Class 12th

|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

I = ∫ (e²? + 2e? - e? - 1)e^ (e? +e? ) dx
I = ∫ (e²? + e? - 1)e^ (e? +e? ) dx + ∫ (e? - e? )e^ (e? +e? ) dx
I = ∫ (e? + 1 - e? )e^ (e? +e? ) dx + e^ (e? +e? )
(e? - e? + 1)dx = du
I = e^ (e? +e? ) + e^ (e? +e? ) = e^ (e? +e? ) (e? + 1) then g (x) = e? + 1
g (0) = 2

Here D = | 2 -4 λ |
| 1 -6 1 | = (λ-3) (3λ+2)
| λ -10 4 |
D = 0 ⇒ λ = 3, -2/3

Let A (α, 0,0), B (0, β, 0), C (0,0, γ), then the centroid is G (α/3, β/3, γ/3) = (1,1,2).
α = 3, β = 3, γ = 6
∴ Equation of plane is x/α + y/β + z/γ = 1
⇒ x/3 + y/3 + z/6 = 1
⇒ 2x + 2y + z = 6
∴ Required line passing through G (1,1,2) and normal to the plane is (x-1)/2 = (y-1)/2 = (z-2)/1.

A = [cosθ, sinθ], [-sinθ, cosθ]
A² = [cos2θ, sin2θ], [-sin2θ, cos2θ]
⇒ A? = [cos4θ, sin4θ], [-sin4θ, cos4θ]
B = [cos4θ, sin4θ], [-sin4θ, cos4θ] + [cosθ, sinθ], [-sinθ, cosθ]
= [cos4θ + cosθ, sin4θ + sinθ], [- (sin4θ + sinθ), cos4θ + cosθ]
det (B) = (cos4θ + cosθ)² + (sin4θ + sinθ)²
= (cos²4θ + sin²4θ) + (cos²θ + sin²θ) + 2 (cos4θcosθ + sin4θsinθ)
= 1 + 1 + 2cos (4θ - θ)
= 2 + 2cos3θ
Given 3θ = 3π/5
|B| = 2 + 2cos (3π/5)
= 2 + 2 (- (√5-1)/4) = 2 - (√5-1)/2 = (4-√5+1)/2 = (5-√5)/2 ∈ (1,2)

Image by objective must be at focus of eye piece.
1/v - 1/u = 1/f?
1/5 - 1/u = 1 ; u = -5/4 cm
Hence, N = 50.

We know, P = √2Em and λ = h/√2Em
For 1st maxima, 2nd sinθ = λ
Put (1) and get the answer
E = 50.47eV.
B = µ? I (200)/2R? R? = 20 cm; R? = 1 cm
e = -dΦ/dt = π (R? )² (500) dB/dt
= π (R? )² (500) (µ? I/2R? ²) (200) (10t-2)
= π (1/10000) (500) (200) (µ? ) (8) (5)/ (2)
= 16π² x 10 x 5 x 10?
= 800π² x 10?
e = 0.7887mv
4/x = 0.8; x = 5.