Class 12th

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

Since spin only magnetic moment is 4.90BM so number of unpaired electrons must be 4, so If the complex is octahedral, then it has to be high spin complex with configuration t? g? e²g¹ in that case
CFSE = 4* (-0.4Δ? ) + 2*0.6Δ? = -0.4Δ?
If the complex is tetrahedral then its electronic configuration will be e? ², t? g² and CFSE will be = 3* (-0.6Δ? ) + 3* (0.4Δ? ) = -0.6Δ?

|A|≠0
For (P): A≠I?
So, A = [1 0; 0 1] or [1; 0 1] or [1 0; 1]
or [1; 1 0]
So (P) is false.
A = [1 0; 1 0] or [1; 0 1] or [1 0; 1]
⇒ tr (A)=2
⇒ Q is true

Two points on the line (say) x/3 = y/2, z=1 are (0,0,1) and (3,2,1)
So dr's of the line is (3,2,0)
Line passing through (1,2,1), parallel to L and coplanar with given plane is r = i+2j+k + t (3i+2j), t∈R (-2,0,1) satisfies the line (for t=-1)
⇒ (-2,0,1) lies on given plane.
Answer of the question is (B)
We can check other options by finding equation of plane
Equation plane: |x-1, y-2, z-1; 1+2, 2-0, 1-1; 2+2, 1-0, 2-1| = 0
⇒ 2 (x-1)-3 (y-2)-5 (z-1)=0
⇒ 2x-3y-5z+9=0