Maths

Truth table for (p → q) ∧ (q → ~p).
| p | q | p → q | ~p | q → ~p | (p → q) ∧ (q → ~p) |
|-|-|-|-|-|-|
| T | T | T | F | F | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | T | T |
The final column is F, T, which is the truth table for ~p.
Therefore, (p → q) ∧ (q → ~p) is equivalent to ~p.

The tangent to the parabola y² = 4ax is y = mx + a/m.

For y² = 4x, a=1. So, the tangent is y = mx + 1/m.
The given line is y = mx + 4.
Comparing the two, 1/m = 4 ⇒ m = 1/4.
The line is y = (1/4)x + 4.
This line is also tangent to x² = 2by.
Substitute y into the parabola equation:
x² = 2b (1/4)x + 4)
x² = ( b/2 )x + 8b
x² - ( b/2 )x - 8b = 0.
For tangency, the discriminant (D) is zero.
D = (-b/2)² - 4 (1) (-8b) = 0.
b²/4 + 32b = 0.
b ( b/4 + 32) = 0.
b = 0 (not possible) or b/4 = -32 ⇒ b = -128.

(3¹/? + 5¹/? )?
General term =? C? (3¹/? )? (5¹/? )? =? C? 3^ (60-r)/4) 5^ (r/8)
Terms are rational for r being a multiple of 8 and (60-r) being a multiple of 4.
If r is a multiple of 8, then 60-r is 60 - 8k. Since 60 is a multiple of 4, 60-8k is also a multiple of 4.
So, we just need r to be a multiple of 8.
r = 0, 8, 16, 24, 32, 40, 48, 56. (Total 8 rational terms)
Total terms are 61.
Number of irrational terms = 61 - 8 = 53 = n.
∴ n - 1 = 52.

g (f (x) = f² (x) + f (x) - 1.
g (f (5/4) = f² (5/4) + f (5/4) - 1.
Given g (f (5/4) = 5/4, let f (5/4) = y.
-5/4 = y² + y - 1 (There appears to be a typo in the image's solution)
y² + y - 1 + 5/4 = 0
y² + y + 1/4 = 0
(y + 1/2)² = 0
y = -1/2.
So, f (5/4) = -1/2.

A = 1/3 [ 1; 1 ω ω² 1 ω² ω ]
A² = A * A = 1/9 [ . ]
(The calculation in the image shows A² is the identity matrix, let's verify)
A² leads to I (Identity matrix).
So A² = I.
A³ = A² * A = I * A = A.
A? = (A²)² = I² = I.
A³? = (A²)¹? = I¹? = I.

y = √ (2cos²α / (sinα cosα) + 1/sin²α)
y = √ (2cotα + cosec²α)
y = √ (2cotα + 1 + cot²α) = √ (1 + cotα)²) = |1 + cotα|.
Given α is in a range where 1+cotα is negative, y = -1 - cotα.
dy/dα = - (-cosec²α) = cosec²α.
At α = 5π/6, dy/dα = cosec² (5π/6) = (1/sin (5π/6)² = (1/ (1/2)² = 2² = 4.

Given Re (z-1)/ (2z+i) = 1, where z = x + iy.
(z-1)/ (2z+i) = [ (x-1) + iy] / [2x + I (2y+1)]
To rationalize, multiply the numerator and denominator by the conjugate of the denominator [2x - I (2y+1)].
Numerator = [ (x-1) + iy] * [2x - I (2y+1)] = 2x (x-1) - I (x-1) (2y+1) + i2xy + y (2y+1)
Real part of the numerator = 2x (x-1) + y (2y+1).
Denominator = (2x)² + (2y+1)².
Re (z-1)/ (2z+i) = [2x (x-1) + y (2y+1)] / [ (2x)² + (2y+1)²] = 1.
2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1.
0 = 2x² + 2y² + 2x + 3y + 1.
So, 2x² + 2y² + 2x + 3y + 1 = 0.

dy/dx + 2y tan (x) = sin (x)
I.F. = e^ (∫2tan (x)dx) = e^ (2ln (sec (x) = sec² (x)
Solution is y sec² (x) = ∫sin (x)sec² (x)dx = ∫sec (x)tan (x)dx
⇒ y sec² (x) = sec (x) + c
y (π/3) = 0 ⇒ 0 * sec² (π/3) = sec (π/3) + c ⇒ 0 = 2 + c ⇒ c = -2.
∴ y = (sec (x) - 2) / sec² (x)
Now let g (t) = (t - 2)/t² = 1/t - 2/t² for |t| ≥ 1.
g' (t) = -1/t² + 4/t³
g' (t) = 0 ⇒ t = 4.
g' (t) = 2/t³ - 12/t? g' (4) < 0, hence maximum.
∴ g (t)max = g (4) = (4 - 2)/4² = 2/16 = 1/8.