দেওয়া আছে,
\(f(x)=2x^2-7x+10\)
\[\therefore Rf^{\prime}(x)=\lim_{h \rightarrow 0+}\frac{f(x+h)-f(x)}{h}\]
\[=\lim_{h \rightarrow 0+}\frac{2(x+h)^2-7(x+h)+10-2x^2+7x-10}{h}\]
\[=\lim_{h \rightarrow 0+}\frac{2x^2+4hx+2h^2-7x-7h-2x^2+7x}{h}\] ➜Note \(\because f(x)=2x^2-7x+10\) \(\Rightarrow f(x+h)=2(x+h)^2-7(x+h)+10\)
\[=\lim_{h \rightarrow 0+}\frac{4hx+2h^2-7h}{h}\]
\[=\lim_{h \rightarrow 0+}\frac{h(4x+2h-7)}{h}\]
\[=\lim_{h \rightarrow 0+}(4x+2h-7)\]
\[=4x-7\]
আবার,
\[Lf^{\prime}(x)=\lim_{h \rightarrow 0-}\frac{f(x+h)-f(x)}{h}\]
\[=\lim_{h \rightarrow 0-}\frac{2(x+h)^2-7(x+h)+10-2x^2+7x-10}{h}\]
\[=\lim_{h \rightarrow 0-}\frac{2x^2+4hx+2h^2-7x-7h-2x^2+7x}{h}\] ➜Note \(\because f(x)=2x^2-7x+10\) \(\Rightarrow f(x+h)=2(x+h)^2-7(x+h)+10\)
\[=\lim_{h \rightarrow 0-}\frac{4hx+2h^2-7h}{h}\]
\[=\lim_{h \rightarrow 0-}\frac{h(4x+2h-7)}{h}\]
\[=\lim_{h \rightarrow 0-}(4x+2h-7)\]
\[=4x-7\]
যেহেতু \(Rf^{\prime}(x)=Lf^{\prime}(x)\) অতএব, \(x\) এর সকল মানের জন্য \(f(x)\) ফাংশনটি \([2, 5]\) বদ্ধ ব্যবধিতে অবিচ্ছিন্ন এবং \(]2, 5[\) খোলা ব্যবধিতে অন্তরীকরণযোগ্য ।
\(\therefore f(x)\) ফাংশন ল্যাগ্রাঞ্জের গড়মান উপপাদ্যের সকল শর্ত সিদ্ধ করে।
এখন,
\(f(x)=2x^2-7x+10 ……(1)\)
\(\therefore f^{\prime}(x)=4x-7\)
তাহলে,
\(f^{\prime}(c)=4c-7\)
আবার,
\((1)\) এর সাহায্যে
\(f(2)=2.2^2-7.2+10\)
\(=2.4-14+10\)
\(=8-4\)
\(=4\)
এবং
\(f(5)=2.5^2-7.5+10\)
\(=2.25-35+10\)
\(=50-25\)
\(=25\)
ল্যাগ্রাঞ্জের গড়মান উপপাদ্য থেকে আমরা জানি,
\(f(5)-f(2)=(5-2)f^{\prime}(c)\)
\(\Rightarrow 25-4=3(4c-7)\) ➜Note\(\because f(5)=25, f(2)=4, f^{\prime}(c)=4c-7\)
\(\Rightarrow 21=12c-21\)
\(\Rightarrow 12c-21=21\)
\(\Rightarrow 12c=21+21\)
\(\Rightarrow 12c=42\)
\(\Rightarrow c=\frac{42}{12}\)
\(\Rightarrow c=\frac{7}{2}\)
\(\because 2\le{\frac{7}{2}}\le{5}\)
\(\therefore 2\le{c}\le{5}\)
যা \([2, 5]\) ব্যবধিতে বিদ্যমান।
সুতরাং প্রদত্ত ফাংশনে ল্যাগ্রাঞ্জের গড়মান উপপাদ্য বৈধ।

দেওয়া আছে,
\(f(x)=x^3-3x^2+18x+15\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(x^3-3x^2+18x+15)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow f^{\prime}(x)=\frac{d}{dx}(x^3)-3\frac{d}{dx}(x^2)+18\frac{d}{dx}(x)+\frac{d}{dx}(15)\) ➜Note\(\because \frac{d}{dx}(f(x))=f^{\prime}(x)\)\(\frac{d}{dx}(u-v+w+..)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)+..\)
\(=3x^2-3.2x+18.1+0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=3(x^2-2x)+18\)
\(=3(x^2-2x+1-1)+18\)
\(=3(x^2-2x+1)-3+18\)
\(=3(x-1)^2+15\)
\(\therefore f^{\prime}(x)=3(x-1)^2+15\)
এখানে,
\(x\) এর সকল বাস্তব মানের জন্য \(f^{\prime}(x)>0\)
\(\therefore \) প্রদত্ত ফাংশনটি ক্রমবর্ধমান।

