Let \(\int _{ 0 }^{ 1 }{ \frac { sin(3{ tan }^{ -1 }x){ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } } dx\)
\(put\quad { tan }^{ -1 }x=t\Rightarrow \frac { 1 }{ 1+{ x }^{ 2 } } dx=dt\)
\(\Rightarrow \int _{ 0 }^{ \frac { \pi }{ 4 } }{ t\quad sin(3t)dt } =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tsin3t } dt\)

\(Let\ u=t\ \ v=sin3tdt\)
\(u'=1\ \ { v }_{ 1 }=\frac { -cos3t }{ 3 } \)
\(u'' = 0 \quad { v }_{ 2 }=\frac { -sin3t }{ 9 } \)
Bernoulli's formula:
\(\int { uvdx } =u{ v }_{ 1 }-u'{ v }_{ 2 }\)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tsin3tdt={ \left[ t\left( \frac { -cos3t }{ 3 } \right) -1\left( \frac { -sin3t }{ 9 } \right) \right] }_{ 0 }^{ \frac { \pi }{ 4 } } } \)
\(=\frac { -1 }{ 9 } { [3tcos3t-sin3t] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { -1 }{ 9 } \left[ \left( 3\frac { \pi }{ 4 } cos\frac { 3\pi }{ 4 } -sin\frac { 3\pi }{ 4 } \right) -(0-0) \right] \)
\(=\frac { -1 }{ 9 } \left[ \frac { 3\pi }{ 4 } \left( -\frac { 1 }{ \sqrt { 2 } } \right) -\frac { 1 }{ \sqrt { 2 } } \right] \)
\(=\frac { -1 }{ 9 } \left[ \frac { -3\pi -4 }{ 4\sqrt { 2 } } \right] =\frac { 3\pi }{ 36\sqrt { 2 } } +\frac { 4 }{ 36\sqrt { 2 } } \)
\(=\frac { \pi }{ 12\sqrt { 2 } } +\frac { 1 }{ 9\sqrt { 9 } } =\frac { 1 }{ \sqrt { 9 } } \left[ \frac { \pi }{ 12 } +\frac { 1 }{ 9 } \right] \)

Given equation of the curve is y = 1 + x³ and the line is x + 12y = 12
Slope of the tangent to the curve
m₁ = \(\frac { dy }{ dx } \) = 3 x² and the
Slope of the line = m₂
= \(\frac{-1}{2}\) \(\left[ \because m=\frac { co-efficient\ of\ x }{ co-efficient\ of\ y } \right] \)
Since the slope of the tangent to the curve and the line are orthogonal, m₁ m₂ = - 1.
∴ 3x²\(\left( \frac { -1 }{ 2 } \right) \) = -1
⇒ \(\frac{x^2}{4}\) = 1
⇒ x² = 4
⇒ x = ±2
When x = 2, y = 1 + 2³ = 9
When x = -2, y = 1+ (-2)³
= 1-8 = -7 
∴ Equation of the tangent at (2, 9) is
y-9 = 12(x-2) [∵ m₁ = 3x² = 3(2)² = 12]
∴ y - 9 = 12x - 24
∴ 12x - y = 15
Equation of the tangent at (-2, -7) is
y+7= 12(x + 2)
⇒ y + 7 = 12x + 24
⇒ 12x-y+17 = 0

Given 2cos α = x+\(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2xcos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α+cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y+\(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β+1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
=\(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cosβ ± i sinβ
\({ x }^{ m }{ y }^{ n }+\cfrac { 1 }{ { x }^{ m }{ y }^{ n } } =2cos(m\alpha +n\beta )\)
x^myⁿ = (cos α + i sin mα) (cos nβ + i sin nβ)
cos(mα+nβ)+i sin(mα+nβ)
\(\frac { 1 }{ { x }^{ m }{ y }^{ n } } \) = cos(mα+nβ)-i sin(mα+nβ)

= 2cos(mα+nβ)

