Maths
Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
a year 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
a year 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
a year agoContributor-Level 10
Using Lagrange's Mean Value Theorem (LMVT) for x ∈ [−7, -1].
[f (-1) - f (-7)] / [-1 - (-7)] ≤ 2
[f (-1) - (-3)] / 6 ≤ 2
f (-1) + 3 ≤ 12
f (-1) ≤ 9
Using LMVT for x ∈ [−7, 0].
[f (0) - f (-7)] / [0 - (-7)] ≤ 2
[f (0) - (-3)] / 7 ≤ 2
f (0) + 3 ≤ 14
f (0) ≤ 11
Therefore, f (0) + f (-1) ≤ 11 + 9 = 20.
New answer posted
a year agoContributor-Level 10
Answer given by NTS is (1) which is wrong.
I = 1/ (a+b) ∫? x [f (x) + f (x+1)]dx . (1)
Using the property x → a + b - x
I = 1/ (a+b) ∫? (a+b-x) [f (a+b-x) + f (a+b+1-x)]dx
Given f (a+b+1-x) = f (x)
I = 1/ (a+b) ∫? (a+b-x) [f (x+1) + f (x)]dx . (2)
Adding (1) and (2):
2I = 1/ (a+b) ∫? (a+b) [f (x) + f (x+1)]dx
2I = ∫? [f (x) + f (x+1)]dx
2I = ∫? f (x)dx + ∫? f (x+1)dx
Let x+1 = t in the second integral, so dx = dt.
When x=a, t=a+1. When x=b, t=b+1.
∫? f (x+1)dx = ∫? ¹ f (t)dt = ∫? ¹ f (x)dx
New answer posted
a year agoContributor-Level 10
c = λ (a x b).
a = I + j - k
b = I + 2j + k
a x b = | I j k |
| 1 -1 |
| 1 2 1 |
= I (1 - (-2) - j (1 - (-1) + k (2-1) = 3i - 2j + k.
c = λ (3i - 2j + k).
Given c ⋅ (i + j + 3k) = 8.
λ (3i - 2j + k) ⋅ (i + j + 3k) = 8
λ (3 - 2 + 3) = 8 => 4λ = 8 => λ = 2.
c = 2 (a x b).
We need to find c ⋅ (a x b).
c ⋅ (a x b) = 2 (a x b) ⋅ (a x b) = 2|a x b|².
|a x b|² = 3² + (-2)² + 1² = 9 + 4 + 1 = 14.
So, c ⋅ (a x b) = 2 * 14 = 28.
New answer posted
a year 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
a year agoContributor-Level 10
f (x) = {x+a, if x<0; |x-1|, if x0}
g (x) = {x+1, if x<0; (x-1)+b, if x0}
g (f (x) must be continuous. The potential points of discontinuity are where the definitions of f (x) and g (f (x) change. This is at x=0 and where f (x)=0.
f (x)=0 when x=-a (if a>0) or when x=1.
Continuity at x = 0:
lim (x→0? ) g (f (x) = lim (x→0? ) g (x+a). Since a could be anything, let's analyze f (0? )=a. So, lim is g (a).
lim (x→0? ) g (f (x) = g (f (0? ) = g (|0-1|) = g (1). Since 1≥0, g (1) = (1-1)²+b = b.
g (f (0) = g (|0-1|) = g (1) = b.
So we need g (a) = b.
Case 1: a < 0. g (a) = a+1. So a+1=b.
Case 2: a ≥ 0. g (a) = (a-1)²+b. So (a-1)²+b=b => (a-1)²=0 => a=1.
Now consider continuity at x=-a (assuming a>0).
l
New answer posted
a year 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.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 67k Colleges
- 1.2k Exams
- 717k Reviews
- 1850k Answers