দেওয়া আছে,
\(f(x)=1-13x+6x^2-x^3\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(1-13x+6x^2-x^3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow f^{\prime}(x)=\frac{d}{dx}(1)-13\frac{d}{dx}(x)+6\frac{d}{dx}(x^2)-\frac{d}{dx}(x^3)\) ➜Note\(\because \frac{d}{dx}(f(x))=f^{\prime}(x)\)\(\frac{d}{dx}(u-v+w-..)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)-..\)
\(=0-13.1+6.2x-3x^2\) ➜Note\(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(=-13+12x-3x^2\)
\(=-3x^2+12x-13\)
\(=-(3x^2-12x+13)\)
\(=-\{3(x^2-4x)+13\}\)
\(=-\{3(x^2-4x+4-4)+13\}\)
\(=-\{3(x^2-4x+4)-12+13\}\)
\(=-\{3(x-2)^2+1\}\)
\(=-3(x-2)^2-1\)
\(\therefore f^{\prime}(x)=-3(x-2)^2-1\)
এখানে,
\(x\) এর সকল বাস্তব মানের জন্য \(0>f^{\prime}(x)\)
\(\therefore \) প্রদত্ত ফাংশনটি ক্রমহ্রাসমান।

দেওয়া আছে,
\(y=x(12-2x)^2\)
\(\Rightarrow y=x\{(12)^2-2.12.2x+(2x)^2\}\)
\(\Rightarrow y=x\{144-48x+4x^2\}\)
\(\Rightarrow y=144x-48x^2+4x^3\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(144x-48x^2+4x^3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=144\frac{d}{dx}(x)-48\frac{d}{dx}(x^2)+3\frac{d}{dx}(x^3)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=144.1-48.2x+4.3x^2\) ➜Note\(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=144-96x+12x^2\)
\(=12x^2-96x+144\)
\(=12(x^2-8x+12)\)
\(=12(x^2-6x-2x+12)\)
\(=12\{x(x-6)-2(x-6)\}\)
\(=12(x-6)(x-2)\)
\(\therefore \frac{dy}{dx}=12(x-6)(x-2)\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 12(x-6)(x-2)=0\) ➜Note\(\because \frac{dy}{dx}=12(x-6)(x-2)\)
\(\Rightarrow (x-6)(x-2)=0\)
\(\Rightarrow x-6=0, x-2=0\)
\(\Rightarrow x=6, x=2\)
\(\therefore x=6, 2\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(144-96x+12x^2)\) ➜Note\(\because \frac{dy}{dx}=12(x-6)(x-2)=144-96x+12x^2\)
\(=\frac{d}{dx}(144)-96\frac{d}{dx}(x)+12\frac{d}{dx}(x^2)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=0-96.1+12.2x\) ➜Note\(\because \frac{d}{dx}(c)=0,\frac{d}{dx}(x)=1\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(=-96+24x\)
\(=24x-96\)
\(\therefore \frac{d^2y}{dx^2}=24(x-4)\)
এখন, \(x=2\) হলে,
\(\frac{d^2y}{dx^2}=24(2-4)\)
\(=24.(-2)\)
\(=-48\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore x=2\) বিন্দুতে ফাংশনটির গুরুমান হবে।
আবার, \(x=6\) হলে,
\(\frac{d^2y}{dx^2}=24(6-4)\)
\(=24.2\)
\(=48\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=6\) বিন্দুতে ফাংশনটির লঘুমান হবে।

দেওয়া আছে,