Let O₁ O₂ be the centres of the two semi circular vents.
First vent with centre O₁ (12, 0) and radius r = 10 yields equation to first semicircle as
(x−12)²+(y− 0)² = 10²
\(\Rightarrow\) x²+y²−24x + 44 = 0, y > 0
Second vent with centre O₂ (34, 0) and radius r = 10 yields equation to second vent as
(x−34)²+ y² = 10² 
x²+y²− 68x  + 1056 = 0, y > 0 

Let z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ I+2i } \right) ^{ 15 }\)
Here, \(\frac { 19+9i }{ 5-3i } =\frac { (19+9i)(5+3i) }{ (5-3i)(5+3i) } \)
= \(\frac { (95-27)+i(45+57) }{ { 5 }^{ 2 }+{ 3 }^{ 2 } } =\frac { 68+102i }{ 34 } \)
= 2 + 3i ................(1)
and \(\frac { 8+i }{ 1+2i } =\frac { (8+i)(1-2i) }{ (1+2i)(1-2i) } \)
= \(\frac { (8+2)+i(1-16) }{ { 1 }^{ 2 }+{ 2 }^{ 2 } } =\frac { 10-15i }{ 5 } \)
= 2 - 3i .............. (2)
Now z =\(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\)
⇒ z = (2 + 3i)¹⁵ - (2 - 3i)¹⁵   (by (1) and (2))
Then by definition, \(\bar { z } =\left( \overline { (2+3i)^{ 15 }-(2-3i)^{ 15 } } \right) \)
= \(\left( \overline { 2+3i } \right) ^{ 15 }-\left( \overline { 2-3i } \right) ^{ 15 }\) (using properties of conjugates)
= (2 - 3i)¹⁵ - (2 + 3i)¹⁵ = -((2 + 3i)¹⁵ - (2-3i)¹⁵)
⇒ \(\\ \overline { z } \) = -z
Therefore, \(\left( \frac { 19+9i }{ 5-3i } \right) ^{ 15 }-\left( \frac { 8+i }{ 1+2i } \right) ^{ 15 }\) is purely imaginary.

Since A² = \(\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] =\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] \).
A² + xA + yI₂ = O₂ ⇒ \(\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] +x\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] +y\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 22+4x+y & 27+3x \\ 18+2x & 31+5x+y \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \).
So, we get 22 + 4x + y = 0, 31 + 5x + y = 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = −9 and y = 14. Then, we get  A² - 9A + 14I₂ = O₂.
Postmultiplying this equation by A^-1, we get A - 9I₂ + 14A^-1 = O₂. Hence, we get
A^-1 = \(\frac { 1 }{ 14 } \) (9I₂ - A) = \(\frac { 1 }{ 14 } \left( 9\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \right) =\frac { 1 }{ 14 } \left[ \begin{matrix} 5 & -3 \\ -2 & 4 \end{matrix} \right] \).

Let x = length of the cut on each side of the little squares.
V = the volume of the folded box.
The length of the base after two cuts along each edge of size x is 12 − 2x. The depth of the box  after folding is x, so the volume is V = x \(\times\) (12 - 2x)² Note that,
when x = 0 or 6, the volume is zero and hence there cannot be a box. Therefore the problem is to maximize, V = x \(\times\) (12 - 2x)², x ∈ (0, 6) 
 \(\frac { dV }{ dx } ={ (12-2x) }^{ 2 }-4x(12-2x)\)
= (12 − 2x)(12 − 6x).

\(\frac { dV }{ dx } \) = 0 gives the stationary points x = 2, 6. Since 6 ∉ (0, 6) the only stationary point is at x = 2∈(0, 6). Further, \(\frac { dV }{ dx } \)- changes its sign from postive to negative when passing through x = 2 .
Therefore at x = 2 the volume V is local maximum. The local maximum volume value is V = 128 units. Hence the maximum cut can only be 2 units.

Let (x) = cos2 x
f^I(x) = 2cos x (- sin x)
= - sin 2x ⇒ f^l(0) = 0
f^Il(x) = - 2 cos 2x ⇒ f^Il(0) = -2
f^III(x) = + 4 sin 2 x ⇒ f^III(0) = 0
f^IV(x) = 8 cos 2 x ⇒ f^IV(0) = 8
f^V(x) = -16 sin 2x ⇒ f^V(0) = 0
f^VI(x) = - 32 cos 2x ⇒ f^VI(0) = -32
∴ Maclaurin's series
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\) .......
∴ cos x = 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { 8{ x }^{ 4 } }{ 4! } +\frac { { 32x }^{ 6 } }{ 6! } \) + ...
= 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { { 2 }^{ 3 }{ x }^{ 4 } }{ 4! } +\frac { { { 2 }^{ 5 }x }^{ 6 } }{ 6! } \)+ ...

