Complex Numbers and Quadratic Equations

Case-I : $a-2>0$

$F(-2)>0$

$4(a-2)+4a+a>0$

$9\mathrm{a}-8>0$

$a>\frac{8}{9}$

$F\left(1\right)>0$

$a-2-2a+a>0$

  $-2>0$  (not possible)

Case-II : $a-2<0;a<2$

$F(-2)<0\mathrm{}\Rightarrow \mathrm{}a<\frac{8}{9}$

$F\left(1\right)<0\mathrm{}\Rightarrow \mathrm{}a\in R$

$-2<\frac{-b}{2a}<1\mathrm{}a<\frac{4}{3}\mathrm{}a\in \left[0,\frac{8}{9}\right)\cup \left\{2\right\}$

$D\ge 0\mathrm{}a\ge 0$

|x|³/? = y
⇒ y² – 26y – 27 = 0 ⇒ y = −1 or 27
⇒ y = 27 ⇒ |x|³/? = 3³
|x| = 3? ⇒ x = ±3?
Product of roots = (3? ) (−3? ) = −3¹?

Let S = z + 2z² + 3z³ + . + 18z¹?
zS = z² + 2z³ + . + 18z¹?
(1 − z)S = z + z² + z³ + . - 18z¹?
(1 − z)S = z (z¹? - 1)/ (z-1) - 18z¹?
Now z = cos 20° + isin 20° ⇒ z¹? = 1
Also |z| = 1
⇒ (1 − z)S = -18z
⇒ S = -18z / (1-z)
|S|? ¹ = | (1-z)/ (-18z)| = |1-z|/18
= (1/18) |1 – cos 20° - isin 20°| = (1/18) |2s

Kindly consider the following figure

β² – 6β + 12 = 0 β − 6 = -12/β
∴ Reqd. Expression is (α – 2)³)? + (-12)¹² / (αβ)¹²) - 1
= (α³ – 8 – 6α (α – 2)? + (12)¹² / (12)¹²) - 1 = (α (α² – 6α + 12) – 8)? = 8? = 2¹²

Let P is a root   $\left(\in I\right)$

Case-I : p is odd

p¹⁰ + ap⁹ + b   $\ne 0$

Because LHS odd

Case-II : p is even

$p^{10}+a{p}^9+b\ne 0$  

because, LHs is odd

tan? ¹ (x) = t
t² - 4t + 3 > 0
t ∈ (-∞, 1) U (3, ∞)
tan? ¹ (x) ∈ (-π/2, 1)
x ∈ (-∞, tan1)
the largest integral x = 1

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 × 4² = 48

z - 2² = z + 2²
⇒ x=0
Hence minimum '0'

  $\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$

Let sin x = t, t  $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$         

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$           

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$           

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$          

Minimum value of a for which solution exist = 9