Maths

Let the equation of circle be
x (x-1/2) + y² + λy = 0
=> x² + y² - x/2 + λy = 0
Radius = √ (1/16 + λ²/4) = 2
=> λ² = 63/4 => (x-1/4)² + (y+λ/2)² = 4
∴ This circle and parabola
y-α = (x-1/4)² touch each other, so
α = -λ/2 + 2 => α-2 = -λ/2 => (α-2)² = λ²/4 = 63/16
(4α–8)² = 63

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side
∴ a = √ (3-1)²+ (6-2)² = √20 and
b/ (a/2) = 4/a => b = 8/√5
Area
ab = 2√5 * 8/√5 = 16

T? = r/ (2r²)²+1)
= r/ (2r²+1)²- (2r)²)
= (1/4) * (4r)/ (2r²+2r+1) (2r²-2r+1)
S? = 1/4 Σ? ¹? [ 1/ (2r²-2r+1) - 1/ (2r²+2r+1) ]
=> S? = (1/4) * (220/221) = 55/221 = m/n
∴ m+n = 276

Required area (above x-axis)
A? = 2∫? (8/2 - x - √x)dx
= 2 [16 - 16/4 - 8/3*2] = 40/3
and A? = 4 (1/2 k²) = 2k²

∴ 27 * (40/3) = 5 * (2k²)
=> k = 6
for above x-axis.

T? =? C? (x¹/³)? (x? ¹/? )? (5¹/²)? (5? ¹)?
for x¹? : (60-r)/3 - r/5 = 10
=> 180 – 3r – 2r = 60
=> r = 24
k = 3 + exponent of 5 in? C?
= 3 + [60/5] + [60/25] – [24/5] – [24/25] – [36/5] – [36/25]
= 3 + (12+2–4–0–7–1)
= 3+2=5

First we arrange 5 red cubes in a row and assume x? , x? , x? , x? , x? and x? number of blue cubes between them
Here, x? + x? + x? + x? + x? + x? = 11
and x? , x? , x? , x? ≥ 2
so x? + x? + x? + x? + x? + x? = 3
No. of solutions =? C? = 56

|adj (adj (A)| = |A|? ¹² = |A|?
|A| = | (14, 28, -14), (-14, 28), (25, -14, 14) |
= (14)² | (1, 2, -1), (-1, 2), (25/14, -1, 1) |
= (14)² (3–2 (–5)–1 (–1)
|A|? = (14)? => |A| = 14²

Let e? = t then equation reduces to
t³ – 11t – 45/t + 81/t² = 0
=> 2t? – 22t³ + 81t – 45 = 0 . (i)
If roots of
e³? – 11e²? – 45e? + 81/2 = 0
=> α? + α? + α? = ln45 => p = 45

f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99