Class 11th

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

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

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

| p | q | p^q | ¬ (p^q) | (¬ (p^q)vq | pvq | p? q | p? (pvq) |
|-|-|-|-|-|-|-|-|
| T | T | T | F | T | T | T | T |
| T | F | F | T | T | T | F | T |
| F | T | F | T | T | T | T | T |
| F | F | F | T | T | F | T | T |
Tautology, Tautology

x²/a² + y²/b² = 1
(-4/√5)²)/a² + (2/√5)²)/b² = 1
=> 32/a² + 9/b² = 1
=> 32/ (5a²) + 9/b² = 1 . (i)
From (i)
6/b² + 9/b² = 1 => b²=15 & a²=16
a²+b² = 15+16=31