\(f(x)=x^3-6x^2+24x+4\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(x^3-6x^2+24x+4)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow f^{\prime}(x)=\frac{d}{dx}(x^3)-6\frac{d}{dx}(x^2)+24\frac{d}{dx}(x)+\frac{d}{dx}(4)\) ➜Note\(\because \frac{d}{dx}(f(x))=f^{\prime}(x)\)\(\frac{d}{dx}(u-v+w+..)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)+..\)
\(=3x^2-6.2x+24.1+0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=3x^2-12x+24\)
\(=3(x^2-4x+8)\)
\(=3(x^2-4x+4+4)\)
\(=3\{(x-2)^2+4\}\)
\(\therefore f^{\prime}(x)=3\{(x-2)^2+2^2\}\) ➜Note একাধিক রাশির বর্গের যোগফল ঋনাত্মক হতে পারে না।
যা, সর্বদাই ধনাত্মক।
\(\therefore x\) এর কোনো বাস্তব মানের জন্য \(f^{\prime}(x)\) কখনো শূন্য হতে পারে না।
\(\therefore \) প্রদত্ত ফাংশনটির কোনো গুরুমান অথবা লঘুমান নেই।

ধরি,
\(y=1+2\sin{x}+3\cos^2{x} ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(1+2\sin{x}+3\cos^2{x})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(1)+2\frac{d}{dx}(\sin{x})+3\frac{d}{dx}(\cos^2{x})\) ➜Note\(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=0+2\cos{x}+3.2\cos{x}(-\sin{x})\) ➜Note\(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\sin{x})=\cos{x}\),\(\frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\therefore \frac{dy}{dx}=2\cos{x}-6\sin{x}\cos{x}\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 2\cos{x}-6\sin{x}\cos{x}=0\) ➜Note\(\because \frac{dy}{dx}=2\cos{x}-6\sin{x}\cos{x}\)
\(\Rightarrow 2\cos{x}(1-3\sin{x})=0\)
\(\Rightarrow \cos{x}=0, 1-3\sin{x}=0\)
\(\Rightarrow \cos{x}=0, -3\sin{x}=-1\)
\(\Rightarrow \cos{x}=0, \sin{x}=\frac{-1}{-3}\)
\(\Rightarrow \cos{x}=0, \sin{x}=\frac{1}{3}\)
\(\therefore x=\frac{\pi}{2}, \sin{x}=\frac{1}{3}\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(2\cos{x}-3\sin{2x})\) ➜Note\(\because \frac{dy}{dx}=2\cos{x}-6\sin{x}\cos{x}=2\cos{x}-3\sin{2x}\)
\(=2\frac{d}{dx}(\cos{x})-3\frac{d}{dx}(\sin{2x})\) ➜Note\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(=-2\sin{x}-3\cos{2x}.2\)
\(=-2\sin{x}-6\cos{2x}\)
\(\therefore \frac{d^2y}{dx^2}=-2\sin{x}-6\cos{2x}\)
এখন, \(x=\frac{\pi}{2}\) হলে,
\(\frac{d^2y}{dx^2}=-2\sin{x}-6\cos{2x}\)
\(=-2\sin{\frac{\pi}{2}}-6\cos{2.\frac{\pi}{2}}\)
\(=-2.1-6\cos{\pi}\)
\(=-2.1-6.(-1)\)
\(=-2+6\)
\(=4\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=\frac{\pi}{2}\) বিন্দুতে ফাংশনটির লঘুমান আছে।
ফাংশনটির লঘুমান \(=1+2\sin{\frac{\pi}{2}}+3\cos^2{\frac{\pi}{2}}\) ➜Note \((1)\) হতে \(\because y=1+2\sin{x}+3\cos^2{x}\)
ফাংশনটির লঘুমান \(=1+2.1+3.0\)
\(=1+2+0\)
\(=3\)
আবার, \(\sin{x}=\frac{1}{3}\) হলে,
\(\frac{d^2y}{dx^2}=-2\sin{x}-6\cos{2x}\)
\(=-2.\frac{1}{3}-6\{1-2.\left(\frac{1}{3}\right)^2\}\)
