Even i was out of my mind when i saw this.......But here is the OE......Let me know if you understand the OE becoz its difficult for me to assume things..........
We are asked whether the expression $_(\frac{x}{2})^2 + (\frac{x}{3})^3$_ is greater than or equal to $_0$_.
Statement 1 tells us that $_x < 0$_. When $_x$_ is negative, the term $_(\frac{x}{2})^2$_ will always be positive because squaring any non-zero number will produce a positive result. However, the term $_(\frac{x}{3})^3$_ will always be negative because cubing any negative number will always produce a negative result.
Therefore, when $_x < 0$_, $_\left(\frac{x}{2}\right)^2 + \left(\frac{x}{3}\right)^3$_ is the sum of a positive number and a negative number. The value of the expression will be positive or negative depending on the magnitudes of the two terms. Testing cases will be helpful here. We choose simple, intuitive values to test.
When $_x = -1$_, the expression becomes $_\left(\frac{-1}{2}\right)^2 + \left(\frac{-1}{3}\right)^3 = \frac14 + \frac{-1}{27}$_. The magnitude of $_\frac14$_ is greater than the magnitude of $_-\frac{1}{27}$_, so the value of the expression is positive.
Let's see if we can find a value of $_x$_ that makes the expression negative-that is, a value of $_x$_ that makes the magnitude of $_\left(\frac{x}{3}\right)^3$_ greater than the magnitude of $_\left(\frac{x}{2}\right)^2$_. Since the second term, $_(\frac{x}{3})^3$_, is being cubed, it will increase in magnitude much faster than $_(\frac{x}{2})^2$_ as $_x$_ increases in magnitude. So, we should test a value where $_x$_ has a large magnitude; $_x = -10$_ is a good candidate to test since it is easy to work with.
When $_x = -10$_, the expression becomes $_(\frac{-10}{2})^2 + (\frac{-10}{3})^3 = \frac{100}{4} + \frac{-1000}{27}$_. The magnitude of the term $_\frac{-1000}{27}$_ is greater than that of $_\frac{100}{4}$_, so the sum will be negative. Thus, we have found a case where the value of the expression is positive and a case where it is negative. Statement 1 alone is insufficient. Eliminate answer choices A and D. The correct answer choice is either B, C, or E.
Statement 2 tells us that $_x > -\frac{27}{4}$_. It would be too difficult to plug that value into the expression, so let's see what we can do to the expression itself.
We can solve the inequality $_(\frac{x}{2})^2 + (\frac{x}{3})^3 \geq 0$_ to see where the expression is positive. First, subtract $_(\frac{x}{3})^3$_ from both sides:
$_(\frac{x}{2})^2 \geq -(\frac{x}{3})^3$_. Apply the exponent to the numerator and denominator of the fraction:
$_\frac{x^2}{4} \geq -\frac{x^3}{27}$_. Cross multiply:
$_27x^2 \geq -4x^3$_. Cancel $_x^2$_ from both sides. Note that because x^2 must be positive, we don't have to worry about flipping the inequality; we can also ignore the case in which $_x=0$_, since $_x=0$_ makes the inequality in the prompt true:
$_27 \geq -4x$_. Divide both sides by $_-4$_, and flip the inequality because we are dividing by a negative value:
$_-\frac{27}{4} \leq x$_.
So, when $_x > -\frac{27}{4}$_, the expression is greater than or equal to $_0$_. Remember that when $_x = 0$_ the value of the expression is $_0$_. Thus, the expression in the prompt must be greater than or equal to $_0$_, and so Statement 2 alone is sufficient.
eagleeye wrote:karthikpandian19 wrote:Is $_(\frac{x}{2})^2 + (\frac{x}{3})^3 \geq 0$_?
$_x < 0$_
$_x > -\frac{27}{4}$_
How to approach this problem....???? What to assume ?
What language is this in? ahahaha. Sorry, I am cracking up at reading this. Could you please post an image or something. Seems like something went wonky here.
