Class 11th
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New answer posted
2 months agoContributor-Level 9
Let z = αi + βj be a complex number & √3i + j = (√3 + i) where I = √-1.
|z| = |z - (√3 + i)| and |z| = |√3 + i| = 2.
This implies z lies on the perpendicular bisector of the segment from 0 to √3+i and on a circle of radius 2 centered at the origin.
z = (√3 + i) ( (1+i)/√2 )
z = (1/√2) [ (√3 - 1) + I (√3 + 1)]
∴ α = (√3 - 1)/√2, β = (√3 + 1)/√2
Area of required triangle = (1/2) * base * height = (1/2) * |α| * |β| * 2 (This seems to be area of triangle with vertices 0, z, and another point)
The provided calculation: Area = (1/2) * | (√3-1)/√2| * | (√3+1)/√2| = (1/2) * (3-1)/2 = 1/2.
New answer posted
2 months agoContributor-Level 10
y = √ (2cos²α / (sinα cosα) + 1/sin²α)
y = √ (2cotα + cosec²α)
y = √ (2cotα + 1 + cot²α) = √ (1 + cotα)²) = |1 + cotα|.
Given α is in a range where 1+cotα is negative, y = -1 - cotα.
dy/dα = - (-cosec²α) = cosec²α.
At α = 5π/6, dy/dα = cosec² (5π/6) = (1/sin (5π/6)² = (1/ (1/2)² = 2² = 4.
New answer posted
2 months agoContributor-Level 10
Given Re (z-1)/ (2z+i) = 1, where z = x + iy.
(z-1)/ (2z+i) = [ (x-1) + iy] / [2x + I (2y+1)]
To rationalize, multiply the numerator and denominator by the conjugate of the denominator [2x - I (2y+1)].
Numerator = [ (x-1) + iy] * [2x - I (2y+1)] = 2x (x-1) - I (x-1) (2y+1) + i2xy + y (2y+1)
Real part of the numerator = 2x (x-1) + y (2y+1).
Denominator = (2x)² + (2y+1)².
Re (z-1)/ (2z+i) = [2x (x-1) + y (2y+1)] / [ (2x)² + (2y+1)²] = 1.
2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1.
0 = 2x² + 2y² + 2x + 3y + 1.
So, 2x² + 2y² + 2x + 3y + 1 = 0.
New answer posted
2 months agoContributor-Level 10
. Let the terms in Arithmetic Progression be a – 2d, a – d, a, a + d, a + 2d.
Sum of terms: (a – 2d) + (a – d) + a + (a + d) + (a + 2d) = 5a.
5a = 25 ⇒ a = 5.
Product of terms: (5 – 2d) (5 – d) (5) (5 + d) (5 + 2d) = 2520.
5 (25 – 4d²) (25 – d²) = 2520.
(25 – 4d²) (25 – d²) = 504.
625 – 25d² – 100d² + 4d? = 504.
4d? – 125d² + 121 = 0.
Factoring the equation: (4d² - 121) (d² - 1) = 0.
So, d² = 1 or d² = 121/4.
d = ±1 or d = ±11/2.
If d = ±1, the terms are 3, 4, 5, 6, 7.
If d = ±11/2, the terms are -6, -1/2, 5, 21/2, 16.
The largest term is 5 + 2d = 5 + 2 (11/2) = 5 + 11 = 16.
New answer posted
2 months agoContributor-Level 10
Given 2ae = 6 and 2a/e = 12.
From these, ae = 3 and a/e = 6.
Multiplying the two equations: (ae) (a/e) = 3 * 6 => a² = 18.
We know that b² = a² (1 - e²) = a² - a²e² = 18 - (ae)² = 18 - 3² = 18 - 9 = 9.
The length of the latus rectum (L.R.) is 2b²/a.
L.R. = 2 * 9 / √18 = 18 / (3√2) = 6/√2 = 3√2.
New answer posted
2 months ago
Contributor-Level 10
1st observation: n? =10, mean x? =2, variance σ? ²=2.
Σx? = n? x? = 20.
σ? ² = (Σx? ² / n? ) - x? ² => 2 = (Σx? ²/10) - 2² => 6 = Σx? ²/10 => Σx? ² = 60.
2nd observation: n? , mean y? =3, variance σ? ²=1. Let n? =n.
Σy? = ny? = 3n.
σ? ² = (Σy? ² / n) - y? ² => 1 = (Σy? ²/n) - 3² => 10 = Σy? ²/n => Σy? ² = 10n.
Combined variance σ² = 17/9. n_total = 10+n.
Combined mean = (Σx? +Σy? )/ (10+n) = (20+3n)/ (10+n).
Combined Σ (squares) = 60+10n.
σ² = (Combined Σsq / n_total) - (Combined mean)²
17/9 = (60+10n)/ (10+n) - [ (20+3n)/ (10+n)]²
Multiply by 9 (10+n)²:
17 (10+n)² = 9 (60+10n) (10+n) - 9 (20+3n)²
17 (100+
New answer posted
2 months agoContributor-Level 10
1/16, a, b are in GP. So, a² = b/16 .
Also, a, b, 1/6 are in AP. So, 2b = a + 1/6.
From the first equation, b = 16a².
Substitute into the second: 2 (16a²) = a + 1/6 => 32a² - a - 1/6 = 0.
192a² - 6a - 1 = 0.
The solution appears to solve a different problem.
New answer posted
2 months agoContributor-Level 10
S? (x) = log? ¹? ²x + log? ¹? ³x + .
This is incorrect; the bases are numbers, not powers of 'a'. Let's assume the bases are 1/2, 1/3, 1/6, 1/11, .
The series is S' = 2, 3, 6, 11, 18, .
The differences are 1, 3, 5, 7, . which is an AP.
The n-th term t? is a quadratic in n.
t? = An² + Bn + C.
t? =A+B+C=2
t? =4A+2B+C=3
t? =9A+3B+C=6
Solving these, we get 3A+B=1 and 5A+B=3, which gives 2A=2, A=1. Then B=-2, C=3.
t? = n² - 2n + 3 = (n-1)² + 2.
The solution confirms this finding t? = 2 + (n-1)².
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