\(=-\frac{2}{3}-6\{1-2\frac{1}{9}\}\)
\(=-\frac{2}{3}-6\{1-\frac{2}{9}\}\)
\(=-\frac{2}{3}-6.\frac{9-2}{9}.\)
\(=-\frac{2}{3}-2.\frac{7}{3}\)
\(=-\frac{2}{3}-\frac{14}{3}\)
\(=\frac{-2-14}{3}\)
\(=\frac{-16}{3}\)
\(=-\frac{16}{3}\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore \sin{x}=\frac{1}{3}\) বিন্দুতে ফাংশনটির গুরুমান আছে।
ফাংশনটির গুরুমান \(=1+2\sin{x}+3\cos^2{x}\) ➜Note \((1)\) হতে \(\because y=1+2\sin{x}+3\cos^2{x}\)
\(=1+2\sin{x}+3(1-\sin^2{x})\)
\(=1+2.\frac{1}{3}+3\{1-\left(\frac{1}{3}\right)^2\}\)
\(=1+\frac{2}{3}+3\{1-\frac{1}{9}\}\)
\(=1+\frac{2}{3}+3.\frac{9-1}{9}\)
\(=1+\frac{2}{3}+3.\frac{8}{9}\)
\(=1+\frac{2}{3}+\frac{24}{9}\)
\(=\frac{39}{9}\)
\(=\frac{13}{3}\)
\(=4\frac{1}{3}\)

\((a)\)
প্রদত্ত সীমা,
\[\lim_{x \rightarrow 0}\frac{1-\cos{2x}}{x}\]
\[\Rightarrow \lim_{x \rightarrow 0}\frac{2\sin^2{x}}{x}\] ➜Note\(\because 1-\cos{2A}=2\sin^2{A}\)
\[= 2\lim_{x \rightarrow 0}\frac{\sin^2{x}}{x}\]
\[= 2\lim_{x \rightarrow 0}\frac{\sin^2{x}}{x^2}\times{x}\]
\[= 2\left(\lim_{x \rightarrow 0}\frac{\sin{x}}{x}\right)^2\times{\lim_{x \rightarrow 0}x}\]
\[= 2(1)^2\times{0}\] ➜Note\[\because \lim_{\theta \rightarrow 0}\frac{\sin{\theta}}{\theta}=1\]
\[= 2\times{1}\times{0}\]
\[=0\]
\((b)\)
ধরি,
\(y=4x(6-x)^2 …….(1)\)
\(\Rightarrow y=4x(6^2-2.6.x+x^2)\)
\(\Rightarrow y=4x(36-12x+x^2)\)
\(\Rightarrow y=144x-48x^2+4x^3\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(144x-48x^2+4x^3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=144\frac{d}{dx}(x)-48\frac{d}{dx}(x^2)+3\frac{d}{dx}(x^3)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=144.1-48.2x+4.3x^2\) ➜Note\(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=144-96x+12x^2\)
\(=12x^2-96x+144\)
\(=12(x^2-8x+12)\)
\(=12(x^2-6x-2x+12)\)
\(=12\{x(x-6)-2(x-6)\}\)
\(=12(x-6)(x-2)\)
\(\therefore \frac{dy}{dx}=12(x-6)(x-2)\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 12(x-6)(x-2)=0\) ➜Note\(\because \frac{dy}{dx}=12(x-6)(x-2)\)
\(\Rightarrow (x-6)(x-2)=0\)
\(\Rightarrow x=6, x=2\)
\(\therefore x=6, 2\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(144-96x+12x^2)\) ➜Note\(\because \frac{dy}{dx}=12(x-6)(x-2)=144-96x+12x^2\)
\(=\frac{d}{dx}(144)-96\frac{d}{dx}(x)+12\frac{d}{dx}(x^2)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=0-96.1+12.2x\) ➜Note\(\because \frac{d}{dx}(c)=0,\frac{d}{dx}(x)=1\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(=-96+24x\)
\(=24x-96\)
\(\therefore \frac{d^2y}{dx^2}=24(x-4)\)
এখন,
\(x=6\) হলে,
\(\frac{d^2y}{dx^2}=24(6-4)\)
\(=24.2\)
