Class 12th

Equation x²/5 + y²/4 = 1 then P (√5cosθ, 2sinθ)
(PQ)² = 5cos²θ + 4 (sinθ+2)² = cos²θ + 16sinθ + 20
= -sin²θ + 16sinθ + 21
= 85 - (sinθ - 8)²
∴ (PQ)²max = 85 - 49 = 36
? (sinθ - 8)² ∈

C? → C? - C?
f (θ) = | -sin²θ -1 |
| -cos²θ -1 |
| 12 -2 -2|
= 4 (cos²θ - sin²θ) = 4 (cos2θ), θ ∈ [π/4, π/2]
f (θ)max = M = 0
f (θ)min = m = -4

f (x) is differentiable then will also continuous then f (π) = -1, f (π? ) = -k?
k? = 1
Now f' (x) = { 2k? (x-π) if x≤π
{ -k? sinx if x>π
then f' (π? ) = f' (π? ) = 0
f' (x) = { 2k? if x≤π
{ -k? cosx if x>π
then 2k? =k?
k? = 1/2

|x + y|² = |x|²
(x+y)· (x+y) = x·x
|x|² + 2x·y + |y|² = |x|²
|y|² + 2x·y = 0 (1)
and (2x + λy)·y = 0
2x·y + λ|y|² = 0 (2)
From (1), 2x·y = -|y|².
Substitute into (2):
-|y|² + λ|y|² = 0
(λ-1)|y|² = 0
Assuming y is a non-zero vector, |y|² ≠ 0, therefore λ=1.

R (-1+2r, 3-2r, -r)
dr's of PR are (2 - 2r, -1+2r, -3+r)
Then 2 (2-2r) + 2 (-1+2r) + 1 (3-r) = 0
9-9r = 0 ⇒ r = 1
R (1,1, -1)
then a+1=2, b+2=2, c-3=-2
a=1, b=0, c=1
∴ a+b+c = 2

|λ-1 3λ+1 2λ|
|λ-1 4λ-2 λ+3| = 0
|2 3λ+1 3 (λ-1)|
R? → R? - R? and R? → R? - R? (from a similar matrix setup, applying operations to simplify)
The provided solution uses a slightly different matrix but let's follow the subsequent steps.
A different matrix from the image is used in the calculation:
|λ-1 3λ+1 2λ|
|0 λ-3 -λ+3|
|3-λ 0 λ-3 |
C? → C? + C?
|3λ-1 3λ+1 2λ |
|3-λ λ-3-λ | = 0
|0 λ-3 |
⇒ (λ-3) [ (3λ-1) (λ-3) - (3λ+1) (3-λ)] = 0
⇒ (λ-3) [ (λ-3) (3λ-1) + (λ-3) (3λ+1)] = 0
⇒ (λ-3)² [3λ-1 + 3λ+1] = 0
⇒ (λ-3)² [6λ] = 0 ⇒ λ = 0, 3
Sum of values of λ = 3

P (A∪B∪C) = P (A) + P (B) + P (C) – P (A∩B) – P (B∩C) – P (C∩A) + P (A∩B∩C)
Given relations lead to: α = 1.4 – P (A∩B) – β ⇒ α + β = 1.4 - P (A∩B)
Again, from P (A∪B) = P (A) + P (B) – P (A∩B), and given values, it is found that P (A∩B) = 0.2.
From (1) and (2), α = 1.2 – β.
Now given 0.85 ≤ α ≤ 0.95
⇒ 0.85 ≤ 1.2 – β ≤ 0.95
⇒ -0.35 ≤ -β ≤ -0.25
⇒ 0.25 ≤ β ≤ 0.35, so β ∈ [0.25, 0.35]

I = ∫ from -π/2 to π/2 (1 / (1+e^ (sin x) dx
I = ∫ from -π/2 to π/2 (e^ (sin x) / (1+e^ (sin x) dx
2I = ∫ from -π/2 to π/2 1dx ⇒ I = 1/2 ∫ from -π/2 to π/2 dx
I = 1/2 [x] from -π/2 to π/2 ⇒ I = π/2