Class 11th
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New answer posted
6 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
6 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
6 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
6 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.
New answer posted
6 months agoContributor-Level 10
The point of intersection of the ellipse x²/16 + y²/b² = 1 and the curve y² = 3x² lies on both.
Substitute y² = 3x² into the ellipse equation:
x²/16 + 3x²/b² = 1
x² (1/16 + 3/b²) = 1
x² (b² + 48) / 16b² = 1
x² = 16b² / (b² + 48).
For a solution to exist, we need x² > 0, which is true if b≠0.
The problem seems to have a condition missing or misinterpreted in the OCR. The provided solution also shows x² + y² = 4b, which might be another curve involved. Assuming the point lies on x²+y²=4b.
x² + 3x² = 4b => 4x² = 4b => x² = b.
Substitute x²=b into the ellipse equation: b/16 + 3b/b² = 1 (assuming y²=3b).
b/16 + 3/b = 1
New answer posted
6 months agoContributor-Level 9
(|x| - 3)|x + 4| = 6
(-x - 3) (- (x + 4) = 6
(x + 3) (x + 4) = 6 ⇒ x² + 7x + 12 = 6 ⇒ x² + 7x + 6 = 0
(x + 1) (x + 6) = 0 ⇒ x = -6 (since x < -4)
Case (ii) -4 ≤ x < 0
(-x - 3) (x + 4) = 6
⇒ -x² - 7x - 12 = 6
⇒ x² + 7x + 18 = 0
The discriminant is D = 7² - 4 (1) (18) = 49 - 72 < 0, so no real solution.
Case (iii) x ≥ 0
(x - 3) (x + 4) = 6
⇒ x² + x - 12 = 6
⇒ x² + x - 18 = 0
x = [-1 ± √ (1² - 4 (1) (-18)] / 2 = [-1 ± √73] / 2
Since x ≥ 0 ⇒ x = (√73 - 1) / 2
Only two solutions.
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