Given f(x) = 2.0 + 3x2 - 12x
The given function is defined and differentiable for all x ∈(-∞,∞)
f'(x) = 6x² + 6x - 12
The stationary points are given by
6x² + 6x - 12 = 0
\(\Rightarrow\) x² +x-2 = 0
\(\Rightarrow\) (x + 2)(x - 1) = 0
\(\Rightarrow\) x = -2, 1
Hence the intervals of monotonicity are
(-∞, - 2), (-2, 1) and (1, ∞).

ஃ f(x) is strictly increasing on (-∞, - 2)
(1, ∞) and strictly decreasing on (-2, 1).
Since f'(x) changes from positive to negative at x = -2, there is a local maximum at x = -2
\(\therefore\) f(-2) = 2 (2)³ + 3 (2)² - 12 (-2) 
= 2(-8) + 3(4) + 24 
= -16+12+24 = 20
Also f'(x) changes from negative to positive at x = 1, there is a local minimum at x = 1.
ஃ f(1) = 2 (1)³ + 3 (1)² - 12(1)
= 5 -12 = -7

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB² = AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
 ⇒ y² = 289 - 64 = 225
 ⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
 Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

Given s (t) = 2t³ − 9t² + 12t ≥ 0

x+y+z-2 = 0, 6x-4y+5z-31= 0, 5x+2y+2z = 13
The matrix form of the system is
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 13 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right| =1\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| -1\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \)
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| & +\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ -4 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 6 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 6 & -4 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(-8-10) & -(12-25) & +(12+20) \\ -(2-2) & +(2-5) & -(2-5) \\ +(5+4) & -(5-6) & +(-4-6) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 13 & 32 \\ 0 & -3 & 3 \\ 9 & 1 & -10 \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ X = A^-1B
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -36 & +0 & +117 \\ 26 & -93 & +13 \\ 64 & +93 & -130 \end{matrix} \right] =\frac { 1 }{ 27 } \left[ \begin{matrix} 81 \\ -54 \\ 27 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ 1 \end{matrix} \right] \)
∴ x = 3, y = -2, z = 1
∴ Solution set is {3, -2, 1}

Here the number of unknowns is 3. So, if the system is consistent and has a non-trivial solution, then the rank of the coefficient matrix is equal to the rank of the augmented matrix and is less than 3.
So the determinant of the coefficient matrix should be 0.
Hence we get
\(\left| \begin{matrix} 3\lambda -8 & 3 & 3 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 or \(\left| \begin{matrix} 3\lambda -2 & 3\lambda -2 & 3\lambda -2 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by applying R₁ ➝ R₁ + R₂ + R₃)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by taking out (3λ − 2) from R₁)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -11 & 3 \\ 3 & 3 & 3\lambda -11 \end{matrix} \right| \) = 0 (by applying R₂ ➝ R₂ - 3R₁, R₃ ➝ R₃ - 3R₁)
or (3λ - 2)(3λ - 11)² 0. So λ = \(\frac { 2 }{ 3 } \) and λ = \(\frac { 11 }{ 3 } \).
We now give an application of system of linear homogeneous equations to chemistry. You are already aware of balancing chemical reaction equations by inspecting the number of atoms present on both sides.

