সৃজনশীলঃ ‘f(x)=|x−3|g(x)=p+qx+rx2…’ বরিশাল বোর্ড ২০১৭ – জটিল সংখ্যা, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=|x-3|$

এখানে, $f\left(x\right)<\frac{1}{7}$

বা, $|x-3|<\frac{1}{7}$....................(1)

বা, $|x-3|+6<\frac{1}{7}+6$ [উভয়পক্ষে 6 যোগ করে]

বা, $|x-3|+|6|<\frac{1+42}{7}$

বা, $|x-3+6|\le |x-3|+|6|<\frac{43}{7}[\because |b+c|\le |b|+|c|]$

বা, $|x-3+6|<\frac{43}{7}$

বা, $|x+3|<\frac{43}{7}$....................(2)

(1) ও (2) নং গুণ করে পাই,

$|x-3|.|x+3|<\frac{1}{7}\cdot \frac{43}{7}$

বা, $\left|\right(x-3\left)\right(x+3)|<\frac{43}{49}[\because |b||c|=|bc|]$
বা, $|x^2-3^2|<\frac{43}{49}$

$\therefore |x^2-9|<\frac{43}{49}$. (প্রমাণিত)

দেওয়া আছে, $g\left(x\right)=p+qx+r{x}^2$

এখন, $p+q+r=0$

a=x=3

এবং $\omega$ এককের কাল্পনিক ঘনমূল।

এখন,

$\begin{array}{cc}g\left(\omega \right)& =p+q\omega +r{\omega }^2\\ g\left({\omega }^2\right)& =p+q{\omega }^2+r{\omega }^4=p+q{\omega }^2+r\omega \cdot {\omega }^3\\ & =p+q{\omega }^2+r\omega \cdot 1[\because {\omega }^3=1]\\ & =p+q{\omega }^2+r\omega \end{array}$

$\begin{array}{cc}\left\{g\right(\omega \left)\right\}+\left\{g\left({\omega }^2\right)\right\}& =p+q\omega +r{\omega }^2+p+q{\omega }^2+r\omega \\ & =2p+q\omega +q{\omega }^2+r\omega +r{\omega }^2\\ & =2p+q(\omega +{\omega }^2)+r(\omega +{\omega }^2)\\ & =2p+q\cdot (-1)+r\cdot (-1)[\because 1+\omega +{\omega }^2=0]\\ & =2p-q-r=2p-(q+r)\\ & =2p-(-p)[\because p+q+r=0]\\ & =2p+p=3p\end{array}$

$\begin{array}{cc}& \left\{g\right(\omega \left)\right\}\cdot \left\{g\left({\omega }^2\right)\right\}=(p+q\omega +r{\omega }^2)(p+q{\omega }^2+r\omega )\\ & =p^2+\mathrm{pq}{\omega }^2+\mathrm{rp}\omega +\mathrm{pq}\omega +q^2{\omega }^3+\mathrm{qr}{\omega }^2+\mathrm{rp}{\omega }^2+\mathrm{qr}{\omega }^4+r^2{\omega }^3\\ & =p^2+\mathrm{pq}{\omega }^2+\mathrm{rp}\omega +\mathrm{pq}\omega +q^2{\omega }^3+\mathrm{qr}{\omega }^2+\mathrm{rp}{\omega }^2+\mathrm{qr}\omega \cdot {\omega }^3+r^2{\omega }^3\\ & =p^2+\mathrm{pq}{\omega }^2+\mathrm{rp}\omega +\mathrm{pq}\omega +q^2\cdot 1+\mathrm{qr}{\omega }^2+\mathrm{rp}{\omega }^2+\mathrm{qr}\omega \cdot 1+r^2\cdot 1[\because {\omega }^3=1]\\ & =p^2+q^2+r^2+\mathrm{pq}\omega +\mathrm{pq}{\omega }^2+\mathrm{qr}\omega +\mathrm{qr}{\omega }^2+\mathrm{pp}\omega +\mathrm{rp}{\omega }^2\\ & =p^2+q^2+r^2+\mathrm{pq}(\omega +{\omega }^2)+\mathrm{qr}(\omega +{\omega }^2)+\mathrm{pp}(\omega +{\omega }^2)\\ & =p^2+q^2+r^2+\mathrm{pq}\cdot (-1)+\mathrm{qr}.(-1)+\mathrm{rp}.(-1)[\because 1+\omega +{\omega }^2=0]\\ & =p^2+q^2+r^2-\mathrm{pq}-\mathrm{qr}+\mathrm{rp}\\ & =p^2+(q+r{)}^2-3\mathrm{qr}-p(q+r)\\ & =p^2+(-p{)}^2-3\mathrm{qr}-p.(-p\left)\right[\because p+q+r=0]\\ & =p^2+p^2-3\mathrm{qr}+p^2=3p^2-3\mathrm{qr}\\ & \end{array}$

বামপক্ষ, 

$\begin{array}{cc}& =\left\{g\right(\omega ){\}}^3+{\left\{g\left({\omega }^2\right)\right\}}^3\\ & =\left[\right\{g\left(\omega \right)\}+\left\{g\left({\omega }^2\right)\right\}]\left[\right\{g\left(\omega \right){\}}^2-\left\{g\right(\omega \left)\right\}\cdot \left\{g\left({\omega }^2\right)\right\}+{\left\{g\left({\omega }^2\right)\right\}}^2]\\ & =\left[\right\{g\left(\omega \right)\}+g\left({\omega }^2\right)\}][{\left[\right\{g\left(\omega \right)\}+\left\{g\left({\omega }^2\right)\right\}]}^2-3\cdot \{g\left(\omega \right)\}\cdot \left\{g\left({\omega }^2\right)\right\}]\\ & =3p\cdot \left\{\right(3p{)}^2-3\cdot (3p^2-3\mathrm{qr})\}\\ & =3p(9p^2-9p^2+9\mathrm{qr})\\ & =3p.9\mathrm{qr}=27\mathrm{pqr}=3^3\mathrm{pqr}\\ & =a^x\mathrm{pqr}[\because a=x=3]\end{array}$

= ডানপক্ষ

$\therefore \left\{g\right(\omega ){\}}^3+{\left\{g\left({\omega }^2\right)\right\}}^3=a^x$ pqr. (প্রমাণিত)