Class 11th

C= (2,3), O= (0,0). r = OC = √13.
Slope of OC = 3/2. Slope of PQ = -2/3.
Let P= (x, y). Vector CP = (x-2, y-3). Vector OC = (2,3).
CP.OC=0 ⇒ 2 (x-2)+3 (y-3)=0 ⇒ 2x+3y=13.
Also (x-2)²+ (y-3)²=13.
From 2x=13-3y, x= (13-3y)/2.
(13-3y)/2-2)²+ (y-3)²=13 ⇒ (9-3y)/2)²+ (y-3)²=13
(9 (y-3)²/4) + (y-3)² = 13 ⇒ (13/4) (y-3)²=13 ⇒ (y-3)²=4 ⇒ y-3=±2.
y=5 or y=1.
If y=5, x=-1. P= (-1,5).
If y=1, x=5. Q= (5,1).

? P? =? P? ⇒ n!/ (n-r)! = n!/ (n-r-1)! ⇒ n-r=1.
? C? =? C? ⇒ r + (r-1) = n ⇒ n = 2r-1.
Substitute n: (2r-1)-r=1 ⇒ r=2.

T? = ¹? C? (xsinα)¹? (acosα/x)? = ¹? C? x¹? ²? sin¹? α a? cos? α
For term independent of x, 10-2r=0 ⇒ r=5.
T? = ¹? C? sin? α a? cos? α = ¹? C? (sin2α/2)? a?
This is the greatest when sin2α=1.
the greatest value = ¹? C? (a/2)? = 10!/ (5!)².
¹? C? = 252.
252 (a/2)? = 252.
(a/2)? = 1 ⇒ a=2.

Coeff of x? in (x²+1/bx)¹¹: T? = ¹¹C? (x²)¹¹? (1/bx)? = ¹¹C? x²²? ³? b?
22-3r=7 ⇒ 3r=15 ⇒ r=5. Coeff is ¹¹C? /b?
Coeff of x? in (x-1/bx²)¹¹: T? = ¹¹C? (x)¹¹? (-1/bx²)? = ¹¹C? x¹¹? ³? (-1)? b?
11-3r=-7 ⇒ 3r=18 ⇒ r=6. Coeff is ¹¹C? (-1)? /b? = ¹¹C? /b?
¹¹C? /b? = ¹¹C? /b? ⇒ b = ¹¹C? /¹¹C? = (11-5+1)/5 = 7/5. This differs from the solution.
Let's check the exponents again.
x? : 22-2r-r=7 => 22-3r=7 => 3r=15 => r=5. Correct.
x? : 11-r-2r=-7 => 11-3r=-7 => 3r=18 => r=6. Correct.
The given solution has b=1.

Reflected point of (2,1) about y-axis is (-2,1).

Reflected ray passes through (-2,1) and (5,3).
Equation: (y-1)/ (x+2) = (3-1)/ (5+2) = 2/7 ⇒ 2x - 7y + 11 = 0.
This is one directrix. Let the other be 2x - 7y + α = 0.
Distance between directrices = 2a/e = |11-α|/√53.
Distance from focus to directrix = a/e - ae = 8/√53.
a/e (1-e²) = 8/√53.
e=1/3. a/e (8/9) = 8/√53 ⇒ a/e = 9/√53.
2a/e = 18/√53 = |11-α|/√53.
|11-α| = 18.
11-α = 18 ⇒ α = -7.
11-α = -18 ⇒ α = 29.
Other directrix: 2x-7y-7=0 or 2x-7y+29=0.

PV = nRT
n = PV/RT = (1 atm * 4*10? L) / (0.083 LatmK? ¹mol? ¹ * 300 K) = 1.6 * 10? mol
Mass = n * Molar Mass = 1.6 * 10? mol * 16 g/mol = 25.6 * 10? g ≈ 26 * 10? g

cotθ = (1+cos2θ)/sin2θ
cot (π/24) = (1+cos (π/12)/sin (π/12)
cos (π/12) = cos (15°) = cos (45-30) = (√3+1)/2√2
sin (π/12) = sin (15°) = sin (45-30) = (√3-1)/2√2
cot (π/24) = (1+ (√3+1)/2√2)/ (√3-1)/2√2) = (2√2+√3+1)/ (√3-1)
= (2√2+√3+1) (√3+1)/2 = (2√6+2√2+3+√3+√3+1)/2
= √6 + √2 + √3 + 2

Energy per second = 1000 J / 10 s = 100 J/s
Energy of one photon E = hc/λ = (6.626*10? ³? * 3*10? ) / (400*10? ) = 4.965 * 10? ¹? J
Number of electrons ejected = Total energy / Energy per photon = 100 / (4.965 * 10? ¹? ) = 20.14 * 10¹? ≈ 2 * 10²?

Truth table analysis shows that (P ∨ Q) ∧ (¬P) is equivalent to Q ∧ ¬P.
Then (Q ∧ ¬P) ⇒ Q. This is a tautology.
The provided solution seems to have an error.
Let's check the options. (P ∨ Q) is a tautology. (P ∧ ¬Q) is a contradiction.
~ (P ⇒ Q) ⇔ P ∧ ¬Q is true.