Class 12th

(a+b+c)² = a²+b²+c²+2 (ab+bc+ca)
1² = a²+b²+c²+2 (2) ⇒ a²+b²+c² = -3.
a²b²+b²c²+c²a² = (ab+bc+ca)² - 2abc (a+b+c) = 2² - 2 (3) (1) = -2.
a? +b? +c? = (a²+b²+c²)² - 2 (a²b²+b²c²+c²a²) = (-3)² - 2 (-2) = 9+4=13.

Δ = |1,1, -1; 1,2, α 2, -1,1| = 1 (2+α)-1 (1-2α)-1 (-1-4) = 2+α+2α-1+5 = 3α+6=0 ⇒ α=-2.
Δ? = |2,1, -1; 1,2, α β, -1,1| = 2 (2+α)-1 (1-αβ)-1 (-1-2β) = 4+2α-1+αβ+1+2β = 4+2α+αβ+2β=0.
4-4-2β+2β=0. This holds.
Δ? = |1,2, -1; 1,1, α 2, β,1| = 1 (1-αβ)-2 (1-2α)-1 (β-2) = 1-αβ-2+4α-β+2 = 1+4α-αβ-β=0.
1-8+2β-β=0 ⇒ -7+β=0 ⇒ β=7.
α+β = -2+7 = 5.

sec y dy/dx = 2sinxcosy.
sec²y dy = 2sinx dx.
tan y = -2cosx + C.
y (0)=0 ⇒ 0=-2+C ⇒ C=2.
tan y = 2-2cosx.
y' = (-2sinx)/sec²y.
5y' (π/2) = 5 (2sin (π/2)/sec² (π/2)
sec²y dy/dx = 2sinx.
y' (π/2)? At x=π/2, tan y = 2. sec²y = 1+tan²y = 5.
5 (2sin (π/2) = 5 (2)=10.

P (x) = a (x-2)² + b (x-2) + c.
lim (x→2) P (x)/sin (x-2) = lim (x→2) P (x)/ (x-2) = P' (2) = 7.
P' (x) = 2a (x-2) + b. P' (2) = b = 7.
P' (x) = 2a.
P (3) = a (1)² + b (1) + c = a+b+c = 9.
Continuity at x=2 means lim f (x) = f (2).
lim (x→2) (a (x-2)²+b (x-2)+c)/ (x-2) = P' (2) = b=7. This is given.
The problem states f (2)=7.
P (x) = (x-2) (ax+b) form used in solution. Let's use this.
lim (x→2) (x-2) (ax+b)/sin (x-2) = lim (x→2) ax+b = 2a+b = 7.
P (3) = (3-2) (3a+b) = 3a+b=9.
Solving: a=2, b=3.
P (x) = (x-2) (2x+3).
P (5) = (5-2) (2*5+3) = 3 * 13 = 39.

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).

(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 =?

Area = ∫? ² (2? - logx)dx = [2? /ln2 - (xlnx-x)]? ²
= (4/ln2 - (2ln2-2) - (2/ln2 - (0-1) = 2/ln2 - 2ln2 + 1.
α=2, β=-2, γ=1.
(α+β-2γ)² = (2-2-2)² = 4.