By definition,
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -1 & 0 \\ 1 & -1 & -4 \end{matrix} \right| =4\hat { i } +4\hat { j } ,\vec { c } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 0 & 3 & -1 \\ 2 & 5 & 1 \end{matrix} \right| =8\hat { i } -2\hat { j } -6\hat { k } \)
\((\vec { a } \times \vec { b } )(\vec { c } \times \vec { d } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & 0 \\ 8 & -2 & -6 \end{matrix} \right| =-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(1)
On the other hand, we have
\([\vec { a }, \vec { b }, \vec { d } ]\vec { c } -[\vec { a } ,\vec { b } ,\vec { c } ]\vec { d } =28(3\vec { j } -\vec { k } )-12(2\hat { i } +5\hat { j } +\hat { k } )=-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(2)
Therefore, from equations (1) and (2), identity (i) is verified.
The verification of identity (ii) is left as an exercise to the reader

Let AB be the rod and P(x₁, y₁) be a point on the rod such that AP = 0.3 m.
Draw PD ⊥ x-axis and PC ⊥ y - axis.

Δ ADP ≅ Δ PCB
∴ \(\frac { PC }{ DA } =\frac { PB }{ AP } =\frac { BC }{ PD } \)
⇒ \(\frac { x_{ 1 } }{ DA } =\frac { 0.9 }{ 0.3 } =\frac { BC }{ { y }_{ 1 } } \)
⇒ \(DA=\frac { 0.3{ x }_{ 1 } }{ 0.9 } =\frac { { x }_{ 1 } }{ 3 } \)
and BC = \(\frac { 0.9{ y }_{ 1 } }{ 0.3 } =\frac { 9 }{ 3 } { y }_{ 1 }=3{ y }_{ 1 }\)
Now OA = OD + DA
= \({ x }_{ 1 }+\frac { { x }_{ 1 } }{ 3 } =\frac { 4{ x }_{ 1 } }{ 3 } \)
OB = OC + BC = y₁ + 3y₁ = 4y₁
But OA² + OB² = AB²
⇒ \({ \left( \frac { 4{ x }_{ 1 } }{ 3 } \right) }^{ 2 }+{ \left( 4{ y }_{ 1 } \right) }^{ 2 }={ \left( 1.2 \right) }^{ 2 }\)
⇒ \(\frac { { { x }_{ 1 } }^{ 2 } }{ 9 } +\frac { { { y }_{ 1 } }^{ 2 } }{ 9 } =\frac { 1.44 }{ 16 } =0.09\) ≅ 1
∴ Locus of (x₁, y₁) is \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 1 } =1\)
Here a² = 9, b² = 1
∴ \(e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 1 }{ 9 } } =\sqrt { \frac { 9-1 }{ 9 } } \)
= \(\sqrt { \frac { 8 }{ 9 } } \)
e = \(\frac { 2\sqrt { 2 } }{ 3 } \) 

The cross section of a nuclear cooling tower is in the shape of a hyperbola.

GIven OC = \(\frac12\) OD and CD = 150 m
Its equation is OC = 50 m & OD = 100 m
\(\frac { { x }^{ 2 } }{ { 30 }^{ 2 } } -\frac { { y }^{ 2 } }{ { 44 }^{ 2 } } =1\)  ......(1)
Let I be the radius of the top of the tower
∴ A(l, 50) is a point on the hyperbola
∴ \(\frac { { l }^{ 2 } }{ { 30 }^{ 2 } } \frac { { 50 }^{ 2 } }{ { 44 }^{ 2 } } =1\)
⇒ \(\frac { { l }^{ 2 } }{ { 30 }^{ 2 } } =1+\frac { { 50 }^{ 2 } }{ { 44 }^{ 2 } } =\frac { { 44 }^{ 2 }+{ 50 }^{ 2 } }{ { 44 }^{ 2 } } \)
⇒ \({ l }^{ 2 }=\frac { { 30 }^{ 2 } }{ { 44 }^{ 2 } } \) (1936 + 2500)
⇒ l² = \(\frac { 30 }{ 40 } \sqrt { 4436 } =\frac { 30 }{ 44 } \)(66.60)
⇒ \(\frac { 30 }{ 40 } \sqrt { 4436 } =\frac { 30 }{ 44 } \) = 45.41 m
Radius of the top of the tower is 45.41 m.
Let h be the radius of the base of the tower.
∴ B(h, 100) is a point on the hyperbola
∴ (1) becomes
\(\frac { { h }^{ 2 } }{ { 30 }^{ 2 } } -\frac { { 100 }^{ 2 } }{ { 44 }^{ 2 } } =1\Rightarrow \frac { h^{ 2 } }{ { 30 }^{ 2 } } =1+\frac { { 100 }^{ 2 } }{ { 44 }^{ 2 } } =\frac { { 44 }^{ 2 }+{ 100 }^{ 2 } }{ { 44 }^{ 2 } } \)
⇒ \({ h }^{ 2 }=\frac { { 30 }^{ 2 } }{ { 44 }^{ 2 } } \) (1936 + 10000)
⇒ \({ h }^{ 2 }=\frac { 30 }{ 44 } \sqrt { 11936 } =\frac { 30 }{ 44 } \)(109.25)
⇒ \(h=\frac { 3277.5 }{ 44 } \) = 74.48 m.
Radius of the base of the tower is 74.48 m.
Diameter of the base = 148.96 m
Diameter of the top and base of the tower are 90.82 m and 148.96 m.

