Decide whether statements about square and cube numbers are always true, sometimes true or never true.
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", "ii) $x^2$ is greater than $x$
", "iii) If $x$ is negative, $x^2$ is negative
", "iv) If $x$ is negative, $x^3$ is negative
", "v) $x^2 = x$
", "vi) $x^2 = - x$
", "vii) $(x+1)^2 \\gt x$
", "viii) $(x+1)^3 \\gt x$
", "ix) $x^3 \\times x = x^2 \\times x^2$
", "x) $x^2$ has the opposite sign to $x$
", "xi) $x^3$ has the opposite sign to $x$
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", "Sometimes
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\nSuppose $x$ is negative, for example $x = -5$.
\n\\[ \\begin{align} x^2 &= (-5)^2 \\\\&= 25 \\\\ x^3 &= (-5)^3 \\\\&= -125 \\\\x^3 &\\lt x^2\\text{.} \\end{align} \\]
\nNow let $x$ be positive, for example $x = 5$.
\n\\[ \\begin{align} x^2 &= 5^2 \\\\&= 25 \\\\ x^3 &= 5^3 \\\\&= 125 \\\\x^3 &\\gt x^2\\text{.}\\end{align} \\]
\nTherefore, $x^3$ is sometimes greater than $x^2$.
\n\n
\n
ii)
\nThis is true for either $x \\gt 1$ or $x \\lt 0$ but false for $0 \\leq x \\leq 1$. For example, let $x=0.5$. Then
\n\\[ \\begin{align} \\text{When } x &= 0.5\\text{,} \\\\x^2 &= 0.25\\text{, so} \\\\ x^2 &\\lt x \\text{.} \\end{align} \\]
\nTherefore, $x^2$ is sometimes greater than $x$.
\n\n
\n
iii)
\nMultiplying two negative numbers gives a positive answer and multiplying two postive numbers gives a positive answer. Therefore, $x^2$ is never negative.
\n\n
\n
iv)
\nMultiplying a negative number by itself an odd number of times always gives a negative answer. For example, let $x = -1$. Then
\n\\[\\begin{align} x^3 &= (-1)^3 \\\\&= -1\\times-1\\times-1 \\\\&=1\\times-1 \\\\&= - 1 \\text{.} \\end{align}\\]
\nTherefore, if $x$ is negative, $x^3$ is always negative.
\n\n
\n
v)
\nThis is true when $x = 1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $x$.
\n\n
\n
vi)
\nThis is true when $x = -1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $-x$.
\n\n
\n
vii)
\nWhen $x$ is positive, for example $x = 5$:
\n\\[ \\begin{align} (x+1)^2 &= (5 + 1)^2 \\\\&= 6^2 \\\\&= 36 \\gt x = 5 \\end{align} \\]
\nWhen $x = 0$:
\n\\[ \\begin{align} (x+1)^2 &= (0 + 1)^2 \\\\&= 1^2 \\\\&= 1 \\gt x = 0 \\end{align} \\]
\nWhen $x$ is negative, such as $x = -4$:
\n\\[ \\begin{align} (x+1)^2 &= (-4 + 1)^2 \\\\&= (-3)^2 \\\\&= 9 \\gt x = -4 \\end{align} \\]
\nTo see the behaviour of $(x+1)^2$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:
\n| $x$ | \n$-3$ | \n$-2$ | \n$-1$ | \n$-0.5$ | \n$0$ | \n$0.5$ | \n$1$ | \n$2$ | \n\n $3$ \n | \n
|---|---|---|---|---|---|---|---|---|---|
| $(x+1)^2$ | \n$4$ | \n$1$ | \n$0$ | \n$0.25$ | \n$1$ | \n$2.25$ | \n$4$ | \n$9$ | \n$16$ | \n
Therefore, $(x+1)^2$ is always greater than $x$.
\n\n\n
viii)
\nWhen $x$ is positive, for example $x = 5$:
\n\\[ \\begin{align} (x+1)^3 &= (5 + 1)^3 \\\\&= 6^3 \\\\&= 216 \\gt x = 5 \\end{align} \\]
\nWhen $x = 0$:
\n\\[ \\begin{align} (x+1)^3 &= (0 + 1)^3 \\\\&= 1^3 \\\\&= 1 \\gt x = 0 \\end{align} \\]
\nWhen $x$ is negative, such as $x = -4$:
\n\\[ \\begin{align} (x+1)^2 &= (-4 + 1)^3 \\\\&= (-3)^3 \\\\&= -27 \\lt x = -4 \\end{align} \\]
\nTo see the behaviour of $(x+1)^3$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:
\n| $x$ | \n$-3$ | \n$-2$ | \n$-1$ | \n$-0.5$ | \n$0$ | \n$0.5$ | \n$1$ | \n$2$ | \n\n $3$ \n | \n
|---|---|---|---|---|---|---|---|---|---|
| $(x+1)^3$ | \n$-8$ | \n$-1$ | \n$0$ | \n$0.125$ | \n$1$ | \n$3.375$ | \n$8$ | \n$27$ | \n$64$ | \n
Therefore, $(x+1)^3$ is sometimes greater than $x$.
\n\n\nix)
\nWe can write
\n\\[x^3\\times x = x \\times x \\times x \\times x = x^2 \\times x^2\\text{.}\\]
\nTherefore, $x^3 \\times x$ always equals $x^2 \\times x^2$.
\n\n\nx)
\nSince $x^2$ is always positive, $x^2$ only has the opposite sign to $x$ when $x$ is negative.
\nTherefore, $x^2$ sometimes has the opposite sign to $x$.
\n\n\nxi)
\nAs seen in part iv), multiplying a negative number by itself an odd number of times always gives a negative answer. It is also true that multiplying a positive number by itself an odd number of times will give a positive answer. So, $x^3$ always has the same sign as $x$, since we have an odd power.
\nTherefore, $x^3$ never has the opposite sign to $x$.
\n\n", "statement": "
For each of the following statements, select one option from \"Always\", \"Sometimes\" or \"Never\".
\nSelect:
\n