LUANDATO wrote:Which one of the following is greatest among the three ?
$$x^3,\ x^2\ and\ 1/x^2$$
$$(1)x>0$$
$$(2)x<1$$
The OA is C.
Can any expert assist me with this DS question, please? I don't understand it. Thanks.
Let's study the values of x^3, x^2 and 1/x^2 at a different range of values of x.
1. Range x < 0: Say x = -2, thus we have x^3 = (-2)^3 = -8; x^2 = (-2)^2 = 4; and 1/x^2 = 1/(-2)^2 = 1/4.
We see that x^3 < 1/x^2 < x^2.
2. Range x = -1: We have x^3 = (-1)^3 = -1; x^2 = (-1)^2 = 1; and 1/x^2 = 1/(-1)^2 = 1.
We see that x^3 < 1/x^2 = x^2.
3. Range -1 < x < 0: Say x = -1/2, thus we have x^3 = (-1/2)^3 = -1/8; x^2 = (-1/2)^2 = 1/4; and 1/x^2 = 1/(-1/2)^2 = 4.
We see that x^3 < x^2 < 1/x^2.
4. At x = 0, 1/x^3 is not defined.
5. Range 0 < x < 1: Say x = 1/2, thus we have x^3 = (1/2)^3 = 1/8; x^2 = (1/2)^2 = 1/4; and 1/x^2 = 1/(1/2)^2 = 4.
We see that x^3 < x^2 < 1/x^2.
6. Range x = 1: We have x^3 = (1)^3 = 1; x^2 = (1)^2 = 1; and 1/x^2 = 1/(1)^2 = 1.
We see that x^3 = 1/x^2 = x^2.
7. Range x > 1: Say x = 2, thus we have x^3 = (2)^3 = 8; x^2 = (2)^2 = 4; and 1/x^2 = 1/(2)^2 = 1/4.
We see that x^3 > x^2 > 1/x^2.
Hope this helps you solve the question.
-Jay
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