সৃজনশীলঃ ‘g(x)=x এবং y=x+1x…’ কুমিল্লা বোর্ড ২০২২ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $g\left(x\right)=x$

ধরি, $y={\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-g\left(x\right)}{1+g\left(x\right)}}\}$
$={\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\}$
$={\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\}$ধরি,$x=\cos \theta$
$={\cos }^4\{{\cot }^{-1}\sqrt{\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}}\}$
$={\cos }^4\{{\cot }^{-1}(\tan \frac{\theta }{2})\}={\cos }^4\{{\cot }^{-1}\cot (\frac{\pi }{2}-\frac{\theta }{2})\}$
$={\cos }^4(\frac{\pi }{2}-\frac{\theta }{2})={\sin }^4\frac{\theta }{2}=\frac{1}{4}{\{2{\sin }^2\frac{\theta }{2}\}}^2=\frac{1}{4}(1-\cos \theta {)}^2$
$=\frac{1}{4}(1-x{)}^2[\because \cos \theta =x]$
$\therefore y=\frac{1}{4}(x-1{)}^2$
$\therefore \frac{dy}{dx}=\frac{d}{dx}\frac{1}{4}(x-1{)}^2=\frac{1}{4}\cdot 2(x-1)\cdot 1=\frac{1}{2}(x-1)$
$\therefore \frac{d}{dx}[{\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-g\left(x\right)}{1+g\left(x\right)}}\}]=\frac{1}{2}(x-1).$ (প্রমাণিত)

দেওয়া আছে,

এখন$\begin{array}{cc}& g\left(x\right)=x\\ & y=\sqrt{4+3g(\sin x)}\\ & \Rightarrow y^2=4+3g(\sin x)\\ & \Rightarrow y^2=4+3\sin x\\ & \Rightarrow \frac{d}{dx}{y}^2=\frac{d}{dx}(4+3\sin x)\\ & \Rightarrow 2y\cdot y_1=3\cos x\\ & \Rightarrow \frac{d}{dx}\left(2y_1\right)=\frac{d}{dx}(3\cos x)\\ & \Rightarrow 2\cdot (y\cdot y_2+y_1\cdot y_1)=-3\sin x\\ & \Rightarrow q{y}_2+2y_1^2=-3\sin x\\ & \text{ এখন, }2{\mathrm{yy}}_1+2y_1^2+y^2-4=-3\sin x+y^2-4\\ & \Rightarrow 2{\mathrm{yy}}_1+2y_1^2+y^2-4=-3\sin x+4+3\sin x-4\\ & =0\left(\text{ poved }\right)\\ & \end{array}$