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New answer posted
a year agoContributor-Level 10
We are given bounds for a function f (t) on two intervals and need to find the range of g (3) = ∫? ³ f (t) dt.
We split the integral: g (3) = ∫? ¹ f (t)dt + ∫? ³ f (t)dt.
For the first interval t ∈ [0, 1], we have 1/3 ≤ f (t) ≤ 1. Integrating from 0 to 1 gives:
∫? ¹ (1/3) dt ≤ ∫? ¹ f (t)dt ≤ ∫? ¹ 1 dt ⇒ 1/3 ≤ ∫? ¹ f (t)dt ≤ 1.
For the second interval t ∈ (1, 3], we have 0 ≤ f (t) ≤ 1/2. Integrating from 1 to 3 gives:
∫? ³ 0 dt ≤ ∫? ³ f (t)dt ≤ ∫? ³ (1/2) dt ⇒ 0 ≤ ∫? ³ f (t)dt ≤ (1/2) (3-1) = 1.
Adding the inequalities for the two parts of the integral:
1/3 + 0 ≤ g (3) ≤ 1
New answer posted
a year agoContributor-Level 10
For a system of linear homogeneous equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.
Δ = | 4 λ 2 |
| 2 -1 | = 0
| μ 2 3 |
To simplify, perform the row operation R? → R? - 2R? :
Δ = | 0 λ+2 0 |
| 2 -1 | = 0
| μ 2 3 |
Expand the determinant along the first row:
- (λ+2) * det (| 2 1 |, | μ 3 |) = 0.
- (λ+2) (2*3 - 1*μ) = 0.
(λ+2) (μ-6) = 0.
This implies that either λ+2 = 0 or μ-6 = 0.
So, the conditions are λ = -2 (for any μ) or μ = 6 (for any λ).
New answer posted
a year 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
a year agoContributor-Level 10
The equation of the tangent to the ellipse x²/27 + y² = 1 at the point (3√3 cosθ, sinθ) is:
x (3√3 cosθ)/27 + y (sinθ)/1 = 1 ⇒ x/ (3√3) cosθ + y sinθ = 1.
To find the intercepts on the axes:
x-intercept (set y=0): x = 3√3 / cosθ = 3√3 secθ.
y-intercept (set x=0): y = 1 / sinθ = cosecθ.
The sum of the intercepts is z (θ) = 3√3 secθ + cosecθ.
To find the minimum value of z, we differentiate with respect to θ and set it to zero:
dz/dθ = 3√3 secθ tanθ - cosecθ cotθ = 0.
3√3 (sinθ/cos²θ) = cosθ/sin²θ.
3√3 sin³θ = cos³θ ⇒ tan³θ = 1/ (3√3).
⇒ tanθ = 1/√3.
Since θ ∈ (0, π/2), the solution
New answer posted
a year 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
a year 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
a year agoContributor-Level 10
The problem provides non-standard formulas for sums S? and S? :
S? = n [a + (2n-1)d]
S? = 2n [a + (4n-1)d]
We are given S? - S? = 1000.
S? - S? = 2n [a + (4n-1)d] - n [a + (2n-1)d]
= (2na + 8n²d - 2nd) - (na + 2n²d - nd)
= na + 6n²d - and = n [a + (6n-1)d].
So, n [a + (6n-1)d] = 1000.
We need to find S? n. Assuming the pattern S? n = kn [a + (2kn-1)d], then S? n would be 3n [a + (6n-1)d].
S? n = 3 * (n [a + (6n-1)d]) = 3 * 1000 = 3000.
New answer posted
a year 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
a year 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.
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