সৃজনশীলঃ ‘M=Nx+xN…’ ঝিনাইদহ ক্যাডেট কলেজ ২০২৪ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $M=N^x+x^N$এবং $N=\tan x$

$\therefore M=(\tan x{)}^x+x^{\tan x}$

ধরি, $y_1=(\tan x{)}^x$বা, $\ln y_1=x\ln (\tan x)$

$\therefore \frac{d}{dx}(\ln y_1)=\frac{x}{\tan x}\cdot {\sec }^{2}x+\ln (\tan x)$

বা, $\frac{1}{y_1}\frac{d{y}_1}{dx}=\ln (\tan x)+x\mathrm{cosec}x\cdot \sec x$

$\therefore \frac{d{y}_1}{dx}=(\tan x{)}^x\{\ln (\tan x)+x\mathrm{cosec}x\cdot \sec x\}$

আবার,

ধরি, $y_2=x^{\tan x}$বা, $\ln y_2=\tan x\ln \left(x\right)$

$\therefore \frac{d}{\mathrm{dx}}(\ln y_2)=\frac{\tan x}{x}+\ln \left(x\right){\sec }^{2}x$

বা, $\frac{1}{y_2}\frac{d{y}_2}{dx}={\sec }^{2}x\ln x+\frac{\tan x}{x}$

$\therefore \frac{d{y}_2}{dx}=x^{\tan x}({\sec }^{2}x\ln x+\frac{\tan x}{x})$

এখন, $\frac{dM}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

$=(\tan x{)}^x\{\ln (\tan x)+x\sec x\cdot \mathrm{cosec}x\}+x^{\tan x}(\ln x\cdot {\sec }^{2}x+\frac{\tan x}{x})$

দেওয়া আছে, $a^4+b^4+c^4+a^2{b}^2=2c^2(a^2+b^2)$

বা, $a^4+b^4+c^4+a^2{b}^2-2c^2{a}^2-2b^2{c}^2=0$

বা, $a^4+b^4+c^4+2a^2{b}^2-2b^2{c}^2-2c^2{a}^2=a^2{b}^2$

বা, ${\left(a^2\right)}^2+{\left(b^2\right)}^2+{(-c^2)}^2+2a^2{b}^2+2b^2(-c^2)+2(-c^2)a^2=a^2{b}^2$.

বা, ${(a^2+b^2-c^2)}^2=a^2{b}^2$বা, $a^2+b^2-c^2=\pm \sqrt{a^2{b}^2}$

বা, $\frac{a^2+b^2-c^2}{2ab}=\pm \frac{ab}{2ab}$বা, $\cos C=\pm \frac{1}{2}$

এখন, $\cos C=\frac{1}{2}$হলে, $\cos C=\cos {60}^{\circ }\therefore C={60}^{\circ }$

আবার, $\cos C=-\frac{1}{2}$হলে, $\cos C=\cos {120}^{\circ }\therefore C={120}^{\circ }$

$\therefore C={60}^{\circ }$অথবা ${120}^{\circ }$ (দেখানো হলো)