// Numbas version: exam_results_page_options {"name": "Always, sometimes or never: square and cube numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Always, sometimes or never: square and cube numbers", "type": "question", "statement": "

For each of the following statements, select one option from \"Always\", \"Sometimes\" or \"Never\".

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Select:

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• \"Always\", if it's true for every possible $x$.
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• \"Sometimes\", if there is an $x$ for which it's true but also an $x$ for which it's false.
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• \"Never\", if there is no $x$ for which it's true.
• \n
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This variable was created solely for the purpose of being able to publish this question.

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i)      $x^3$ is greater than $x^2$

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ii)      $x^2$ is greater than $x$

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iii)      If $x$ is negative, $x^2$ is negative

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iv)      If $x$ is negative, $x^3$ is negative

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v)      $x^2 = x$

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vi)      $x^2 = - x$

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vii)      $(x+1)^2 \\gt x$

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viii)      $(x+1)^3 \\gt x$

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ix)      $x^3 \\times x = x^2 \\times x^2$

", "

x)      $x^2$ has the opposite sign to $x$

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xi)      $x^3$ has the opposite sign to $x$

Always

", "

Sometimes

", "

Never

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Decide whether statements about square and cube numbers are always true, sometimes true or never true.

"}, "preamble": {"css": "", "js": ""}, "advice": "

i)

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Suppose $x$ is negative, for example $x = -5$.

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\\\begin{align} x^2 &= (-5)^2 \\\\&= 25 \\\\ x^3 &= (-5)^3 \\\\&= -125 \\\\x^3 &\\lt x^2\\text{.} \\end{align} \

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Now let $x$ be positive, for example $x = 5$.

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\\\begin{align} x^2 &= 5^2 \\\\&= 25 \\\\ x^3 &= 5^3 \\\\&= 125 \\\\x^3 &\\gt x^2\\text{.}\\end{align} \

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Therefore, $x^3$ is sometimes greater than $x^2$.

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ii)

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This 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

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\\\begin{align} \\text{When } x &= 0.5\\text{,} \\\\x^2 &= 0.25\\text{, so} \\\\ x^2 &\\lt x \\text{.} \\end{align} \

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Therefore, $x^2$ is sometimes greater than $x$.

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iii)

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Multiplying two negative numbers gives a positive answer and multiplying two postive numbers gives a positive answer. Therefore, $x^2$ is never negative.

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iv)

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Multiplying a negative number by itself an odd number of times always gives a negative answer. For example, let $x = -1$. Then

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\\\begin{align} x^3 &= (-1)^3 \\\\&= -1\\times-1\\times-1 \\\\&=1\\times-1 \\\\&= - 1 \\text{.} \\end{align}\

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Therefore, if $x$ is negative, $x^3$ is always negative.

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v)

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This is true when $x = 1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $x$.

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vi)

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This is true when $x = -1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $-x$.

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vii)

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When $x$ is positive, for example $x = 5$:

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\\\begin{align} (x+1)^2 &= (5 + 1)^2 \\\\&= 6^2 \\\\&= 36 \\gt x = 5 \\end{align} \

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When $x = 0$:

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\\\begin{align} (x+1)^2 &= (0 + 1)^2 \\\\&= 1^2 \\\\&= 1 \\gt x = 0 \\end{align} \

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When $x$ is negative, such as $x = -4$:

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\\\begin{align} (x+1)^2 &= (-4 + 1)^2 \\\\&= (-3)^2 \\\\&= 9 \\gt x = -4 \\end{align} \

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To see the behaviour of $(x+1)^2$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $x$ $(x+1)^2$ $-3$ $-2$ $-1$ $-0.5$ $0$ $0.5$ $1$ $2$ \n$3$\n $4$ $1$ $0$ $0.25$ $1$ $2.25$ $4$ $9$ $16$
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Therefore, $(x+1)^2$ is always greater than $x$.

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viii)

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When $x$ is positive, for example $x = 5$:

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\\\begin{align} (x+1)^3 &= (5 + 1)^3 \\\\&= 6^3 \\\\&= 216 \\gt x = 5 \\end{align} \

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When $x = 0$:

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\\\begin{align} (x+1)^3 &= (0 + 1)^3 \\\\&= 1^3 \\\\&= 1 \\gt x = 0 \\end{align} \

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When $x$ is negative, such as $x = -4$:

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\\\begin{align} (x+1)^2 &= (-4 + 1)^3 \\\\&= (-3)^3 \\\\&= -27 \\lt x = -4 \\end{align} \

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To see the behaviour of $(x+1)^3$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $x$ $(x+1)^3$ $-3$ $-2$ $-1$ $-0.5$ $0$ $0.5$ $1$ $2$ \n$3$\n $-8$ $-1$ $0$ $0.125$ $1$ $3.375$ $8$ $27$ $64$
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Therefore, $(x+1)^3$ is sometimes greater than $x$.

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ix)

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We can write

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\$x^3\\times x = x \\times x \\times x \\times x = x^2 \\times x^2\\text{.}\$

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Therefore, $x^3 \\times x$ always equals $x^2 \\times x^2$.

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x)

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Since $x^2$ is always positive, $x^2$ only has the opposite sign to $x$ when $x$ is negative.

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Therefore, $x^2$ sometimes has the opposite sign to $x$.

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xi)

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As 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.

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Therefore, $x^3$ never has the opposite sign to $x$.

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