2x+3y = 9, 7x+3y-5z = 8, 2x+3y+⋋z = μ
The matrix form of the system is AX = B where
A =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \)
Applying elementary row operations augmented matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 2 & 3 & 5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 8 \\ 9 \\ \mu \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-\frac { 2 }{ 7 } { R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & \frac { 15 }{ 7 } & \frac { 45 }{ 7 } \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ \frac { 45 }{ 7 } \\ 4-9 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 7 }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ 47 \\ \mu -9 \end{matrix} \right] \)
Case (i): when λ = 5
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ -4 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B]
Hence the system is inconsistent and has no solution
Case (ii) : When λ ≠ 5, μ ≠ 9
[A|B] =\(\\ \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} -8 \\ 47 \\ not\quad zero \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
∴ \(\rho \)(A) = \(\rho \)[A|B] = 3 = number of unknowns
Hence, the system is consistent with solution
Case (iii) : When λ = 5, μ = 9
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2
∴ The system is consistent and has infinite number of solutions.

Now, \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =tan^{ -1 }\left[ \frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \right] =\frac { \pi }{ 4 } \)
Thus, \(\frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \) = 1, which on simplification gives 2x²−4 = −3
Thus, x² = \(\frac{1}{2}\)gives x = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

Given A =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \), B=\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+3+6 & -5+2+3 & -10+1+9 \\ 7+3-10 & 7+2-3 & 14+1-15 \\ 1-3+2 & 1-2+1 & 2-1+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃
BA =\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+7+2 & 1+1-2 & 3-5+2 \\ -15+14+1 & 3+2-1 & 9-10+1 \\ -10+7+3 & 2+1-3 & 6-5+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃.
So, we get AB = BA = 4. I₃
⇒ \(\left( \frac { 1 }{ 4 } A \right) B=B\left( \frac { 1 }{ 4 } A \right) =1\)
⇒ B^-1 = \(\frac { 1 }{ 4 } \) = 1
Writing the given set of equations in matrix form we get,
\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(B=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ={ B }^{ -1 }\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] =\left[ \frac { 1 }{ 4 } A \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5+7+6 \\ 7+7-10 \\ 1-7+2 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 8 \\ 4 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ -1 \end{matrix} \right] \)
∴ x = 2, y = 1, z = -1
Hence, the solution set is {2, 1, - 1}.

The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \),X = \(\left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] \),B = \(\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] \).
We find |A| = \(\left| \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right| \) = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 10 + 15 + 15 = 40 ≠ 0.
So, A⁻¹ exists and
A^-1 = \(\frac { 1 }{ \left| A \right| } \) (adj A) = \(\frac { 1 }{ 40 } { \left[ \begin{matrix} +\left( 4+1 \right) & -\left( -2-3 \right) & +\left( -1+6 \right) \\ -\left( -6+3 \right) & +\left( -4-9 \right) & -\left( -2-9 \right) \\ +\left( 3+6 \right) & -\left( 2-3 \right) & +\left( -4-3 \right) \end{matrix} \right] }^{ T }=\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \)
Then, applying X = A⁻¹B, we get
\(\left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 40 \\ 80 \\ -40 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] \)
So, the solution is (x₁ = 1, x₂ = 2, x₃ = -1).

