Maths

Using L'Hopital Rule:
lim (x→2) (2xf (2) - 4f' (x)/1 = 2 (2)f (2) - 4f' (2) = 4 (4) - 4 (1) = 12.

I = ∫? π/? π/? dx/ (1+e^ (xcosx) (sin? x+cos? x). Using ∫? f (x)dx = ∫? f (a+b-x)dx. a+b=0.
I = ∫? π/? π/? dx/ (1+e? ) (sin? x+cos? x) = ∫? π/? π/? e? dx/ (e? +1) (sin? x+cos? x).
2I = ∫? π/? π/? dx/ (sin? x+cos? x) = 2∫? π/? dx/ (sin? x+cos? x).
I = ∫? π/? sec? xdx/ (tan? x+1). Let t=tanx.
I = ∫? ¹ (t²+1)dt/ (t? +1) = ∫? ¹ (1+1/t²)dt/ (t²-√2t+1) (t²+√2t+1). No, this is hard.
I = ∫? ¹ (1+1/t²)dt/ (t-1/t)²+2). Let u=t-1/t. I = ∫ du/ (u²+2) = (1/√2)tan? ¹ (u/√2).
= π/ (2√2).

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.

(2? -2) is multiple of 3.
If n is odd, 2? ≡ (-1)? =-1 (mod 3). 2? -2 ≡ -1-2 = -3 ≡ 0 (mod 3).
If n is even, 2? ≡ (-1)? =1 (mod 3). 2? -2 ≡ 1-2 = -1 (mod 3).
So n must be odd.
2-digit numbers are 10-99 (90 numbers).
Odd numbers are 11,13, .,99. Number of terms = (99-11)/2 + 1 = 45.
Probability = 45/90 = 1/2.

For x>2, f (x) = ∫? ¹ (5+1-t)dt + ∫? ² (5+t-1)dt + ∫? (5+t-1)dt
= ∫? ¹ (6-t)dt + ∫? ² (4+t)dt + ∫? (4+t)dt
= [6t-t²/2]? ¹ + [4t+t²/2]? ² + [4t+t²/2]?
= (6-1/2) + (8+2 - (4+1/2) + (4x+x²/2 - (8+2)
= 5.5 + 5.5 + 4x+x²/2 - 10 = 4x+x²/2 + 1.
f (2? ) = 8+2+1 = 11. f (2? ) = 5 (2)+1 = 11. Continuous.
f' (x) = 4+x for x>2. f' (2? ) = 6.
For x<2, f' (x)=5. f' (2? )=5.
Not differentiable at x=2.

Circle: (x-1)² + (y-3)² = 4. C= (1,3), r=2.
Length of tangent from P (-1,1) is L = √ (-1)²+1²-2 (-1)-6 (1)+6) = √ (1+1+2-6+6) = √4 = 2.

Area of quadrilateral PACB = 2 * Area (PAC) = 2 * (1/2 * L * r) = 2*2=4.
AB is chord of contact. T=0 => -x+y- (x-1)-3 (y-1)+6=0 => -2x-2y+10=0 => x+y=5.
Distance of C from AB = |1+3-5|/√2 = 1/√2.
Length of AB = 2√ (r²-d²) = 2√ (4-1/2) = 2√ (7/2) = √14.
Area of ABD =?

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.