Class 12th

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.

log (x + √ (x²+1) is an odd function.
f (-x) = log (-x + √ (x²+1) = log (1/ (x+√ (x²+1) = -log (x+√ (x²+1) = -f (x).
Integral of an odd function over a symmetric interval is 0.

xdy - ydx = x³cosxdx
(xdy-ydx)/x² = xcosxdx
d (y/x) = xcosxdx
y/x = ∫xcosxdx = xsinx - ∫sinxdx = xsinx + cosx + c
y = x²sinx + xcosx + cx
y (π) = 0 + π (-1) + cπ = 0 ⇒ c = 1
y = x²sinx + xcosx + x
y (π/2) = (π/2)² (1) + 0 + π/2 = π²/4 + π/2

lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
= (1 + 2ln6) - (0 + 2ln4) = 1 + 2ln (6/4) = 1 + 2ln (3/2).

1st sample: n? =100, x? =15, σ? =3. Σx? = 1500. Σx? ² = n? (σ? ²+x? ²) = 100 (9+225) = 23400.
Whole group: n=250, x? =15.6, σ²=13.44. Σx = 250*15.6 = 3900.
2nd sample: n? =150. Σy? = 3900 - 1500 = 2400. y? = 2400/150 = 16.
Σ (x+y)² = n (σ²+x? ²) = 250 (13.44+15.6²) = 250 (13.44+243.36) = 64200.
Σy? ² = 64200 - 23400 = 40800.
σ? ² = Σy? ²/n? - y? ² = 40800/150 - 16² = 272 - 256 = 16.
σ? = 4.

If (gof)? ¹ exist then gof is a bijective function. For gof to be bijective, f must be one-one and g must be onto.