5 Marks Questions

Question 15 Marks

If $\text{x}^4+\frac{1}{\text{x}^4}=119,$ find the valu of $\text{x}^3-\frac{1}{\text{x}^3}.$

Answer

In the given problem, we have to find the value of $\text{x}^3-\frac{1}{\text{x}^3}$
Given $\text{x}^4+\frac{1}{\text{x}^4}=119$
We shall use the idntity $(x + y)^2 = x^2 + y^2 + 2xy$
Here putting $\text{x}^4+\frac{1}{\text{x}^4}=119,$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2\times\not\text{x}^2\times\frac{1}{\not\text{x}^2}$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=119+2$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=121$
$\text{x}^2+\frac{1}{\text{x}^2}=\sqrt{11\times11}$
$\text{x}^2+\frac{1}{\text{x}^2}=\pm11$
In order to find $\Big(\text{x}-\frac{1}{\text{x}}\Big)$ we are using identity $(x - y)^2 = x^2 + y^2 - 2xy$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2\times\text{x}\times\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=11-2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=9$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)=\sqrt9$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)=\sqrt{3\times3}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)=\pm3$
In order to find $\Big(\text{x}^3-\frac{1}{\text{x}^3}\Big)$ we are using identity $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}+\text{x}\times\frac{1}{\text{x}}\Big)$
Here $\text{x}^2+\frac{1}{\text{x}^2}=11$ and $\Big(\text{x}-\frac{1}{\text{x}}\Big)=3$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}+\not\text{x}\times\frac{1}{\not\text{x}}\Big)$
$=3(11+ 1)$
$=3\times12$
$=36$
Hence the value of $\text{x}^3-\frac{1}{\text{x}^3}$ is $36.$

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Question 25 Marks

If $\text{x}+\frac{1}{\text{x}}=3,$ calculate $\text{x}^2+\frac{1}{\text{x}^2},\ \text{x}^3+\frac{1}{\text{x}^3}$ and $\text{x}^4+\frac{1}{\text{x}^4}.$

Answer

In the given problem, we have to find the value of $\text{x}^2+\frac{1}{\text{x}^2},\ \text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^4+\frac{1}{\text{x}^4}$
Given $\text{x}+\frac{1}{\text{x}}=3,$
We shall use the identity $(x + y)^2 = x^2 + y^2 + 2xy$
Here putting $\text{x}+\frac{1}{\text{x}}=3,$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}$
$(3)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\not\text{x}\times\frac{1}{\not\text{x}}$
$9=\text{x}^2+\frac{1}{\text{x}^2}+2$
$9-2=\text{x}^2+\frac{1}{\text{x}^2}$
$7=\text{x}^2+\frac{1}{\text{x}^2}$
Again squaring on both sides we get,
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(7)^2$
We shall use the identity $(x + y)^2 = x^2 + y^2 + 2xy$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$(7)^2=\text{x}^4+\frac{1}{\text{x}^4}+2\times\not\text{x}^2\times\frac{1}{\not\text{x}^2}$
$49=\text{x}^4+\frac{1}{\text{x}^4}+2$
$49-2=\text{x}^4+\frac{1}{\text{x}^4}$
$47=\text{x}^4+\frac{1}{\text{x}^4}$
Again cubing on both sides we get,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=(3)^3$
We shall use the identity $(a + b)^3 = a^3 + b^3 + 2ab$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$(3)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\times\not\text{x}\times\frac{1}{\not\text{x}}\times3$
$27=\text{x}^3+\frac{1}{\text{x}^3}+9$
$27-9=\text{x}^3+\frac{1}{\text{x}^3}$
$18=\text{x}^3+\frac{1}{\text{x}^3}$
Hence the value of $\text{x}^2+\frac{1}{\text{x}^2},\ \text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^4+\frac{1}{\text{x}^4}$ is $7, 18, 47$ respectively.

