\( |z|=2 \)
\(|z+6+8 i|=|z|+|6+8 i| \)
\(=2+\sqrt{6^{2}+8^{2}} \)
\(=2+\sqrt{100} \)
= 2 + 10
= 12
\( \therefore|z+6+8 i| \leq 12 \) ............. (1)
\(|z+6+8 i| \geq|| z|-|-6-8 i|| \)
\(=|2-10| \)
\(=|-8|\)
= 8
\(|z+6+8 i| \geq 8\) .............(2)
From 1 and 2 we get
\(8 \leq|z+6+8 i| \leq 12\)
Hence proved

\(sin\left( { tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)

Let \({ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\Rightarrow \frac { 1 }{ 2 } =tanx\)
\(cosx=\frac { adj }{ hyp } =\frac { 2 }{ \sqrt { 5 } } \)
\(\therefore sinx= \frac { 1 }{ \sqrt { 5 } } \)

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =y\Rightarrow \frac { 4 }{ 5 } =cosy\)
\(siny=\frac { opp }{ hyp } =\frac { 3 }{ 5 } \)
\(\therefore sin\left( { tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) -{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \)
= sin (x - y) = sin x cos y - cos x sin y
= \(\frac { 1 }{ \sqrt { 5 } } .\frac { 4 }{ 5 } -\frac { 2 }{ \sqrt { 5 } } .\frac { 3 }{ 5 } \)
= \(\frac { 4 }{ 5\sqrt { 5 } } =\frac { 6 }{ 5\sqrt { 5 } } =\frac { -2 }{ 5\sqrt { 5 } }\)

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

\({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }(-1)\)
Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\) and -1 = sin y
\(\Rightarrow cosx=\frac { 1 }{ 2 } =cos\frac { \pi }{ 3 } \)
\(x=\frac { \pi }{ 3 } \)
\(siny=-1=-sin\left( \frac { -\pi }{ 2 } \right) \)
= \(-sin\left( \frac { -\pi }{ 2 } \right) \)
\(\left[ \because sin\left( -\theta \right) =-sin\ \theta\ and\ \frac { \pi }{ 2 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\Rightarrow y=\frac { -\pi }{ 2 } \)
\(\therefore { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( -1 \right) =x+y\)
= \(\frac { \pi }{ 3 } -\frac { \pi }{ 2 } =\frac { 2\pi -3\pi }{ 6 } =-\frac { \pi }{ 6 } \)
\(\therefore { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }(-1)=-\frac {\pi }{ 6 } \)

Consider the equation |z+2−i| = 2.
This can be written as |z−(−2+i) | = 2.
The above equation represents the circle with centre z₀ = -2+i and radius r = 2. Therefore z + 2−i<2 represents all points inside the circle with centre at −2+i and radius 2 as shown in figure.

\(c\ is\frac { 2\pi }{ 9 } +c\ is\frac { 4\pi }{ 9 } +c\ is\frac { 6\pi }{ 9 } +c\ is\frac { 8\pi }{ 9 } +c\ is\left( \frac { 10\pi }{ 9 } \right) +c\ is\frac { 12\pi }{ 9 } +c\ is\frac { 14\pi }{ 9 } +c\ is\frac { 16\pi }{ 9 } \)
=\(\ c\ is\left( \frac { 2\pi }{ 9 } +\frac { 4\pi }{ 9 } +\frac { 6\pi }{ 9 } +\frac { 8\pi }{ 9 } +\frac { 10\pi }{ 9 } +\frac { 12\pi }{ 9 } +\frac { 14\pi }{ 9 } +\frac { 16\pi }{ 9 } \right) \)

= \(\left[ \because 1+2+3+....+n=\frac { n(n+1) }{ 2 } \right] \)
= c is 8π = [cos(8π)+i sin 8π]
= -1 + i(0) [∴ cos8π = -1 and sin 8π = 0 = -1

sin(cos⁻¹(1-x))
we know that \({ cos }^{ -1 }x={ sin }^{ -1 }\left( \sqrt { 1-{ x }^{ 2 } } \right) \text {if}\ 0\le x\le 1\)
\(\therefore { cos }^{ -1 }\left( 1-x \right) ={ sin }^{ 1 }\sqrt { 1-\left( 1-x \right) ^{ 2 } } \left[ \because 0\le x\le 1 \right] \)
= \({ sin }^{ -1 }\left( \sqrt { 1-\left( 1+{ x }^{ 2 }-2x \right) } \right) \)
= \({ sin }^{ -1 }\left( \sqrt { 1-1-{ x }^{ 2 }+2x } \right) ={ sin }^{ -1 }\left( \sqrt { 2x-{ x }^{ 2 } } \right) \)
\(\therefore sin\left( { cos }^{ -1 }\left( 1-x \right) \right) =sin\left( { sin }^{ -1 }\left( \sqrt { 2x-{ x }^{ 2 } } \right) \right) \)
= \(\sqrt { 2x-{ x }^{ 2 } } \)

Let cot−1\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) =\alpha \). Then, cot \(\alpha =\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \) and α is acute.
We construct a right triangle with the given data.
From the triangle, sec\(\alpha=\frac{x}{1}=x\). Thus, \(\alpha\) = sec^-1x
Hence, cot⁻¹\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ={ sec }^{ -1 }x,|x|>1\)

By definition, cot^-1x\(\in(0,\pi)\)
Therefore, cot^-1\((\frac{1}{7})\) = \(\theta\) implies \(\theta \in(0,\pi)\)
But cot^-1\((\frac{1}{7})\) = \(\theta\) implies cot \(\theta\) = \(\frac{1}{7}\) and hence tan \(\theta\) = 7 and \(\theta\) is acute.
Using tan \(\theta\) = \(\frac{7}{1}\), We construct a right triangle as shown .
Then, we have, cos \(\theta\) = \(\frac{1}{5\sqrt2}\).