\(=48\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=6\) বিন্দুতে ফাংশনটির লঘুমান আছে।
আবার,
\(x=2\) হলে,
\(\frac{d^2y}{dx^2}=24(2-4)\)
\(=24.(-2)\)
\(=-48\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore x=2\) বিন্দুতে ফাংশনটির গুরুমান আছে।
ফাংশনটির গুরুমান \( =4.2(6-2)^2\) ➜Note \((1)\) হতে \(\because y=4x(6-x)^2\)
\(=8.4^2\)
\(=8.16\)
\(=128\)
\((c)\)
দেওয়া আছে,
\(f(x)=e^{\tan^{-1}{x}} …..(1)\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(e^{\tan^{-1}{x}})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow f^{\prime}(x)=e^{\tan^{-1}{x}}\frac{d}{dx}(\tan^{-1}{x})\) ➜Note \(\tan^{-1}{x}\) কে \(x\) মনে করে সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow f^{\prime}(x)=f(x)\frac{1}{1+x^2}\) ➜Note\(\because f(x)=e^{\tan^{-1}{x}}, \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)f^{\prime}(x)=f(x)\)
\(\Rightarrow \frac{d}{dx}\{(1+x^2)f^{\prime}(x)\}=\frac{d}{dx}\{f(x)\}\) ➜Note আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}\{f^{\prime}(x)\}+f^{\prime}(x)\frac{d}{dx}(1+x^2)=f^{\prime}(x)\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1+x^2)f^{\prime\prime}(x)+f^{\prime}(x)(0+2x)=f^{\prime}(x)\)
\(\Rightarrow (1+x^2)f^{\prime\prime}(x)+2xf^{\prime}(x)=f^{\prime}(x)\)
\(\Rightarrow (1+x^2)f^{\prime\prime}(x)+2xf^{\prime}(x)-f^{\prime}(x)=0\)
\(\therefore (1+x^2)f^{\prime\prime}(x)+(2x-1)f^{\prime}(x)=0\)
(Proved)

দেওয়া আছে,
\(f(x)=x^3-3x^2+3x\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(x^3-3x^2+3x)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow f^{\prime}(x)=\frac{d}{dx}(x^3)-3\frac{d}{dx}(x^2)+3\frac{d}{dx}(x)\) ➜Note\(\because \frac{d}{dx}(f(x))=f^{\prime}(x)\)\(\frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=3x^2-3.2x+3.1\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1\)
\(=3x^2-6x+3\)
\(\therefore f^{\prime}(x)=3x^2-6x+3\)
এখান,
\(x=2\)হলে,
\(f^{\prime}(2)=3.2^2-6.2+3\)
\(=3.4-12+3\)
\(=12-12+3\)
\(=3>0\)
\(\therefore f^{\prime}(2)>0\)
\(\therefore \) প্রদত্ত ফাংশনটি \(x=2\) বিন্দুতে বৃদ্ধি পায়।

\((a)\)
দেওয়া আছে,
\(f(x)=17-15x+9x^2-x^3 …..(1)\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(17-15x+9x^2-x^3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow f^{\prime}(x)=\frac{d}{dx}(17)-15\frac{d}{dx}(x)+9\frac{d}{dx}(x^2)-\frac{d}{dx}(x^3)\) ➜Note\(\because \frac{d}{dx}(f(x))=f^{\prime}(x)\)\(\frac{d}{dx}(u-v+w-..)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)-..\)
\(=0-15.1+9.2x-3x^2\) ➜Note\(\because \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x)=1\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(=-15+18x-3x^2\)