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Question 35 Marks

Simplify the following products:
$\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big(\frac{2}{5}-\frac{\text{x}}{2}\Big)-\text{x}^2+2\text{x}$

Answer

In the given problem, we have to find product of $\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big(\frac{2}{5}-\frac{\text{x}}{2}\Big)-\text{x}^2+2\text{x}$
On rearranging we get
$\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big(\frac{2}{5}-\frac{\text{x}}{2}\Big)-\text{x}^2+2\text{x}\\=\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big[-\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big]-\text{x}^2+2\text{x}$
$\Rightarrow\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big(\frac{2}{5}-\frac{\text{x}}{2}\Big)-\text{x}^2+2\text{x}\\=-\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)^2-\text{x}^2+2\text{x}$
We shall use the identity $(\text{x}-\text{y})^2=\text{x}^2-2\text{xy}+\text{y}^2$
By substituting $\text{x}=\frac{\text{x}}{2},\ \text{y}=\frac{2}{5}$
$\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big(\frac{2}{5}-\frac{\text{x}}{2}\Big)-\text{x}^2+2\text{x}\\=\Big(\frac{\text{x}^2}{4}+\frac{4}{25}-\frac{2\times\text{x}}{\not2}\times\frac{\not2}{5}\Big)-\text{x}^2+2\text{x}$
$=-\Big(\frac{\text{x}^2}{4}+\frac{4}{25}-\frac{2\text{x}}{5}\Big)-\text{x}^2+2\text{x}$
$=-\frac{\text{x}^2}{4}-\frac{4}{25}+\frac{2\text{x}}{5}-\text{x}^2+2\text{x}$
$=\Big[-\frac{\text{x}^2}{4}-\text{x}^2\Big]-\frac{4}{25}+\Big[\frac{2\text{x}}{5}+2\text{x}\Big]$
$=\Big[-\frac{\text{x}^2}{4}-\frac{\text{x}^2}{1}\frac{\times4}{\times4}\Big]-\frac{4}{25}+\Big[\frac{2\text{x}}{5}+\frac{2\text{x}}{1}\frac{\times5}{\times5}\Big]$
$=\Big[-\frac{\text{x}^2}{4}-\frac{4\text{x}^2}{4}\Big]-\frac{4}{25}+\Big[\frac{2\text{x}}{5}+\frac{10\text{x}}{5}\Big]$
$=\Big[\frac{-\text{x}^2-4\text{x}^2}{4}\Big]-\frac{4}{25}+\Big[\frac{2\text{x}+10\text{x}}{5}\Big]$
$=\frac{-5\text{x}^2}{4}-\frac{4}{25}+\frac{12\text{x}}{5}$
Hence the value of $\Big(\frac{\text{x}}{2}-\frac{2}{5}\Big)\Big(\frac{2}{5}-\frac{\text{x}}{2}\Big)-\text{x}^2+2\text{x}$ is $\frac{-5\text{x}^2}{4}-\frac{4}{25}+\frac{12\text{x}}{5}.$

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Question 45 Marks

If $\text{x}+\frac{1}{\text{x}}=3, $ then find the value of $\text{x}^6+\frac{1}{\text{x}^6}.$

Answer

We have to find the value of $\text{x}^6+\frac{1}{\text{x}^6}$
Given $\text{x}+\frac{1}{\text{x}}=3$
Using identity $(a + b)^2 = a^2 + 2ab + b^2$
Here $\text{a} = \text{x},\ \text{b}=\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+2\times\text{x}\times\frac{1}{\text{x}}+\Big(\frac{1}{\text{x}}\Big)^2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+2\times\not\text{x}\times\frac{1}{\not\text{x}}+\frac{1}{\text{x}}\times\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+2+\frac{1}{\text{x}^2}$
By substituting the value of $\text{x}+\frac{1}{\text{x}}=3$ We get,
$(3)^2=\text{x}^2+2+\frac{1}{\text{x}^2}$
$3\times3=\text{x}^2+2+\frac{1}{\text{x}}^2$
By transposing $+ 2$ to left hand side, we get
$9-2=\text{x}^2+\frac{1}{\text{x}^2}$
$7=\text{x}^2+\frac{1}{\text{x}^2}$
Cubing on both sides we get,
$(7)^3=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)^3$
Using identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
Here $\text{a}=\text{x}^2,\ \text{b}=\frac{1}{\text{x}^2}$
$343=(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3\times\text{x}^2\times\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
$343=\text{x}^6+\frac{1}{\text{x}^6}+3\times\text{x}^2\times\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
Put $\text{x}^2+\frac{1}{\text{x}^2}=7$ we get
$343=\text{x}^6+\frac{1}{\text{x}^6}+3\times7$
$343=\text{x}^6+\frac{1}{\text{x}^6}+21$
By transposing $21$ to left hand side we get ,
$343-21=\text{x}^6+\frac{1}{\text{x}^6}$
$322=\text{x}^6+\frac{1}{\text{x}^6}$
Hence the value of $\text{x}^6+\frac{1}{\text{x}^6}$ is $322.$

