Class 12th

 Δ = |1 1 1| = 0
|1 2 3|
|3 2 λ|
⇒ 1(2λ - 6) - 1(λ - 9) + 1(-4) = 0
⇒ λ = 1
Δx = |6 1 1|
|10 2 3| = 0
|μ 2 λ|
⇒ 2λ + μ = 16
⇒ μ = 14
μ - λ² = 14 - 1 = 13

Let tan?¹x = θ ⇒ x=tanθ ⇒ sinθ = x/√(1+x²)
y = (x/√(1+x²)) + (1/√(1+x²)) = (x+1)/√(1+x²). This is not f(x).
Let's follow the solution:
y = (x+1)²/(1+x²) - 1 = (2x)/(1+x²) = f(x)
Now dy/dx = (1/2√y) * f'(x) = .
The solution seems to take y as a different function. Let's assume y = (x/(√(1+x²))) + (1/√(1+x²)) - 1. No.
Let's assume y's derivative is taken w.r.t to f(x).
y = -tan?¹x + c
given y(√3)=π/6 ⇒ π/6 = -π/3 + c ⇒ c=π/2
y = cot?¹x. Now y(-√3) = cot?¹(-√3) = 5π/6

f(3)=f(4) ⇒ α=12
f'(x) = (x²-12)/(x(x²+12))
∴ f'(c)=0 ⇒ c=√12
∴ f''(c) = 1/12

2(cos²θ/sin²θ) - 5/sinθ + 4 = 0
(2sinθ - 1)(sinθ - 2) = 0
sinθ = 1/2 only
∴ θ = π/6, 5π/6
↓↓
θ?, θ?
∫(π/6 to 5π/6) cos²3θdθ = ∫(π/6 to 5π/6) (1+cos6θ)/2 dθ = π/3

6.00
b·a = c·a
|a+b-c|² = |a|²+|b|²+|c|²+2(a·b - b·c - a·c)
= 4+16+16+2(a·b - 0 - a·b) = 36
⇒ |a+b-c| = 6

Yes, the phase of the reactant (solid, liquid or gas) actually affects the rate of a reaction. Gases and Liquids tend to move more freely and effectively as compared to solids, which increaes the frequency of particles colliding with each other.

0.50
 y = x²-3x+2, x+y=a, x-y=b
x?=2 x?=1
y? = 4-6+2 = 0 y?=0
(2,0)
(1,0)
b=2
a=1
∴ a/b = 1/2 = 0.5

This is possible because these catalysts have a lower activation energy, which allows a comparatively higher number of particles to react at a faster rate. More number of particles will mean more frequent collisions which speeds up the process.

Probability (at most two machines will be out of service) = (3/4)³ . k
⇒ ?C?(1/4)?(3/4)? + ?C?(1/4)¹(3/4)? + ?C?(1/4)²(3/4)³ = (3/4)³ . k
⇒ 17/8 (3/4)³ = (3/4)³ . k
⇒ k = 17/8

f(x) = (3x²+ax-2-a)e?
f'(x) = (3x²+ax-2-a)e? + e?(6x+a)
= e?(3x²+(a+6)x-2)
∴ x=1 is a critical point
∴ f'(1)=0
∴ 3+a+6-2=0
a = -7
∴ f'(x) = e?(3x²-x-2)
= e?(3x+2)(x-1)

∴ maxima at x = -2/3
∴ minima at x = 1