Class 12th

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)

System of equations can be written as
[2 3 6; 1 2 a; 3 5 9] [x; y; z] = [8; 5; b]
R? →R? -1/2R? , R? →R? -3/2R?
. the system will have no solution if 3-a=0 and b-13≠0.
i.e. for a=3 & b≠13.

e? - e? - 2e³? - 12e²? + e? + 1 = 0
e³? - 2 + e? ³? - e? [e³? + 12e? - 1] = 0
Let y=e? y? -y? -2y³-12y²+y+1=0.
The solution breaks the equation into (e³? - 4e? - 1) (e³? - 1 + 3e? ) = 0
Either e³? = 4e? + 1 (One Solution)
OR e³? = 1 - 3e? (One Solution)
∴ the equation has total 2 solutions.

Three balls can be given to B? in? C? ways. Now remaining 6 balls can be distributed into 3 boxes in 3? ways.
Total no. of favourable events =? C? * 3? = 28 * 3?
Total no. of events = 9 balls distributed into 4 boxes in 4? ways.
probability = 28 * 3? /4? = 28/9 * (3/4)? ⇒ k = 28/9
k ∈ |x-3|<1

{ (x, y) ∈ R*R, x ≥ 0, 2x² ≤ y ≤ 4 − 2x}.
Required area = ∫? ¹ (4 - 2x - 2x²)dx
= [4x - x² - (2/3)x³]? ¹ = 4-1-2/3 = 7/3

This is the type of reaction which bheavies like a first order reaction inspite of being a higher order reaction (second or third). This happens due to a particular reactant being present in an excessive quantity thorughout the chemical reaction.

These are the important highlights of integrated rate equations:

• predict the rate of a reaction
• calculate order of reaction (zero, first, second)
• understand mechanism of a reaction
• compute half life
• calculate the value of k

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4

 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