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Question 55 Marks

Simplify the following: $\Big(\text{x}+\frac{2}{\text{x}}\Big)^3+\Big(\text{x}-\frac{2}{\text{x}}\Big)^3$

Answer

Given $\Big(\text{x}+\frac{2}{\text{x}}\Big)^3+\Big(\text{x}-\frac{2}{\text{x}}\Big)^3$
We shall use the identity $a^3 + b^3 = (a - b)(a^2 + b^2- ab)$
Here $\text{a}=\Big(\text{x}+\frac{2}{\text{x}}\Big),\ \text{b}=\Big(\text{x}-\frac{2}{\text{x}}\Big)$
By applying identity we get
$=\Big(\text{x}+\frac{2}{\text{x}}+\text{x}-\frac{2}{\text{x}}\Big)\Bigg[\Big(\text{x}+\frac{2}{\text{x}}\Big)^2+\Big(\text{x}-\frac{2}{\text{x}}\Big)^2$
$-\bigg(\Big(\text{x}+\frac{2}{\text{x}}\Big)\times\Big(\text{x}-\frac{2}{\text{x}}\Big)\bigg)\Bigg]$
$=\Big(\text{x}+\frac{2}{\text{x}}+\text{x}-\frac{2}{\text{x}}\Big)\bigg[\Big(\text{x}\times\text{x}+\frac{2}{\text{x}}\times\frac{2}{\text{x}}+2\times\text{x}\times\frac{2}{\text{x}}\Big)$
$+\Big(\text{x}\times\text{x}+\frac{2}{\text{x}}\times\frac{2}{\text{x}}-2\times\text{x}\times\frac{2}{\text{x}}\Big)-\Big(\text{x}^2+\frac{4}{\text{x}^2}\Big)\bigg]$
$=(2\text{x})\bigg[\Big(\text{x}^2+\frac{4}{\text{x}^2}+\frac{4\text{x}}{\text{x}}\Big)+\Big(\text{x}^2+\frac{4}{\text{x}^2}-\frac{4\text{x}}{\text{x}}\Big)-\Big(\text{x}^2-\frac{4}{\text{x}^2}\Big)\bigg]$
$=(2\text{x})\bigg[\text{x}^2+\frac{4}{\text{x}^2}+\frac{4\text{x}}{\text{x}}+\text{x}^2+\frac{4}{\text{x}^2}-\frac{4\text{x}}{\text{x}}-\text{x}^2+\frac{4}{\text{x}^2}\bigg]$
By rearranging the variable we get,
$=(2\text{x})\Big[\text{x}^2+\frac{4}{\text{x}^2}+\frac{4}{\text{x}^2}+\frac{4}{\text{x}}^2\Big]$
$=2\text{x}\times\Big[\text{x}^2+\frac{12}{\text{x}^2}\Big]$
$=2\text{x}^2+\frac{24}{\text{x}}$
Hence the simplified value of $\Big(\text{x}+\frac{2}{\text{x}}\Big)^3+\Big(\text{x}-\frac{2}{\text{x}}\Big)^3$ is $2\text{x}^2+\frac{24}{\text{x}}.$

