }{ z } \) = 0
⇒ z⁴+2\(\sqrt { 2 } \) = 0
If has 4 non-zero solutions
Hence from (1) and (2), z³+2\(\overline { z } \) has 5 solutions.

Given \(\left| z-\frac { 2 }{ z } \right| \) = 2
Consider |z| = \(\left| z-\frac { 2 }{ z } +\frac { 2 }{ z } \right| \)
≤ \(\left| z-\frac { 2 }{ z } \right| +\left| \frac { 2 }{ z } \right| \)   [Triangle law of in equality]
|z| ≤ \(\frac { 2|z|+2 }{ |z| } \)  [∵ \(\left| z-\frac { 2 }{ z } \right| \) = 2]
≤ \(\frac { 2|z|+2 }{ |z| } \) ⇒ |z|² ≤ 2|z|+2 ⇒ |z|²-2|z|≤ 2
Adding 1 both sides.
|z|²-2|z|+1 ≤ 2+1
⇒ [|z|-1]² ≤ 3 ⇒ |z|-1 ≤ ±\(\sqrt { 3 } \)
⇒ |z| ≤ ±\(\sqrt { 3 } \)+1
∴ The greatest value of |z| is \(\sqrt { 3 } \)+1 and the least value pf |z| is 1-\(\sqrt { 3 } \) respectively.

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
A =\(\left[ \begin{matrix} 1 \\ 2 \\ 5 \end{matrix}\begin{matrix} 1 \\ -1 \\ -1 \end{matrix}\begin{matrix} 1 \\ 3 \\ 7 \end{matrix}\begin{matrix} 3 \\ 4 \\ 11 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { { R } }_{ 2 }-2{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 5 \end{matrix}\begin{matrix} 1 \\ -3 \\ -1 \end{matrix}\begin{matrix} 1 \\ 1 \\ 7 \end{matrix}\begin{matrix} 3 \\ -2 \\ 11 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-5{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -6 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\begin{matrix} 3 \\ -2 \\ -4 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { { R } }_{ 3 }-2{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} 3 \\ -2 \\ 0 \end{matrix} \right] \)
The last equivalent matrix is in row echelon form it ha two non-zero row \(\rho \)(A) = 2

From figure the midpoint of line joining foci is the centre C(0, 0).
Transverse axis is y-axis 
AA′ = 2a \(\Rightarrow \) 2a = 8,
SS′ = 2c = 12, c = 6
a = 4
b² = c²−a² = 36−16 = 20
Hence the equation of the required hyperbola is \(\frac { { y }^{ 2 } }{ 16 }- \frac { { x }^{ 2 } }{ 20 } =1\)

Area of the circle = 9π sq.units

πr² = 9π ⇒ r² ⇒ 9 ⇒ r ⇒ 3
Diameters are x + y = 5 ................(1)
and x - y = 1 .................(2)
We know that centre is the point of intersection of diameters.
∴ To find the centre, solve (1) and (2).

⇒ 2x = 6
⇒ x = 3
∴ (1) ⇒ 3 + y = 5
⇒ y = 5 - 3 = 2
∴ Centre is (3, 2)
Hence, equation of the circle is
(x - h)² + (y - k)² = r2
⇒ (x - 3)² + (y - 2)² = 32
\(\Rightarrow x^{2}-6 x+9+y^{2}-4 y+4=9\)
⇒ x² + y² − 6x − 4y + 4 = 0

