Maths

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

(1 + x)? = C? + C? x + C? x² + . C? x?
Put x = 1
2? = C? + C? + C? + C? + … C?
Put x = ω
(1 + ω)? = C? + C? ω + C? ω² + . C? (ω)?
Put x = ω²
(1 + ω²)? = C? + C? ω² + C? ω² + … C? (ω²)?
∴ 1 + ω + ω² = 0
Now add all three equation (s)
2? + (1 + ω)? + (1 + ω²)? = 3 [C? + C? + C? + …]
C? + C? + C? + C? + …
= (2? + (1+ω)? + (1+ω²)? ) / 3
= (2? + (1+ω)? + (-ω)? ) / 3 = (2? - (-1)? ) / 3

Negation of (p → q) ∨ (p ∨ q)
~ [ (p → q) ∨ (p ∨ q)]
≡ ~ (p → q) ∧ ~ (p ∨ q)
[∴ ~ (p → q) ≡ p ∧ ~q]
≡ (p ∧ ~q) ∧ (~p ∧ ~q)
= F (contradiction)

I = ∫? ² [log? x] dx
By using graph
I = ∫? 0 dx + ∫? ² 1 dx
I = 0 + e² - e

According to question
aα² + bα + c = 0 . (i)
−aβ² + bβ + c = 0 . (ii)
a/2 γ² + bγ + c = 0 . (iii)
bα + c = -aα² . (iv)
bβ + c = aβ² . (v)
bγ + c = -a/2 γ² . (vi)
By observing (v) and (vi), we get β > γ
By observing (iv) and (vi), we get α > γ
So α < <

By using condition of tangency,
c² = a² (m² + 1)
⇒ c² = 5 [ (2)² + 1]
⇒ c² = 25
⇒ c = ±5

y = |x − 1|, y = 3 – |x|

(A graph is shown with vertices A (1, 0), B (2, 1), C (0, 3), D (-1, 2). The lines are y = x - 1, y = 3 - x, y = 3 + x, and y = -x + 1)

AB = √2, BC = 2√2
⇒ Area = 4 sq. units

f' (0) = 0   

h' (x) = f' (x) – 2f (x).f' (x) + 3 (f (x)².f' (x)
= f' (x)/3 (f (x) - 1/3)² + 2/9)
∴ h' (x) ≥ 0 ⇔ f' (x) ≥ 0