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Question 65 Marks

Simplify: $(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$

Answer

Given $(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$
By using identity $(x + y + z)^2$
$ = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$ we have
$(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$
$=\big[\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\big]$
$+\big[\text{a}^2+(-\text{b})^2+\text{c}^2+2a\text{}(-\text{b})+2(-\text{b})(\text{c})\big]$
$+\big[\text{a}^2+(-\text{b})^2+\text{c}^2+2\text{ab}+2\text{b}(-\text{c})+2(-\text{c})\text{a}\big]$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}+\text{a}^2+\text{b}^2+\text{c}^2$
$-2\text{ab}-2\text{bc}+2\text{ca}+\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}-2\text{bc}-2\text{ca}$
$\big(\text{a}+\text{b}+\text{c}\big)+\big(\text{a}-\text{b}+\text{c}\big)+\big(\text{a}+\text{b}-\text{c}\big)$
$=\text{a}^2+\text{a}^2+\text{a}^2+\text{b}^2+\text{b}^2+\text{b}^2+\text{c}^2+\text{c}^2+\text{c}^2$
$+2\text{ab}+2\text{ab}-2\text{ab}-2\text{bc}+2\text{bc}-2\text{bc}+2\text{ca}+2\text{ca}-2\text{ca}$
$=3\text{a}^2+3\text{b}^2+3\text{c}^2+2\text{ab}-2\text{bc}+2\text{ca}$
Talking $3$ as a common factor we get
$\big(\text{a}+\text{b}+\text{c})^2 +\big(\text{a}-\text{b}+\text{c}\big)^2 +\big(\text{a}+\text{b}-\text{c}\big)^2$
$=3\big(\text{a}^2+\text{b}^2+\text{c}^2\big)+2\text{ab}-2\text{bc}+2\text{ca}$
Hence the value of $\big(\text{a}+\text{b}+\text{c})^2 +\big(\text{a}-\text{b}+\text{c}\big)^2 +\big(\text{a}+\text{b}-\text{c}\big)^2$ is $3\big(\text{a}^2+\text{b}^2+\text{c}^2\big)+2\text{ab}-2\text{bc}+2\text{ca}.$

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Question 75 Marks

Simplify the following expressions:
$\big(\text{x}+\text{y}+\text{z}\big)^2+\Big(\text{x}+\frac{\text{y}}{2}+\frac{\text{z}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}+\frac{\text{z}}{4}\Big)^2$

Answer

Expanding, we get
$zx=\big[\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{}\big]\\+\Big[\text{x}^2+\frac{\text{y}^2}{4}+\frac{\text{z}^2}{9}+2\text{x}\frac{\text{y}}{2}+2\frac{\text{zx}}{3}+\frac{\text{yz}}{3}\Big]\\-\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{\text{z}^2}{10}+\frac{\text{xy}}{3}+\frac{\text{yz}}{6}+\frac{\text{xz}}{4}\Big]$
$\big[\therefore\big(\text{x}+\text{y}+\text{z}\big)^2=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{zx}\big]$
$=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{zx}+\text{x}^2+\frac{\text{y}^2}{4}+\frac{\text{z}^2}{9}\\+2\text{x}\frac{\text{y}}{2}+\frac{\text{xy}}{3}+\frac{2\text{zx}}{3}-\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}-\frac{\text{z}^2}{10}-\frac{\text{xy}}{3}-\frac{\text{yz}}{6}-\frac{\text{xz}}{4}$
Rearranging coefficients,
$=\frac{8\text{x}^2-\text{x}^2}{4}+\frac{36\text{y}^2+9\text{y}^2-4\text{y}^2}{36}+\frac{144\text{z}^2+16\text{z}^2-9\text{z}^2}{144}\\+\frac{6\text{xy}+3\text{xy}-\text{xy}}{3}\frac{13\text{yz}}{6}+\frac{29\text{xz}}{12}$
$=\frac{7\text{x}^2}{4}+\frac{41\text{y}^2}{36}+\frac{151\text{z}^2}{144}+\frac{8\text{xy}}{3}+\frac{13\text{yz}}{6}+\frac{29\text{zx}}{12}$
$\big(\text{x}+\text{y}+\text{z}\big)^2+\Big(\text{x}+\frac{\text{y}}{3}+\frac{\text{z}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}+\frac{\text{z}}{4}\Big)^2$
$=\frac{7\text{x}^2}{4}+\frac{41\text{y}^2}{36}+\frac{151\text{z}^2}{144}+\frac{8\text{xy}}{3}+\frac{13\text{yz}}{6}+\frac{29\text{zx}}{12}$

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