Class 12th

2x-y+2z=2
x-2y+λz=-4
x+λy+z=4
For no solution:
D=|2, -1,2; 1, -2, λ 1,1,1|=0
⇒ 2 (-2-λ²)+1 (1-λ)+2 (λ+2)=0
⇒ -2λ²+λ+1=0
⇒ λ=1, -1/2
D? =|2, -1,2; -4, -2, λ 4,1,1|
=2 (-2-λ)+1 (-4-4λ)+2 (-4+8)
=2 (1+λ) which is not equal to zero for λ=1, -1/2

Based on theory.

The vapour pressure of solution will be less than vapour pressure of pure solvent, so some vapour molecules will get condensed to maintain new equilibrium.

Energy Density = 1/2 B²/µ?
B = √ (2 * µ? * Energy density)
B = √ (2 * 4π * 10? * 1.02 * 10? ) = 160 * 10? = 160nT

f (x) = {ae? +be? , -1For continuity at x=1
Lim (x→1? )f (x) = Lim (x→1? )f (x)
⇒ ae+be? ¹=c ⇒ b=ce-ae²
For continuity at x=3
Lim (x→3? )f (x) = Lim (x→3? )f (x)
⇒ 9c=9a+6c ⇒ c=3a
f' (0)+f' (2)=e
(ae? -be? ) at x=0 + (2cx) at x=2 = e
⇒ a-b+4c=e
From (1), (2) and (3)
a-3ae+ae²+12a=e
⇒ a (e²+13-3e)=e
⇒ a=e/ (e²-3e+13)

Since p (x) has relative extreme at x=1 and 2
so p' (x)=0 at x=1 and 2
⇒ p' (x)=A (x-1) (x-2)
⇒ p (x)=∫A (x²-3x+2)dx
p (x)=A (x³/3 - 3x²/2 + 2x)+C
P (1)=8
From (1)
8=A (1/3-3/2+2)+C
⇒ 8=5A/6+C ⇒ 48=5A+6C
P (2)=4
⇒ 4=A (8/3-6+4)+C
⇒ 4=-2A/3+C ⇒ 12=-2A+3C
From 3 and 4, C=-12
So P (0)=C=-12

f (x)=sin (|x|+5)/ (x²+1)
For domain:
-1 ≤ (|x|+5)/ (x²+1) ≤ 1
Since |x|+5 and x²+1 is always positive
So (|x|+5)/ (x²+1) ≥ 0 ∀x∈R
So for domain:
(|x|+5)/ (x²+1) ≤ 1
⇒ |x|+5 ≤ x²+1
⇒ 0 ≤ x²-|x|-4
⇒ 0 ≤ (|x|- (1+√17)/2) (|x|- (1-√17)/2)
⇒ |x| ≥ (1+√17)/2 or |x|≤ (1-√17)/2 (Rejected)
⇒ x∈ (-∞, - (1+√17)/2] ∪ [ (1+√17)/2, ∞)
So, a = (1+√17)/2

Bredig's Arc Method is used for preparation of colloidal sol's of less reactive metal like Au, Ag, Pt.

(2+sin x)/ (y+1) dy/dx = -cosx, y>0
⇒ dy/ (y+1) = -cosx/ (2+sinx) dx
By integrating both sides:
ln|y+1| = -ln|2+sinx|+lnK
⇒ y+1 = K/ (2+sinx) (y+1>0)
⇒ y (x) = K/ (2+sinx) - 1
Given y (0)=1 ⇒ 1=K/2-1 ⇒ K=4
So, y (x)=4/ (2+sinx)-1
a=y (π)=1
b=dy/dx|x=π = -cosx/ (y (x)+1)|x=π = 1
So, (a, b)= (1,1)

Slope of tangent to the curve y=x+siny
at (a, b) is (a-2)/ (b-3)=1
⇒ dy/dx|x=a = 1
dy/dx = 1+cos y dy/dx (from equation of curve)
⇒ dy/dx|x=a and y=b = 1+cosb=1
⇒ cosb=0
⇒ sinb=±1
Now, from curve y=x+siny
b=a+sinb
⇒ |b-a|=|sinb|=1