Given A =\(\left[ \begin{matrix} 1 & -1 \\ 2 & 0 \end{matrix} \right] \), B =\(\left[ \begin{matrix} 3 & -2 \\ 1 & 1 \end{matrix} \right] \) and C =\(\left[ \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right] \)
Also,  A X B = C
Premultiply by A^-1 we get,
(A^-1 A) X B = A^-1 C
⇒ XB = A^-1C    [∵ A^-1a = 1]
Post Multiply by B^-1 we get
(X B) B^-1 = (A^-1 C) B^-1
⇒ X = (A^-1 C) B^-1
|A| = \(\left| \begin{matrix} 1 & -1 \\ 2 & 0 \end{matrix} \right| \) = 0 + 2 = 2 ≠ 0
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 1 \\ -2 & 1 \end{matrix} \right] \)
|B| = \(\left| \begin{matrix} 3 & -2 \\ 1 & 1 \end{matrix} \right| \) = 3 + 2 = 5 ≠ 0
∴ B^-1 = \(\frac { 1 }{ |B| } adjB=\frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right] \)
A^-1C = \(\\ \frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 2 } \left[ \begin{matrix} 0+2 & 0+2 \\ -2+2 & -2+2 \end{matrix} \right] \)
=\(\frac { 1 }{ 2 } \left[ \begin{matrix} 2 & 2 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 2 } (2)\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \)
∴ X = Z(A^-1C).B^-1
= \(\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 1-1 & 2+3 \\ 0+0 & 0+0 \end{matrix} \right] \)
= \(\frac { 1 }{ 5 } \left[ \begin{matrix} 0 & 5 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 5 } (5)\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 10 }\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] \)
∴ X = \(\frac { 1 }{ 10 } \left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] \).

Equation of tangent to the circle at P(x₁, y₁ ) is xx₁ yy₁ = a²
That is, x(−3) + y(4) = 25
−3x + 4y = 25
Equation of normal is xy₁ - yx₁ = 0
That is, 4x + 3y = 0 

Given v = 3-4i, w = 4+3i and \(\frac { 1 }{ u } =\frac { 1 }{ v } +\frac { 1 }{ w } \)
∴ \(\frac { 1 }{ u } =\frac { 1 }{ 3-4i } +\frac { 1 }{ 4+3i } \)
= \(\frac { 3+4i }{ (3-4i)(3+4i) } +\frac { 4-3i }{ (4+3i)(4-3i) } \)
= \(\\ \frac { 3+4i }{ 9-(4i)^{ 2 } } +\frac { 4-3i }{ 16-(3i)^{ 2 } } =\frac { 3+4i }{ 9+16 } +\frac { 4-3i }{ 16+9 } \)
= \(\frac { 3+4i }{ 25 } +\frac { 4-3i }{ 25 } =\frac { 3+4i+4-3i }{ 25 } \)
\(\frac { 1 }{ u } =\frac { 7+i }{ 25 } \)
∴ u = \(\frac { 25 }{ 7+i } \times \frac { 7-i }{ 7-i } =\frac { 25(7-i }{ 7^{ 2 }-({ i }^{ 2 }) } \) 
= \(\frac { 25(7-i) }{ 49+1 } =\frac { 25(7-i) }{ 50 } =\frac { 1 }{ 2 } \)(7-i)
∴ u = \(\frac { 1 }{ 2 } \)(7-i) or \(\frac { 7 }{ 2 } \) - \(\frac { i }{ 2 } \)

Given \(\frac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \)
\((cos2\frac { \pi }{ 2 } <3x-1\))
\(\Rightarrow 0<3x-1<-1\)
\(0+1<3x<-1+1\)
\(1<3x<0\)
This inequality is true, only when 0 < x < \(\frac { 1 }{ 3 }\)

Since tan(tan^-1 x) = x, x ∈ R,
We have tan(tan^-1(2019)) = 2019

Let f(x) = 8x³- 2x² - 7x +3 = 0
Here sum of the co-efficients of odd terms = 8 - 7 = 1
and sum of the co-efficients of even terms = -2 +3 = 1
Hence, x = - 1 is a root of f(x)
Let us divide f(x) by (x + 1)

∴ The other factor is 8x² - 10x + 3
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-4(8)(3) } }{ 2\times 8 } \)
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-96 } }{ 16 } \Rightarrow x=\frac { 10\pm 2 }{ 16 } \)
\(\Rightarrow x=\frac { 12 }{ 16 } \ or \)
\(x=\frac { 8 }{ 16 } \Rightarrow x=\frac { 3 }{ 4 } ,\frac { 1 }{ 2 } \)
∴ The roots are -1, \(\frac{1}{2},\frac{3}{4}\)