\(=-3(x^2-6x+5)\)
\(\therefore f^{\prime}(x)=-3(x^2-6x+5)\)
চরমবিদুর জন্য \(f^{\prime}(x)=0\)
\(\therefore -3(x^2-6x+5)=0\) ➜Note\(\because f^{\prime}(x)=-3(x^2-6x+5)\)
\(\Rightarrow x^2-6x+5=0\)
\(\Rightarrow x^2-5x-x+5=0\)
\(\Rightarrow x(x-5)-1(x-5)=0\)
\(\Rightarrow (x-5)(x-1)=0\)
\(\Rightarrow x-5=0, x-1=0\)
\(\Rightarrow x=5, x=1\)
\(\therefore x=5, 1\)
\((1)\) হতে,
যখন, \(x=5\)
\(\therefore f(5)=17-15.5+9.5^2-5^3\)
\(=17-75+9.25-125\)
\(=17-75+225-125\)
\(=242-200\)
\(=42\)
আবার,
যখন, \(x=1\)
\(\therefore f(1)=17-15.1+9.1^2-1^3\)
\(=17-15+9.1-1\)
\(=17-15+9-1\)
\(=26-16\)
\(=10\)
\(\therefore \) প্রদত্ত ফাংশনের চরমবিন্দু \((1, 10); (5, 42)\)
\((b)\)
\((a)\) হতে প্রাপ্ত
\( x=5, 1\) মানদ্বয় সকল বাস্তব সংখ্যাকে \(x>5, 5>x>1\) এবং \(1>x\) ব্যবধিতে বিভক্ত করে।
এখন,
\(1>x\) এর জন্য \(0>(x-1)\) ও \(0>(x-5)\), কাজেই \(0>f^{\prime}(x)\)
\(\therefore 1>x>-\infty\) ব্যবধিতে \(f(x)\) ফাংশন হ্রাস পায়।
আবার,
\(5>x>1\) এর জন্য \((x-1)>0\) ও \(0>(x-5)\), কাজেই \(f^{\prime}(x)>0\)
\(\therefore 5>x>1\) ব্যবধিতে \(f(x)\) ফাংশন বৃদ্ধি পায়।
অপরপক্ষে \(x>5\) এর জন্য \((x-1)>0\) ও \((x-5)>0\), কাজেই \(f^{\prime}(x)>0\)
\(x>5\) ব্যবধিতে \(f(x)\) ফাংশন হ্রাস পায়।
\((c)\)
\((a)\) হতে প্রাপ্ত
\(f^{\prime}(x)=-3(x^2-6x+5)\)
\(\Rightarrow \frac{d}{dx}\{f^{\prime}(x)\}=-3\frac{d}{dx}(x^2-6x+5)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow f^{\prime\prime}(x)=-3\frac{d}{dx}(x^2)+18\frac{d}{dx}(x)-3\frac{d}{dx}(5)\)
\(=-3.2x+18.1-3.0\)
\(\therefore f^{\prime\prime}(x)=-6x+18\)
\(\Rightarrow f^{\prime\prime}(1)=-6.1+18\)
\(=-6+18\)
\(=12>0\)
\(\therefore f^{\prime\prime}(x)>0\)
\(\therefore x=1\) বিন্দুতে \(f(x)\) এর সর্বনিম্ন মান আছে।
\(\therefore f(x)\) এর সর্বনিম্ন মান \(f(1)=17-15.1+9.1^2-1^3\) ➜Note \((1)\) হতে, \(\because f(x)=17-15x+9x^2-x^3\)
\(=17-15+9-1\)
\(=26-16\)
\(=10\)
আবার,
\(\Rightarrow f^{\prime\prime}(5)=-6.5+18\)
\(=-30+18\)
\(=-12<{0}\)
\(\therefore f^{\prime\prime}(x)<{0}\)
\(\therefore x=5\) বিন্দুতে \(f(x)\) এর সর্বোচ্চ মান আছে।
\(\therefore f(x)\) এর সর্বোচ্চ মান \(f(5)=17-15.5+9.5^2-5^3\) ➜Note \((1)\) হতে, \(\because f(x)=17-15x+9x^2-x^3\)
\(=17-75+9.25-125\)
\(=17-75+225-125\)
\(=242-200\)
\(=42\)
\((d)\)
ফাংশনটির লেখচিত্র।

ধরি,
\(y=f(x)=2x^3-21x^2+36x-20 ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(2x^3-21x^2+36x-20)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=2\frac{d}{dx}(x^3)-21\frac{d}{dx}(x^2)+36\frac{d}{dx}(x)-\frac{d}{dx}(20)\) ➜Note\(\because \frac{d}{dx}(u-v+w-..)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)-..\)
