Maths

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

g (x) = (f (nf (x) – n)?
g' (x) = n (f (nf (x) – n)? ¹ . f' (nf (x) – n) . n . f' (x)
∴ g' (0) = 0
⇒ 4 = n (f (nf (0) – n)? ¹ . f' (nf (0) – n) . nf' (0)
⇒ 4 = n (f (0)? ¹ . f' (0) . nf' (0)
⇒ 4 = n . 1 . (-1) . n (-1)
n² = 4
⇒ n = 2

There are 7 letters permutation.
The total number of 4 letters
Out of 2A's, 2K's and 3 different letters O, L, T.
= Coefficient of x? in 4! (1 + x + x²/2!)² (1 + x)³
= Coefficient of x? in 4! (1 + x + x²/2)² (1 + x)³
= Coefficient of x? in 4! [ (1 + x)? + (1 + x)? x² + (1 + x)³ + x? /4]
= 4! [? C? +? C? + ³C? /4] = 4! (5 + 6 + 1/4)
= 24 [11 + 1/4]
= 264 + 6 = 270