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New answer posted
7 months agoContributor-Level 10
The general term Tr+1 in the expansion of (a/x + x)^n (assuming from context, as it is not explicitly stated) is:
Tr+1 = nCr * (a/x)^r * x^ (n-r) = nCr * a^r * x^ (n-2r)
(The text shows (a/x^2) leading to x^ (n-3r)
Assuming the term is (x + a/x^2)^n:
Tr+1 = nCr * x^ (n-r) * (a/x^2)^r = nCr * a^r * x^ (n-3r)
T3 = T (2+1) = nC2 * a^2 * x^ (n-6)
T4 = T (3+1) = nC3 * a^3 * x^ (n-9)
T5 = T (4+1) = nC4 * a^4 * x^ (n-12)
The ratio of the coefficients is given:
(nC2 * a^2) / (nC3 * a^3) = 12/8 = 3/2
(n (n-1)/2) * a^2) / (n (n-1) (n-2)/6) * a^3) = 3/2
(3 / (n-2) * (1/a) = 3/2 => a (n-2) = 2 (i)
(nC3 * a^3) / (nC4 * a^4) = 8/3
(n (n-1) (n-2)/6) * a^3)
New answer posted
7 months agoContributor-Level 9
In Hoffmann bromamide reaction, hypobromite ion react with amide and in this reaction carbonyl group is lost as CO? ²? in form of Na? CO?
New answer posted
7 months agoContributor-Level 10
To find the values of a and b, we evaluate the left-hand limit (LHL) and right-hand limit (RHL) at x=0 and equate them.
LHL: lim (x→0) sin (a+3)/2 * x / x = (a+3)/2. The full limit evaluates to (a+3)/2. So, (a+3)/2 = b.
This gives the relation a - 2b + 3 = 0 — (I)
RHL: lim (x→0) [√ (x+bx³) - √x] / (bx²).
Rationalize the numerator:
lim (x→0) [ (x+bx³) - x] / [bx² (√ (x+bx³) + √x)]
= lim (x→0) bx³ / [bx² (√x (√ (1+bx²) + √x)]
= lim (x→0) x / [√x (√ (1+bx²) + 1)] = lim (x→0) √x / (√ (1+bx²) + 1) = 0/2 = 0.
So, b = 0.
Substituting b=0 into equation (I): a - 2 (0) + 3 = 0 ⇒ a = -3.
New answer posted
7 months agoContributor-Level 10
Let X1, ., X2n be the first 2n observations and Y1, ., Yn be the last n observations.
Given:
ΣXi / 2n = 6 => ΣXi = 12n (i)
ΣYi / n = 3 => ΣYi = 3n (ii)
Combined mean: (ΣXi + ΣYi) / 3n = 5 => ΣXi + ΣYi = 15n. This is consistent with (i) and (ii).
Combined variance: (ΣXi^2 + ΣYi^2) / 3n - (mean)^2 = 4
(ΣXi^2 + ΣYi^2) / 3n - 5^2 = 4
ΣXi^2 + ΣYi^2 = (4 + 25) * 3n = 87n.
New observations are Xi + 1 and Yi - 1.
New mean = (Σ (Xi + 1) + Σ (Yi - 1) / 3n = (ΣXi + 2n + ΣYi - n) / 3n = (15n + n) / 3n = 16n / 3n = 16/3.
New variance k:
k = (Σ (Xi + 1)^2 + Σ (Yi - 1)^2) / 3n - (new mean)^2
k = (Σ (Xi^2 + 2Xi + 1) + Σ (Yi^2 - 2Yi + 1) / 3
New answer posted
7 months agoContributor-Level 10
The differential equation is rearranged to dt/dx - xt = -e? ²/², where t = 1/ (y+1).
This is a linear first-order differential equation. The integrating factor (I.F.) is e^ (∫-x dx) = e? ²/².
The solution is t * (I.F.) = ∫ Q (x) * (I.F.) dx + c.
t * e? ²/² = ∫ -e? ²/² * e? ²/² dx + c = ∫ -1 dx = -x + c.
Substituting t = 1/ (y+1) back: e? ²/² / (y+1) = -x + c.
Using the initial condition y (2) = 0:
e? ²/ (0+1) = -2 + c ⇒ c = e? ² + 2.
The solution is e? ²/² / (y+1) = 2 + e? ² - x.
New answer posted
7 months agoContributor-Level 10
Given f (x) = ax^2 + bx + c.
f (-1) = a - b + c = 2
f' (x) = 2ax + b, so f' (-1) = -2a + b = 1
f' (x) = 2a, so f' (-1) = 2a = 1/2
From 2a = 1/2, we get a = 1/4.
Substituting a into -2a + b = 1: -2 (1/4) + b = 1 => -1/2 + b = 1 => b = 3/2.
Substituting a and b into a - b + c = 2: 1/4 - 3/2 + c = 2 => -5/4 + c = 2 => c = 13/4.
So, f (x) = (1/4)x^2 + (3/2)x + 13/4 = (1/4) (x^2 + 6x + 13).
We need to find f (1):
f (1) = (1/4) (1^2 + 6 (1) + 13) = (1/4) (1 + 6 + 13) = (1/4) (20) = 5.
New answer posted
7 months agoContributor-Level 10
PES University conducts PESSAT for admission into BBA programme. The minimum eligibility for the programme is passing Class 12 with atleast 50% of aggregate marks. Scoring 60% in Class 12 makes the candidate eligible for admission into BBA. However, the candidate must qualify PESSAT and the subsequent personal interview round to get a confirmed seat in the course at the university.
New answer posted
7 months agoContributor-Level 9
CuSO? 5H? O is blue vitriol which is represented as [Cu (H? O)? ]. H? O.SO? Here, secondary valencies are 4 H? O and 1H? O molecule outside coordination sphere is involved in hydrogen bonding.
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