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New answer posted
2 months agoContributor-Level 9
Given f (g (x) is defined piecewise:
f (g (x) =
x³ + 2 ; x < 0
x? ; 0? x < 1
(3x - 2)² ; x? 1
fog (x) is discontinuous at x = 0.? non differentiable.
fog (x) is not differentiable at x = 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 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
2 months 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
2 months 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
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
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).
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