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, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

(i) 
Rearrange the terms as,
(x-5) (x+4) (x-7) (x+6) = 504
{-2, 3, -7, 8}
⇒ (x² -x - 20) (x² -x - 42) = 504
Put x²- x = y
⇒ (y-20)(y-42) = 504
⇒ y²-62y+840-504 = 0
⇒ y²-62y+336 = 0
⇒ (y - 56) (y - 6) = 0
⇒ y = 56, 6
Case (i)
When y = 56, x² - x = 56

⇒ x² - x - 56 = 0
⇒ (x - 8)(x + 7) = 0
⇒ x = 8, -7
Case (ii) 
When y = 6,
x²- x = 6
x²- x - 6 = 0
⇒ (x - 3)(x + 2) = 0
⇒ x = 3, - 2

Hence the roots are -2, 3, 8, -7

Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 ........... (1)
(1) passes through (1, 0)
⇒ 1 + 0 + 2g(1) + 2f(0) + c = 0
⇒ 2g + c = -1 ................(2)
(1) passes through (-1, 0)
⇒ (-1)² + 0 + 2g(-1) + 2f(0)+ c = 0
⇒ -2g + c = -1 ..............(3)
Also (1) passes through (0, 1)
⇒ 0 + 1²+ 2g(0) +2f(1) + c = 0
⇒ 2f + c = -1 .................(4)
(2) + (3) ⇒ 2c = -2
⇒ c = -1
Substituting c = -1 in (2), we get
2g-1 = -1
⇒ 2g = 0
⇒ g = 0
Substituting c = -1 in (4) we get,
2f -1 = -1
⇒ 2f  = 0
⇒ f = 0
∴ The required equation of the circle 
x² + y² - 1 = 0

Let ∝, β, ૪ be the roots of the equation 
Given \(\frac { \alpha }{ \beta } =\frac { 3 }{ 2 } \Rightarrow 2\alpha =3\beta \Rightarrow \alpha =\frac { 3 }{ 2 } \beta \)
\(\therefore \frac { 3 }{ 2 } \beta ,\beta ,\gamma \) are the roots of the given equation
Then by Vieta's formula,
\(\frac { 3 }{ 2 } \beta +\beta +\gamma =\frac { -b }{ a } =\frac { -(-9) }{ 1 } =9\)
\(\frac { 5 }{ 2 } \beta +\gamma =9\Rightarrow \gamma =9-\frac { 5 }{ 2 } \beta \)
\(\Rightarrow \gamma =\frac { 18-5\beta }{ 2 } ...(2)\)
Also \(\frac { 3 }{ 2 } \beta (\beta )+\beta \gamma +\left( \frac { 3 }{ 2 } \beta \right) \gamma =\frac { c }{ a } =\frac { 14 }{ 1 } =14\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 5 }{ 2 } \beta \left( \frac { 18-5\beta }{ 2 } \right) =14\ [using\ (2)]\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 90\beta }{ 4 } -\frac { 25{ \beta }^{ 2 } }{ 4 } =14\)
Multiplying by \(4,6{ \beta }^{ 2 }+90{ \beta }-25{ \beta }^{ 2 }=56\)
\(19{ \beta }^{ 2 }-90{ \beta }+56=0\)
\(\Rightarrow ({ \beta }-4)(19{ \beta }-14)=0\)
\(\Rightarrow { \beta }=4\)
\({ \beta }=\frac { 14 }{ 19 } \)
When \(\beta\) = 4, the other roots are \(\frac { 3 }{ 2 } (4),4,\frac { 18-5 }{ 2 } (4)\)
\(\Rightarrow 6,4,-1\)

When \(\\ \beta =\frac { 14 }{ 19 } ,\) the other roots are \(\frac { 3 }{ 2 } \beta ,\beta \frac { 18-5\beta }{ 2 } [by(2)]\)
\(\Rightarrow \frac { 3 }{ 2 } \left( \frac { 14 }{ 19 } \right) ,\frac { 14 }{ 19 } ,\frac { 18-5\left( \frac { 14 }{ 19 } \right) }{ 2 } \Rightarrow \frac { 21 }{ 19 } ,\frac { 14 }{ 19 } ,\frac { 136 }{ 19 } \)