Class 12th
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New answer posted
2 months agoContributor-Level 10
Given f(x) = e^x sin(x).
Let F(x) = ∫[0 to x] f(t) dt.
By the Fundamental Theorem of Calculus, F'(x) = f(x) = e^x sin(x).
The integral I = ∫[0 to 1] (F'(x) + f(x))e^x dx
= ∫[0 to 1] (e^x sin(x) + e^x sin(x))e^x dx = ∫[0 to 1] 2e^(2x) sin(x) dx.
The text computes I = ∫[0 to 1] 2 sin(x) dx = [-2cos(x)] from 0 to 1 = -2cos(1) - (-2cos(0)) = 2(1 - cos(1)). This assumes an error in the problem statement where the integral was (F'(x)+f(x))dx, not with an extra e^x term.
Using the series expansion for cos(1) = 1 - 1/2! + 1/4! - .
2(1 - cos(1)) = 2(1 - (1 - 1/2 + 1/24 - .)) = 1 - 1/12 + . ≈ 11/12 ≈ 0.916.
The inequality 330/360 < I < 331/360 (i.e., 0.9166 < I < 0.9194) is checked
New answer posted
2 months agoContributor-Level 10
sin?¹(x² + 1/3) + cos?¹(x² - 2/3) = x²
The domains of sin?¹ and cos?¹ require:
-1 ≤ x² + 1/3 ≤ 1 ⇒ -4/3 ≤ x² ≤ 2/3. Since x² ≥ 0, we have 0 ≤ x² ≤ 2/3.
-1 ≤ x² - 2/3 ≤ 1 ⇒ -1/3 ≤ x² ≤ 5/3.
The intersection of these domains is 0 ≤ x² ≤ 2/3.
The range of sin?¹ is [-π/2, π/2] and cos?¹ is [0, π].
Let A = sin?¹(x² + 1/3) and B = cos?¹(x² - 2/3).
The equation is A + B = x².
The LHS, A+B, is a sum of angles, while the RHS, x², is in the range [0, 2/3]. This suggests no solution. The provided solution states that LHS = {π}, which is incorrect. A proper analysis would involve checking if any x in
New answer posted
2 months agoContributor-Level 10
Limit (n→∞) [[r] + [2r] + . + [nr]] / n²
We know that x - 1 < [x] x.
Summing from k=1 to n for [kr]:
Σ(kr - 1) < [kr] (kr)
rΣk - Σ1 < [kr] rk
r(n(n+1)/2) - n < [kr] r(n(n+1)/2)
Divide by n²:
(r/2)(1 + 1/n) - 1/n < ([kr])/n (r/2)(1 + 1/n)
As n → ∞, both the left and right sides approach r/2.
By the Squeeze Theorem, the limit is r/2.
New answer posted
2 months agoContributor-Level 10
cos(x)(3sin(x) + cos(x) + 3)dy = (1 + ysin(x)(3sin(x) + cos(x) + 3))dx
This seems mistyped. A more likely form is:
dy/dx - (sin(x)/(cos(x)))y = 1 / (cos(x)(3sin(x) + cos(x) + 3))
dy/dx - tan(x)y = sec(x) / (3sin(x) + cos(x) + 3)
The integrating factor (I.F.) is:
I.F. = e^∫(-tan(x))dx = e^(ln|cos(x)|) = cos(x).
Multiplying by I.F.:
d(y*cos(x))/dx = 1 / (3sin(x) + cos(x) + 3)
y*cos(x) = ∫ dx / (3sin(x) + cos(x) + 3)
Using Weierstrass substitution, let t = tan(x/2):
sin(x) = 2t/(1+t²), cos(x) = (1-t²)/(1+t²), dx = 2dt/(1+t²)
∫ (2dt/(1+t²)) / (3(2t/(1+t²)) + (1-t²)/(1+t²) + 3)
= ∫ 2dt / (6t + 1 - t² + 3 + 3t²) = ∫ 2dt / (2t² + 6t
New answer posted
2 months agoContributor-Level 9
For the complex K? [Cr (oxalate)? ], the central metal ion is Cr³?
Electronic configuration of Cr (24) is [Ar] 4s¹3d?
Electronic configuration of Cr³? is [Ar] 4s?3d³.
The number of unpaired electrons in Cr³? is 3.
New answer posted
2 months agoContributor-Level 10
Given the function:
f(x) = { x(2 - sin(1/x)), if x ≠ 0
{ 0, if x = 0
For x < 0: f(x) = x(2 - sin(1/x))
For x > 0: f(x) = x(2 - sin(1/x))
The derivative f'(x) for x ≠ 0 is:
f'(x) = 1*(2 - sin(1/x)) + x*(-cos(1/x))*(-1/x²) = 2 - sin(1/x) + (1/x)cos(1/x)
The text calculates the derivative differently:
For x < 0: f'(x) = -2 + sin(1/x) - (1/x)cos(1/x)
For x > 0: f'(x) = 2 - sin(1/x) + (1/x)cos(1/x)
To check if f'(0) is defined, we would need to use the limit definition of the derivative at x=0. As x approaches 0, the term (1/x)cos(1/x) oscillates and does not approach a finite limit. Therefore, f'(0) is undefined.
New answer posted
2 months agoContributor-Level 9
Given K_f = 1.85 K kg mol? ¹ for a solution with molality of 2 m.
ΔT_f = I * K_f * m
3.885 = I * 1.85 * 2
The van't Hoff factor, I = 1.05.
i = 1 + (n-1)α. For an electrolyte dissociating into 2 ions, n=2.
1.05 = 1 + (2-1)α.
The degree of dissociation, α = 0.05 or 50 * 10? ³.
New answer posted
2 months agoContributor-Level 9
For an acidic buffer solution, pH = pKa + log ( [Base]/ [Acid]).
Given pH = 5.74 and pKa = 4.74.
5.74 = 4.74 + log ( [Base]/1).
1 = log ( [Base]).
[Base] = 10M.
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