Solve the Following Question.(5 Marks)

Question 15 Marks

Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$\left|x^2-9\right|+\left|x^2-4\right|=5$

Answer

$\left|x^2-9\right|+\left|x^2-4\right|=5$
$\therefore|(x-3)(x+3)|+|(x-2)(x+2)|=5$
Case l: $x<-3$
Also, $x<-2, x<2, x<3$
$\therefore(x-3)(x+3)>0 \text { and }(x-2)(x+2)>0$
Equation (i) reduces to
$\mathrm{x}^2-9+\mathrm{x}^2-4=5$
$\therefore 2 \mathrm{x}^2=18$
$\therefore \mathrm{x}=-3 \text { or } 3 \text { (both rejected as } \mathrm{x}<-3 \text { ) }$
Case II: $-3 \leq \mathrm{x}<-2$
As $x<-2, x<3$
$\therefore(x-3)(x+3)<0,(x-2)(x+2)>0$
Equation (i) reduces to
$-\left(x^2-9\right)+x^2-4=5$
$\therefore 5=5 \text { (true) }$
$-3 \leq x<-2 \text { is a solution ....(ii) }$
Case III: $-2 \leq x<2$ As $x>-3, x<3$
$ \therefore(x-3)(x+3)<0,$
$(x-2)(x+2)<0 $ Equation (i) reduces to $ 9-\mathrm{x}^2+4-\mathrm{x} 2=5$
$\therefore 2 \mathrm{x}^2=13-5$
$\therefore \mathrm{x}^2=4$
$\therefore \mathrm{x}=-2 \text { is a solution } $
Case IV: $2 \leq x<3$ As $x>-3, x>-2$ $ \therefore(x-3)(x+3)<0,(x-2)(x+2)>0 $
Equation (i) reduces to $ 9-\mathrm{x}^2+\mathrm{x}^2-4=5$
$\therefore 5=5 \text { (true) }$
$\therefore 2 \leq \mathrm{x}<3 \text { is a solution } $
Case V: $3 \leq x$ As $x>-3, x>-2, x>2$
$ \therefore(x+3)(x-3)>0,$
$(x-2)(x+2)>0 $
Equation (i) reduces to
$ \mathrm{x}^2-9+\mathrm{x}^2-4=5$
$\therefore 2 \mathrm{x}^2=18$
$\therefore \mathrm{x}^2=9$
$\therefore \mathrm{x}=3 \ldots .(\mathrm{v})$
$(\mathrm{x}=-3 \text { rejected as } \mathrm{x} \geq 3) $
From (ii), (iii), (iv), (v), we get $ \therefore \text { Solution set }=[-3,-2] \cup[2,3] $

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Question 25 Marks

Solve: $\sqrt{\log _2 x^4}+4 \log _4 \sqrt{\frac{2}{x}}=2$

Answer

$ \sqrt{\log _2 x^4}+4 \log _4 \sqrt{\frac{2}{x}}=2$
$\therefore \quad \sqrt{\log _2 x^4}+4 \log _4\left(\frac{2}{x}\right)^{\frac{1}{2}}=2$
$\therefore \quad \sqrt{4 \log _2 x}+\frac{4}{2} \log _4\left(\frac{2}{x}\right)=2$
$\therefore \quad 2 \sqrt{\log _2 x}+2 \log _4\left(\frac{2}{x}\right)=2$
$\therefore \quad \sqrt{\log _2 x}+\log _4\left(\frac{2}{x}\right)=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{\log _2\left(\frac{2}{x}\right)}{\log _2 4}=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{\log _2\left(\frac{2}{x}\right)}{\log _2(2)^2}=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{\log _2 2-\log _2 x}{2 \log _2 2}=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{1-\log _2 x}{2(1)}=1 \quad \ldots\left[\because \log _a a=1\right]$
Let $\log _2 x=\mathrm{a}$
$\therefore \quad \sqrt{\mathrm{a}}+\frac{1-\mathrm{a}}{2}=1$
Multiplying throughout by 2 , we get
$ 2 \sqrt{a}+1-a=2$
$\therefore 2 \sqrt{a}=a+1$
Squaring on both sides, we get
$ 4 a=(a+1)^2$
$\therefore 4 a=a^2+2 a+1$
$\therefore a^2-2 a+1=0$
$\therefore (a-1)^2=0$
$\therefore a-1=0$
$\therefore a=1 $
Since $\log _2 x=\mathrm{a}$,
$\log _2 x=1$
$\therefore \quad x=2^1$
$\therefore \quad x=2$

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Maths Solve the Following Question.(5 Marks)

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