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Chapter 7: Problem 18

The least natural number a for which \(x+a x^{-2}>2, \forall x \in\) $(0, \infty)$ is (A) 1 (B) 2 (C) 5 (D) None of these

Short Answer

Expert verified
Answer: The least natural number a is 2.

Step by step solution

01

Rewrite the inequality as an inequality of quadratic equations

Note that the given expression can also be written as a quadratic with respect to \(x^{-1}\). Rewrite the inequality as \(x + a (x^{-1})^2 > 2\) which is equivalent to \(ax^{-2} + x > 2\).
02

Multiply the inequality by \(x^2\)

To further simplify the expression, multiply the whole inequality by \(x^2\), which gives us \(ax^2(x^{-2})+x^3>2x^2\). This simplifies to \(a+x^3>2x^2\).
03

Rearrange terms and factorize

Now, it's time to rearrange the inequality and gather all the terms on one side: \(x^3-2x^2+a > 0\). The inequality has an \(x^2\) that can be factored out: \(x^2(x-2)+a>0\).
04

Consider the function for all values of \(x\) in \((0, \infty)\)

Next, we need to analyze the function \(f(x) = x^2(x-2)+a\) for all values of \(x\) in \((0, \infty)\). Notice that this function is a cubic polynomial with a positive leading coefficient, meaning it goes from \(-\infty\) to \(\infty\) as \(x\) goes from \(0\) to \(\infty\). Thus, the function has a local minimum at the point where its derivative is equal to \(0\).
05

Determine the minimum value of the function

To find the point of the local minimum, we will differentiate \(f(x)\) with respect to \(x\) and set it equal to \(0\): $$\frac{d}{dx}\left(x^2(x-2)+a\right) = 3x^2-4x.$$ Setting this equal to \(0\), we get $$3x^2-4x = 0$$ $$x(3x-4) = 0.$$ So \(x=0\) or \(x = \frac{4}{3}\), but we only consider \(x=\frac{4}{3}\) because \(x\) must be in the interval \((0, \infty)\). Thus, the minimum value of the function occurs at \(x=\frac{4}{3}\).
06

Plug in the value of x and find the smallest integer a

Since we want the inequality to hold true for all positive \(x\), we need the function to be strictly positive at \(x=\frac{4}{3}\). So \(f\left(\frac{4}{3}\right)>0\), which gives us \(\left(\frac{4}{3}\right)^2\left(\frac{4}{3}-2\right)+a>0\). Simplifying, we get \(\frac{16}{9}\left(\frac{-2}{3}\right)+a > 0\). This simplifies further to \(-\frac{32}{27} + a > 0\), thus \(a > \frac{32}{27}\). Therefore, the smallest integer value for \(a\) that makes the inequality hold true for all positive values of \(x\) is \(\boxed{2}\), which corresponds to the answer choice (B).

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