Class 12th
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New answer posted
10 months agoContributor-Level 10
z = √ [R² + (X? - X? )²] = √ [6² + (4-10)²] = 6√2 Ω
Power factor = cosφ = R/z = 6/ (6√2) = 1/√2
New answer posted
10 months agoContributor-Level 10
Given matrices A = [[a, b], [c, d]] and B = [[α], [β]] where B ≠ [[0], [0]].
The product AB is:
AB = [[a, b], [c, d]] * [[α], [β]] = [[aα + bβ], [cα + dβ]]
From the problem statement AB = B, we have:
aα + bβ = α (i)
cα + dβ = β (ii)
Rearranging these equations:
(a - 1)α + bβ = 0
cα + (d - 1)β = 0
For this system of linear equations to have a non-trivial solution (since B is not the zero matrix), the determinant of the coefficient matrix must be zero.
det([[a-1, b], [c, d-1]]) = 0
(a - 1)(d - 1) - bc = 0
ad - a - d + 1 - bc = 0
ad - bc = a + d - 1
The provided text jumps to the conclusion ad - bc = 2020.
New answer posted
10 months agoContributor-Level 10
Given the integral In = ∫(log|x|)^n / x^19 dx.
Let t = log|x|, which implies x = e^t and dx = e^t dt.
The integral becomes:
In = ∫ e^(-20t) * t^n dt
Using integration by parts, where u = t^n and dv = e^(-20t) dt:
In = [t^n * e^(-20t) / -20] - ∫ n*t^(n-1) * e^(-20t) / -20 dt
In = e^(-20) / -20 - (n / -20) * In-1
20 * In = -e^(-20) + n * In-1
For n = 10: 20 * I10 = e^20 - 10 * I9 (Note: There seems to be a sign inconsistency in the original document's derivation vs. standard integration by parts, the document states e^20 instead of -e^(-20) and proceeds with e^20).
For n = 9: 20 * I9 = e^20 - 9 * I8
From these two equations, we can express e
New answer posted
10 months agoContributor-Level 10
I? = I? + I? ⇒ I? /I? = 1 + I? /I? ⇒ 1/α = 1 + 1/β = (β+1)/β ⇒ α = β/ (1+β)
⇒ 1/β = 1/α - 1 = (1-α)/α ⇒ β = α/ (1-α)
New answer posted
10 months agoContributor-Level 10
Find the number of solutions for 2tan(x) = π/2 - x in [0, 2π].
This is equivalent to finding the number of intersection points of the graphs y = tan(x) and y = (π/4) - x/2.
Let's sketch the graphs:
y = tan(x) has vertical asymptotes at x = π/2, 3π/2.
y = (π/4) - x/2 is a straight line with a negative slope.
At x=0, y=π/4.
At x=π/2, y=0.
At x=π, y=-π/4.
At x=2π, y=-3π/4.
By observing the graphs, there will be one intersection in (0, π/2), one in (π/2, 3π/2), and one in (3π/2, 2π].
Total number of solutions is 3.
New answer posted
10 months ago
Contributor-Level 10
θ? > I = θ ⇒ sinθ? > sinθ ⇒ 1/μ > sinθ
⇒ μ < 1/sin
Note: If we assume θ = 45° ⇒ μ < 1.414, then red colour light ray will come out from face PR of prism.
New answer posted
10 months agoContributor-Level 10
Given that x, y, z are in A.P., so 2y = x + z.
The determinant is:
| 3 4√2 x |
| 4 5√2 y | = 0
| 5 k z |
Apply the operation R? → R? + R? - 2R?:
The first row becomes:
(3 + 5 - 24) (4√2 + k - 25√2) (x + z - 2y)
= 0 (k - 6√2) (0)
So the determinant becomes:
| 0 k-6√2 0 |
| 4 5√2 y | = 0
| 5 k z |
Expanding along the first row:
-(k - 6√2)(4z - 5y) = 0.
This implies k - 6√2 = 0 or 4z - 5y = 0.
k = 6√2 or y = 4z/5.
The condition y = 4z/5 is stated as not possible.
Therefore, k = 6√2, which means k² = (6√2)² = 36 * 2 = 72.
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