Maths

|Z|max + |Z|min
= 6 + 0
= 6

? + b? = λ? c? (i)
b? + c? = λ? (ii)
Form (i) – (ii),
? – c? = λ? c? – λ?
(1 + λ? )? = (1 + λ? )c?
? ? and c? are non – collinear
⇒ 1 + λ? = 0, 1 + λ? = 0
λ? = λ? = -1
⇒? + b? + c? = 0

f (x) = x√ (1 – x²)
∴ 1 - x² ≥ 0
⇒ x ∈ [-1, 1]
Put x = cos θ
⇒ f (θ) = cos θ|sin θ|
If sin θ > 0, f (θ) = sin θ cos θ
= (sin 2θ)/2, f (θ) ∈ [-1/2, 1/2]
if sin θ < 0, f () = sin cos
= - (sin 2θ)/2, f (θ) ∈ [-1/2, 1/2]

5 red, 3 yellow
There are 3 points in sample space either both are red or both are yellow or one is red other is yellow.
⇒ P (E) = (3.5+5.3) / (3.5+5.3+5.4+3.2) = 15/28

I = ∫ (cos 4x-1)/ (cot x-tan x) dx = ∫ (2 cos² 2x-1-1)/ (cos²x-sin²x)/ (sin x cos x) dx
= 2 ∫ (cos² 2x-1) sin x cos x) / (cos 2x) dx = ∫ (cos² 2x-1) sin 2x) / (cos 2x) dx
Put cos 2x = t
-2 sin 2x dx = dt
I = -1/2 ∫ (t²-1)/t dt = 1/2 ∫ (1/t - t) dt
= 1/2 (ln|t| - t²/2) + c
= 1/2 ln|cos 2x| - 1/4 cos² (2x) + c

| 1 -1 -1 |
| 1 -k | = 0
| k 2 1 |

⇒ 1 (1 + 2k) + 1 (1 + k²) – 1 (2 – k) = 0
2k + 1 + 1 + k² − 2 + k = 0
k² + 3k = 0
k = 0, -3

A = [1]
[0 1]
Now A² = [1] [1] = [1 2]
[0 1] [0 1] [0 1]
A³ = A².A = [1 2] [1] = [1 3]
[0 1] [0 1] [0 1]
Similarly A²? ¹¹ = [1 2011]
[0 1]

x = t² - t + 1 . (i)
y = t² + t + 1 . (ii)
(i) + (ii)
x + y = 2 (t² + 1) . (iii)
(i) – (ii)
x - y = -2t . (iv)
from (iii) and (iv)
x² – 2xy + y² – 2x – 2y + 4 = 0
Here H² = ab and Δ≠ 0
This is parabola, so e = 1

Length of normal
k = y √ (1 + (dy/dx)²)

⇒ ∫ (y dy) / √ (k² - y²) = ∫ dx
Let k² – y² = t²
-2y dy = 2t dt
-√ (k² - y²) = x + c
It passes through (0, k)
x² + y² = k²

| 1+x² |
| x 1+x | = − (x - α? ) (x − α? ) (x – α? ) (x – α? )
| x² x 1+x |

Put x = 0
| 1 0 |
| 0 1 0 | = - (-α? ) (-α? ) (-α? ) (-α? )
| 0 1 |
α ⇒? α? α? α? = −1