সৃজনশীলঃ ‘f(x)=1−tan2⁡x21+tan2⁡x2…’ চট্টগ্রাম কলেজ ২০২৬ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\cos (2\cdot \frac{x}{2})=\cos x$

বামপক্ষ $=\frac{d}{dx}[{\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-f\left(x\right)}{1+f\left(x\right)}}\}]$
$=\frac{d}{dx}{\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}\}$
$=\frac{d}{dx}{\cos }^4\{{\cot }^{-1}\sqrt{\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}}\}$
$=\frac{d}{dx}{\cos }^4\{{\cot }^{-1}\tan \frac{x}{2}\}=\frac{d}{dx}{\cos }^4\{\frac{\pi }{2}-{\tan }^{-1}\tan \frac{x}{2}\}$
$=\frac{d}{dx}{\cos }^4\{\frac{\pi }{2}-\frac{x}{2}\}=\frac{d}{dx}{\sin }^4\frac{x}{2}$= $\frac{d}{dx}{({\sin }^2\frac{x}{2})}^2=\frac{d}{dx}\frac{1}{4}{(2{\sin }^2\frac{x}{2})}^2$
$=\frac{d}{dx}\frac{1}{4}(1-\cos x{)}^2=\frac{1}{4}\cdot 2(1-\cos x{)}^1\cdot \sin x$
$=\frac{1}{2}(1-\cos x)\cdot \sin x=\frac{1}{2}(1-\cos x)\cdot \sqrt{1-{\cos }^{2}x}$
$=\frac{1}{2}(1-f(x\left)\right)\sqrt{1-\left(f\right(x){)}^2}=$ ডানপক্ষ
$\therefore \frac{d}{dx}[{\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-f\left(x\right)}{1+f\left(x\right)}}\}]=\frac{1}{2}(1-f(x\left)\right)\sqrt{1-\left(f\right(x){)}^2}$(প্রমাণিত)

দেওয়া আছে, $g\left(x\right)=(1+f(x)\sqrt{1-\left(f\right(x){)}^2}$
বা, $g\left(x\right)=(1+\cos x)\sqrt{1-{\cos }^{2}x}$ ['খ' হতে প্রাপ্ত $f\left(x\right)=\cos x$ ]
বা, $g\left(x\right)=(1+\cos x)\cdot \sin x=\sin x+\sin x\cdot \cos x$
বা, $g\left(x\right)=\sin x+\frac{1}{2}\cdot 2\sin x\cdot \cos x=\sin x+\frac{1}{2}\sin 2x$
বা, $g^{\prime }\left(x\right)=\cos x+\frac{1}{2}\cdot 2\cos 2x=\cos x+\cos 2x$
বা, $g^{\prime \prime }\left(x\right)=-\sin x-2\sin 2x$
চরমমানের জন্য, $g^{\prime }\left(x\right)=0$
বা, $\cos x+\cos 2x=0$
বা, $\cos x+{\cos }^{2}x-{\sin }^{2}x=0$
বা, $\cos x+{\cos }^{2}x-(1-{\cos }^{2}x)=0$.
বা, $2{\cos }^{2}x+\cos x-1=0$
বা, $2{\cos }^{2}x+2\cos x-\cos x-1=0$
বা, $2\cos (\cos x+1)-1(\cos x+1)=0$
বা, $(\cos x+1)(2\cos x-1)=0$

হয়, $\cos x+1=0$
বা, $\cos x=-1$
বা, $\cos x=\cos {180}^{\circ }$

$\therefore x={180}^{\circ }$

অথবা, $2\cos -1=0$
বা, $2\cos x=1$
বা, $\cos x=\frac{1}{2}$
$\cos x=\cos {60}^{\circ }$

$\therefore x={60}^{\circ }$

$0<x<\pi$ ব্যবধিতে $x={60}^{\circ }$

$g^{\prime \prime }\left({60}^{\circ }\right)=-\sin {60}^{\circ }-2\sin {120}^{\circ }=\frac{-\sqrt{3}}{2}-2\frac{\sqrt{3}}{2}=\frac{-3\sqrt{3}}{2}<0$

$x={60}^{\circ }$ এর জন্য $g\left(x\right)$ এর গুরুমান বিদ্যমান ।
গুরুমান, $g\left({60}^{\circ }\right)=\sin {60}^{\circ }+\frac{1}{2}\sin (2\times {60}^{\circ })$

$=\frac{\sqrt{3}}{2}+\frac{1}{2}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}=\frac{2\sqrt{3}+\sqrt{3}}{4}=\frac{3\sqrt{3}}{4}$

নির্ণেয় চরমমান $\frac{3\sqrt{3}}{4}$.