Class 12th

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.

dy/dx = e^ (3x+4y) = e³? e?
e? dy = e³? dx
∫e? dy = ∫e³? dx
-e? /4 = e³? /3 + C
y (0)=0 ⇒ -1/4 = 1/3 + C ⇒ C = -7/12.
-e? /4 = e³? /3 - 7/12
e? = (7 - 4e³? )/3
y = (-1/4)ln (7-4e³? )/3)
x = -2/3 ln2 = ln (2? ²/³) = ln (1/4¹/³)
e³? = e^ (ln (1/4) = 1/4.
y = (-1/4)ln (7-1)/3) = (-1/4)ln2.
α = -1/4.

CrCl? ·3NH? ·3H? O gives 3 moles of AgCl precipitate. This means all three Cl? are outside the coordination sphere.
The complex is [Cr (NH? )? (H? O)? ]Cl?
The 3 chloride ions satisfy only primary valency.
secondary valency satisfied by chloride ion = 0

Mean = Σx? p? = 0 (1/2) + Σ? ∞ j (1/3)? = (1/3)/ (1-1/3)² = (1/3)/ (4/9) = 3/4
P (X is positive and even) = P (X=2) + P (X=4) + .
= (1/3)² + (1/3)? + . = (1/9)/ (1-1/9) = (1/9)/ (8/9) = 1/8

A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.

L.H.L = lim (x→0? ) (1 + |sin x|)³? /|sin x| = lim (h→0) (1 + sinh)³? /sinh = e³?
R.H.L = lim (x→0? ) e^ (cot 4x / cot 2x) = lim (x→0? ) e^ (tan 2x / tan 4x) = e¹/².
f (0) = b.
For continuity, e³? = e¹/² = b.
3a = 1/2 ⇒ a = 1/6. b = e¹/².
6a + b² = 6 (1/6) + (e¹/²)² = 1 + e