\(=2.3x^2-21.2x+36.1-0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=6x^2-42x+36\)
\(=6(x^2-7x+6)\)
\(=6(x^2-6x-x+6)\)
\(=6\{x(x-6)-1(x-6)\}\)
\(=6(x-6)(x-1)\)
\(\therefore \frac{dy}{dx}=6(x-6)(x-1)\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 6(x-6)(x-1)=0\) ➜Note\(\because \frac{dy}{dx}=6(x-6)(x-1)\)
\(\Rightarrow (x-6)(x-1)=0\)
\(\Rightarrow x-6=0, x-1=0\)
\(\Rightarrow x=6, x=1\)
\(\therefore x=6, 1\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(6x^2-42x+36)\) ➜Note\(\because \frac{dy}{dx}=6(x-6)(x-1)=6x^2-42x+36\)
\(=6\frac{d}{dx}(x^2)-42\frac{d}{dx}(x)+\frac{d}{dx}(36)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=6.2x-42.1+0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\) \(\frac{d}{dx}(x)=1,\frac{d}{dx}(c)=0\),
\(=12x-42\)
\(\therefore \frac{d^2y}{dx^2}=12x-42\)
এখন, \(x=1\) হলে,
\(\frac{d^2y}{dx^2}=12.1-42\)
\(=12-42\)
\(=-30\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore x=1\) বিন্দুতে ফাংশনটির গরিষ্ঠমান আছে।
ফাংশনটির গরিষ্ঠমান \(=2.1^3-21.1^2+36.1-20 \) ➜Note \((1)\) হতে \(\because y=2x^3-21x^2+36x-20\)
\(=2-21+36-20\)
\(=38-41\)
\(=-3\)
আবার, \(x=6\) হলে,
\(\frac{d^2y}{dx^2}=12.6-42\)
\(=72-42\)
\(=3-\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=6\) বিন্দুতে ফাংশনটির লঘিষ্ঠমান আছে।
ফাংশনটির লঘিষ্ঠমান \(=2.6^3-21.6^2+36.6-20\) ➜Note \((1)\) হতে \(\because y=2x^3-21x^2+36x-20\)
\(=432-756+216-20\)
\(=648-776\)
\(=-128\)

ধরি,
\(y=f(x)=x^3-3x^2+6x+3 ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3-3x^2+6x+3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)-3\frac{d}{dx}(x^2)+6\frac{d}{dx}(x)-\frac{d}{dx}(3)\) ➜Note\(\because \frac{d}{dx}(u-v+w+..)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)+..\)
\(=3x^2-3.2x+6.1+0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=3x^2-6x+6\)
\(=3(x^2-2x)+6\)
\(=3(x^2-2x+1-1)+6\)
\(=3(x^2-2x+1)-3+6\)
\(=3(x-1)^2+3\)
\(=3\{(x-1)^2+1\}\)
\(\therefore \frac{dy}{dx}=3\{(x-1)^2+1^2\}\) ➜Note একাধিক রাশির বর্গের যোগফল ঋনাত্মক হতে পারে না।
যা, সর্বদাই ধনাত্মক।
\(\therefore x\) এর কোনো বাস্তব মানের জন্য \(\frac{dy}{dx}\) কখনো শূন্য হতে পারে না।
\(\therefore \) প্রদত্ত ফাংশনটির কোনো সর্বোচ্চ মান অথবা সর্বনিম্ন মান নেই।

ধরি,
\(y=3\sin^2{x}+4\cos^2{x} ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(3\sin^2{x}+4\cos^2{x})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=3\frac{d}{dx}(\sin^2{x})+4\frac{d}{dx}(\cos^2{x})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(=3.2\sin{x}\cos{x}+4.2\cos{x}(-\sin{x})\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\),\(\frac{d}{dx}(\sin{x})=\cos{x}\), \(\frac{d}{dx}(\cos{x})=-\sin{x}\)