Comparing the given equation with equation of a straight line \(\vec { r } =\vec { a } +t\vec { b } \), we have \(\vec { a } =3\hat { i } -2\hat { j } +6\hat { k } \) and \(\vec { b } =3\hat { i } -2\hat { j } +6\hat { k } \). Therefore,
(i) If \(\vec { b } ={ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \), then direction ratios of the straight line are \({ b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 }\). Therefore, direction ratios of the given straight line are proportional to 2, −1,3, and hence the direction cosines of the given straight line are \(\frac { 2 }{ \sqrt { 14 } } ,\frac { -1 }{ \sqrt { 14 } } ,\frac { 3 }{ \sqrt { 14 } } \)
(ii) vector equation of the straight line in non-parametric form is given by \((\vec { r } -\vec { a } )\times \vec { b } =\vec { 0 } \), Therefore, \((\vec{r}-(3\hat { i } -2\hat { j } +6\hat { k } )\times(2\hat { i } -\hat { j } +3\hat { k } )=\vec{0}\)
(iii) Here (x₁, y₁ ,z₁ ) = (3, −2,6) and the direction ratios are proportional to 2, −1,3. Therefore, Cartesian equations of the straight line are \(\frac { x-3 }{ 2 } =\frac { y+2 }{ -1 } =\frac { z-6 }{ 3 } \)

\(LHS={ tan }^{ -1 }x+{ tan }^{ -1 }\frac { 2x }{ 1-{ x }^{ 2 } } \)
= \({ tan }^{ -1 }\left( \frac { x+\frac { 2x }{ 1-{ x }^{ 2 } } }{ 1-x\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x(1-{ x }^{ 2 })+2x }{ 1-{ x }^{ 2 } } }{ \frac { 1-{ x }^{ 2 }-2{ x }^{ 2 } }{ 1-{ x }^{ 2 } } } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { x-{ x }^{ 3 }+2x }{ 1-{ x }^{ 2 } } }{ \frac { 1-3x^{ 2 } }{ 1-{ x }^{ 2 } } } \right) \quad \left[ \because |x|<\frac { 1 }{ \sqrt { 3 } } \right] \)
= \({ tan }^{ -1 }\left( \frac { 3x-{ x }^{ 3 } }{ 1-{ x }^{ 2 } } \times \frac { 1-{ x }^{ 2 } }{ 1-{ 3x }^{ 2 } } \right) \)
If 3x² < 1
⇒ \( |x|<\frac { 1 }{ \sqrt { 3 } }\)

Equation of the hyperbola is 12x²- 9y² = 108
\(\div 108\) we get, \(\frac { { 12x }^{ 2 } }{ 108 } -\frac { 9{ y }^{ 2 } }{ 108 } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 12 } =1\)
∴ a² = 9, b² = 12
Parametric equation of tangent to the hyperbola is \(\frac { x \ sec\ \theta }{ a } -\frac { y \ tan \ \theta }{ b } =1\)
When \(\theta =\frac { \pi }{ 3 } \), the equation is
\(\frac { xsec\frac { \pi }{ 3 } }{ 3 } -\frac { ytan\frac { \pi }{ 3 } }{ 2\sqrt { 3 } } =1\)

⇒ \(\frac { 4x-3y }{ 6 } =1\) ⇒ 4x - 3y - 6 = 0 is the required equation of tangent.
Parametric equation of normal to the hyperbola is
\(\frac { ax }{ sec\theta } +\frac { by }{ tan\theta } ={ a }^{ 2 }+{ b }^{ 2 }\)
At \(\theta =\frac { \pi }{ 3 } ,\frac { 3x }{ sec\frac { \pi }{ 3 } } +\frac { 2\sqrt { 3 } }{ tan\frac { \pi }{ 3 } } =9+12\)
\(\Rightarrow \frac{3 x}{2}+\frac{2 \sqrt{\not 3} y}{\sqrt{\not 3}}=21\)
⇒ \(\frac { 3x }{ 2 } \) + 2y = 21 ⇒ 3x + 4y = 42
⇒ 3x + 4y - 42 = 0 is the required equation of normal.

