2(4x - 1) < 4(3x+1), where x ∈ {set of all negative Integers}

Mathematics

1 answer:

alexgriva [62]3 years ago

3 0

$\large\underline{\sf{Solution-}}$

Given inequation:

$\sf 2(4x-1)$

$\sf\longmapsto 8x-2$

$\sf\longmapsto 8x-2-4$

$\sf\longmapsto 8x-6$

$\sf\longmapsto 8x-6-8x$

$\sf\longmapsto -6$

$\sf\longmapsto 4x>-6$

$\sf\longmapsto \dfrac{4x}{4}>\dfrac{-6}{4}$

$\sf\longmapsto x >-1\dfrac{1}{2}$

Therefore:

$\sf\longmapsto\boxed{ x=\{-1\}}\ \ Ans.$

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Let f be a continuous function defined on the interval [0,1] such that f(0) = f(1). Show that there exists a number 0 <= a &l

Leni [432]

Explanation:

In order to prove that affirmation, we define the function g over the interval [0, 1/2] with the formula $g(x) = f(x+1/2)-f(x) .$

If we evaluate g at the endpoints we have

g(0) = f(1/2)-f(0) = f(1/2) - f(1) (because f(0) = f(1))

g(1/2) = f(1) - f(1/2) = -g(0)

Since g(1/2) = -g(0), we have one chance out of three

g(0) > 0 and g(1/2) < 0
g(0) < 0 and g(1/2) > 0
g(0) = g(1/2) = 0

We will prove that g has a zero on [0,1/2]. If g(0) = 0, then it is trivial. If g(0) ≠ 0, then we are in one of the first two cases, and therefore g(0) * g(1/2) < 0. Since f is continuous, so is g. Bolzano's Theorem assures that there exists c in (0,1/2) such that g(c) = 0. This proves that g has at least one zero on [0,1/2].

Let c be a 0 of g, then we have

$0 = g(c) = f(c+1/2)-f(c)$