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 < \(\frac { 1 }{ 3 }\)

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.

Given that the height of the tree is 12m.
Let x m be the standing part and (12 -x)m be the broken part.
Given \(x=\sqrt [ 3 ]{ 12-x } \)
\(\Rightarrow x=(12-x)^{ \frac { 1 }{ 3 } }\)


Taking power 3 both sides, we get 
⇒ x³ = 12-x
⇒ x³+ x -12 = 0
which is the required mathematical problem.

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

Given ∝, β and ૪ are the roots of ax³ + bx² + cx + d = 0
\(\therefore \alpha +\beta +\gamma =\frac { -b }{ a } \)
\(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \)
\(\alpha \beta \gamma =\frac { -d }{ a } \)
Now, \(\sum { \frac { \alpha }{ \beta \gamma } } =\frac { \alpha }{ \beta \gamma } +\frac { \beta }{ \gamma \alpha } +\frac { \gamma }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } }{ \alpha \beta \gamma } \)
\(=\frac { { (\alpha +\beta +\gamma ) }^{ 2 }-2(\alpha \beta +\beta \gamma +\gamma \alpha ) }{ \alpha \beta \gamma } \)
\(=\frac { { \left( -\frac { b }{ a } \right) }^{ 2 }-2\left( \frac { c }{ a } \right) }{ -\frac { d }{ a } } \)
\(=\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } }{ -\frac { d }{ a } } \Rightarrow \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \times \frac { -a }{ d } \)
\(\therefore \frac { \sum { \alpha } }{ \beta \gamma } =-\frac { \left( { b }^{ 2 }-2ac \right) }{- ad } =\frac { 2ac-{ b }^{ 2 } }{ ad } \)

Let α, β, γ and δ be the roots of ax⁴+ bx³+ cx²+ dx + e = 0
Σ₁ = α + β + γ + δ = -\(\frac { b }{ a } \),
Σ₂ = αβ + αγ + αδ + βγ + βδ + γδ = \(\frac { c }{ a } \),
Σ₃ = αβγ + αβδ +  αγδ +  βγδ = -\(\frac { d }{ a } \)
Σ₄ = αβγδ =\(\frac { e }{ a } \)
We have to find α² + β² + γ² + δ² 
Applying the algebraic identity
(a+b+c+d)² ≡ a²+b²+c²+d²+2(ab+ac+ad+bc+bd+cd),
we get
α² + β² + γ² + δ² = (α + β + γ + δ)²-2(αβ + αγ + αδ + βγ + βδ + γδ)
= \(\left( \frac { b }{ a } \right) ^{ 2 }-2\left( \frac { c }{ a } \right) \)
= \(\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \).

The roots of x³+2x2+3x+4 = 0 are ∝, β, ૪
∴ ∝+β+૪ = -co-efficient of x² = -2 ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3 ....(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4 ...(3)
Form a cubic equation whose roots are 2∝, 2β, 2૪
2∝+2β+2૪ = 2(∝+β+૪) = 2(-2) = -4 [from (1)]
4∝β+4β૪+4૪∝ = 4(∝β+β૪+୪∝) = 4(3) = 12 [from (2)]
(2∝)(2β)(2૪) = 8(∝β૪) = 8(-4) = -32 [from (3)]
∴ The required cubic equation is
x³-(2∝+2β+2૪)x² + (2∝β+2β૪+2୪∝)x - (2∝)(2β)(2૪) = 0
⇒ x³+(-4)x²+12x+32 = 0
⇒ x³+4x²+12x+32 = 0

Given roots are 1, 2 and 3
Here a = 1, β = 2 and ૪ = 3
A cubic polynomial equation whose roots are α, β, ૪ is
x³-(α+β+૪)+x²(αβ+β૪+૪α)x-∝β૪ = 0
⇒ x³-(1+1+2)x²(2+6+3)x-6 = 0
⇒ x³-6x²+11x-6 = 0

The length and breadth of the cuboid are x + 1.
x + 2 and x + 3
[∵ they are increased by 1, 2, 3 units]
Also volume = V + 52 .
[since V is increased by 52]
∴ V + 52 = (x +1)(x + 2)(x + 3) .........(1)
⇒ V = (x + 1) (x + 2) (x + 3) - 52
Here a = -1, β = -2, ૪ = -3
⇒ V = x³-x²(α+β+૪)+x(αβ+β૪+૪α)-αβ૪ = 52
⇒ V = x³-x²(-1-2-3)+x(2+6+3)-(-1)(-2)(-3) = 52
⇒ V = x³-x²(-6)+x(11)+6-52
⇒ x³ = x²+6x²+11x+6-52
⇒ 6x²+11x-46 = 0
⇒ (6x+23)(x-2) = 0
⇒ (6x+23)(x-2) = 0
⇒ x = 2

∴ Volume of the cube = x³ = 2³ = 8.
Volume of a cuboid = 52 + 8 = 60
[∵ x = \(\frac{-23}{6}\) is not possible as x represents the side of the cube]

z₁(z₂ + z₃) = z₁z₂ + z₁z₃
Given z₁= 3, z₂ = -7i, z₃ = 5+4i
LHS = z₁(z₂ + z₃)
= 3 [-7i + 5 + 4i]
= 3[5-3i]
= 15-9i
RHS = z₁z₂ + z₁z₃
= 3(-7i) + 3(5 + 4i)
= -21i +15 +12i
= -9i +15
= 15-9i
LHS = RHS
∴ z₁(z₁ + z₃) = z₁z₂ + z₁z₃
Hence proved