Equation of theparabola is y² = 16x
∴ 4a = 16 ⇒ a = 4
Let y = mx + c ... (1)
be a tangent to the parabola
Since the tangent is perpendicular to
2x + 2y+ 3 =0,
m = \(\frac{-1}{Slope \ of \ the \ line \ 2 x + 2y + 3 = 0}\)
∴ m = \(\frac { -1 }{ \frac { -2 }{ 2 } } =\frac { -1 }{ -1 } \) = 1
[∵ \(Slope=\frac{Co - efficient of \ x}{Co - efficient of \ y}=\frac{-2}{2}\) = -1]
The condition for the line y = mx + c to be a tangent to the parabola is c = \(\frac{a}{m}\)
∴ c = \(\frac14\)= 4
From (1), y = 1(x) + 4 ⇒ x - y + 4 = 0 is the required tangent.

Let the amount of 50% acid be x and the amount of 25% acid be y litre
By the given data, x + y = 10 ..............(1)
and \(x\left( \frac { 50 }{ 100 } \right) +y\left( \frac { 25 }{ 100 } \right) =10\left( \frac { 40 }{ 100 } \right) \)
⇒ 50x + 25y = 400 ⇒ 2x + y = 16 ...............(2)
The matrix from of the equation is \(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \end{matrix} \right] ,B=\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ X = A^-1N |A| = \(\left| \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right| \) = 1 - 2 = -1
⇒ X = \(\frac { 1 }{ |A| } \)adj A.B
⇒ X= \(-1\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
= -\(\left[ \begin{matrix} 10-16 \\ -20+16 \end{matrix} \right] \)
⇒ X = -\(\left[ \begin{matrix} -6 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 6 \\ 4 \end{matrix} \right] \)
Thus, the amount of 50% acid is 6 litre and the amount of 25% acid is 4 litre = 10 litres of 40% acid solution.

Since sum of two of its roots vanishes, let the roots be ∝, -∝ and β 
\(\alpha -\alpha +\beta =\frac { -b }{ a } =\frac { 1 }{ 2 } \)
\(\Rightarrow \ \beta =\frac { 1 }{ 2 } \)
Also, \(\alpha \beta \gamma =\frac { -d }{ a } \)
\(\Rightarrow \alpha (-\alpha )(\frac { 1 }{ 2 } )=\frac { -9 }{ 2 } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ 2 } =\frac { 9 }{ 2 } \Rightarrow { \alpha }^{ 2 }=9\Rightarrow \alpha =\pm 3\)
\(\Rightarrow \alpha =3\)
∴ The roots are 3, -3 and \(\frac{1}{2}\).

Let A be the origin, \(\vec { b } \) be the position vector of B and \(\vec {c } \) be the position vector of C .
Now D is the midpoint of BC , and so the position vector of D \(\frac{\vec{b}+\vec{c}}{2}\). There, we get

\({ \left| \vec { AD } \right| }^{ 2 }=\vec { AD } .\vec { AD } \)= \(\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) .\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) \)= \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )\)....(1)
Now, \(\vec { BD } =\vec { AD } -\vec { AB } \) = \(\frac { \vec { b } +\vec { c } }{ 2 } -\vec { b }=\frac { \vec {c } -\vec { b} }{ 2 }\)
Then, we get, =\({ \left| \vec { BD } \right| }^{ 2 }=\vec { BD } .\vec {BD } \) = \(\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) .\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) \) = \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )\)....(2)
Now, adding (1) and (2), we get
Therefore, \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )+\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )=\frac { 1 }{ 2 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 })\)
⇒ \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 2 } ({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 })\)
Hence, \({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 }=2({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 })\)

We observe that the sum of the coefficients of the odd powers and that of the even powers are equal.
Hence −1 is a root of the equation.
To find other roots, we divide 2x³+11x²-9x-18 by x+1 and get 2x²+9x-18 as the quotient.
Solving this we get \(\frac{3}{2}\) and -6 as roots.
Thus -6, -1, \(\frac{3}{2}\) are the roots or solutions of the given equation.