Maths Continuity and Differentiability
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New answer posted
a month agoContributor-Level 10
The functional equation f (x+y) = f (x)f (y) implies f (x) = a? for some constant a.
Then f' (x) = a? ln (a).
Given f' (0) = 3, we have a? ln (a) = 3 ⇒ ln (a) = 3 ⇒ a = e³.
So, f (x) = e³?
We need to evaluate the limit: lim (x→0) (f (x)-1)/x = lim (x→0) (e³? -1)/x.
Using the standard limit lim (u→0) (e? -1)/u = 1, we can write:
lim (x→0) 3 * (e³? -1)/ (3x) = 3 * 1 = 3.
New answer posted
a month agoContributor-Level 10
To find the values of a and b, we evaluate the left-hand limit (LHL) and right-hand limit (RHL) at x=0 and equate them.
LHL: lim (x→0) sin (a+3)/2 * x / x = (a+3)/2. The full limit evaluates to (a+3)/2. So, (a+3)/2 = b.
This gives the relation a - 2b + 3 = 0 — (I)
RHL: lim (x→0) [√ (x+bx³) - √x] / (bx²).
Rationalize the numerator:
lim (x→0) [ (x+bx³) - x] / [bx² (√ (x+bx³) + √x)]
= lim (x→0) bx³ / [bx² (√x (√ (1+bx²) + √x)]
= lim (x→0) x / [√x (√ (1+bx²) + 1)] = lim (x→0) √x / (√ (1+bx²) + 1) = 0/2 = 0.
So, b = 0.
Substituting b=0 into equation (I): a - 2 (0) + 3 = 0 ⇒ a = -3.
New answer posted
a month agoContributor-Level 10
The problem provides an equation involving the coordinates (α, β, γ) of a point P:
((α + β + γ) / √3)^2 + ((α - nγ) / √(l^2 + n^2))^2 + ((α - 2β + γ) / √6)^2 = 9
The locus of P(α, β, γ) is given by replacing (α, β, γ) with (x, y, z):
((x + y + z) / √3)^2 + ((lx - nz) / √(l^2 + n^2))^2 + ((x - 2y + z) / √6)^2 = 9
This represents the equation of an ellipsoid. The text proceeds by comparing coefficients. By expanding the equation, the coefficients of x^2, y^2, z^2, and cross-product terms are collected. From the given conditions:
Coefficient of x^2: 1/3 + l^2 / (l^2 + n^2) + 1/6 = 1
Coefficient of y^2: 1/3 + 0 + 4/6
New answer posted
a month agoContributor-Level 10
cos(x)(3sin(x) + cos(x) + 3)dy = (1 + ysin(x)(3sin(x) + cos(x) + 3))dx
This seems mistyped. A more likely form is:
dy/dx - (sin(x)/(cos(x)))y = 1 / (cos(x)(3sin(x) + cos(x) + 3))
dy/dx - tan(x)y = sec(x) / (3sin(x) + cos(x) + 3)
The integrating factor (I.F.) is:
I.F. = e^∫(-tan(x))dx = e^(ln|cos(x)|) = cos(x).
Multiplying by I.F.:
d(y*cos(x))/dx = 1 / (3sin(x) + cos(x) + 3)
y*cos(x) = ∫ dx / (3sin(x) + cos(x) + 3)
Using Weierstrass substitution, let t = tan(x/2):
sin(x) = 2t/(1+t²), cos(x) = (1-t²)/(1+t²), dx = 2dt/(1+t²)
∫ (2dt/(1+t²)) / (3(2t/(1+t²)) + (1-t²)/(1+t²) + 3)
= ∫ 2dt / (6t + 1 - t² + 3 + 3t²) = ∫ 2dt / (2t² + 6t
New answer posted
a month agoContributor-Level 10
-5 ≤ [x/2] < 5
I ⇒ [x/2] = -5, -4, -3, -2, -1,0,1,2,3,4
Hence, function is discontinues at = -4, -3, -2, -1,1,2,3,4 Number of values is 8.
New answer posted
a month agoContributor-Level 10
-5? [x/2] < 5
I? [x/2] = -5, -4, -3, -2, -1,0,1,2,3,4
Hence, function is discontinues at = -4, -3, -2, -1,1,2,3,4 Number of values is 8.
New answer posted
a month agoContributor-Level 10
f (x) is differentiable then will also continuous then f (π) = -1, f (π? ) = -k?
k? = 1
Now f' (x) = { 2k? (x-π) if x≤π
{ -k? sinx if x>π
then f' (π? ) = f' (π? ) = 0
f' (x) = { 2k? if x≤π
{ -k? cosx if x>π
then 2k? =k?
k? = 1/2
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
Since, lim (x→0) f (x)/x exist ⇒ f (0) = 0
Now, f' (x) = lim (h→0) (f (x+h)-f (x)/h = lim (h→0) (f (h)+xh²+x²h)/h (take y = h)
= lim (h→0) f (h)/h + lim (h→0) (xh) + x²
⇒ f' (x) = 1 + 0 + x²
⇒ f' (3) = 10
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