\(=3.2\sin{x}\cos{x}-4.2\sin{x}\cos{x}\)
\(=3\sin{2x}-4\sin{2x}\)
\(\therefore \frac{dy}{dx}=-\sin{2x}\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow -\sin{2x}=0\) ➜Note\(\because \frac{dy}{dx}=-\sin{2x}\)
\(\Rightarrow \sin{2x}=0\)
\(\Rightarrow 2x=0, \frac{\pi}{2}\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(-\sin{2x})\) ➜Note\(\because \frac{dy}{dx}=-\sin{2x}\)
\(=-\cos{2x}.2\)
\(=-2\cos{2x}\)
\(\therefore \frac{d^2y}{dx^2}=-2\cos{2x}\)
এখন, \(x=0\) হলে,
\(\frac{d^2y}{dx^2}=-2\cos{2.0}\)
\(=-2\cos{0}\)
\(=-2.1\)
\(=-2\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore x=\frac{\pi}{2}\) বিন্দুতে ফাংশনটির সর্বোচ্চ মান আছে।
ফাংশনটির সর্বোচ্চ মান \(=3\sin^2{0}+4\cos^2{0}\) ➜Note \((1)\) হতে \(\because y=3\sin^2{x}+4\cos^2{x}\)
ফাংশনটির লঘুমান \(=3.0+4.1\)
\(=0+4\)
\(=4\)
আবার, \(x=\frac{\pi}{2}\) হলে,
\(\frac{d^2y}{dx^2}=-2\cos{2.\frac{\pi}{2}}\)
\(=-2\cos{\pi}\)
\(=-2\times{-1}\)
\(=2\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore \sin{x}=\frac{1}{3}\) বিন্দুতে ফাংশনটির সর্বনিম্ন মান আছে।
ফাংশনটির সর্বনিম্ন মান \(=3\sin^2{\frac{\pi}{2}}+4\cos^2{\frac{\pi}{2}}\) ➜Note \((1)\) হতে \(\because y=3\sin^2{x}+4\cos^2{x}\)
ফাংশনটির লঘুমান \(=3.1^2+4.0^2\)
\(=3.1+4.0\)
\(=3+0\)
\(=3\)

ধরি,
\(y=f(x)=x^3-18x^2+96x ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3-18x^2+96x)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)-18\frac{d}{dx}(x^2)+96\frac{d}{dx}(x)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=3x^2-18.2x+96.1\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=3x^2-36x+96\)
\(=3(x^2-12x+32)\)
\(=3\{x^2-8x-4x+32\}\)
\(=3\{x(x-8)-4(x-8)\}\)
\(=3(x-8)(x-4)\)
\(\therefore \frac{dy}{dx}=3(x-8)(x-4)\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 3(x-8)(x-4)=0\) ➜Note\(\because \frac{dy}{dx}=3(x-8)(x-4)\)
\(\Rightarrow 3(x-8)(x-4)=0\)
\(\Rightarrow (x-8)(x-4)=0\)
\(\Rightarrow x-8=0, x-4=0\)
\(\Rightarrow x=8, x=4\)
\(\therefore x=8, 4\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(3x^2-36x+96)\) ➜Note\(\because \frac{dy}{dx}=3(x-8)(x-4)=3x^2-36x+96\)
\(=3\frac{d}{dx}(x^2)-36\frac{d}{dx}(x)+\frac{d}{dx}(96)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=3.2x-36.1+0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\) \(\frac{d}{dx}(x)=1,\frac{d}{dx}(c)=0\),
\(=6x-36\)
\(\therefore \frac{d^2y}{dx^2}=6x-36\)
এখন, \(x=4\) হলে,
\(\frac{d^2y}{dx^2}=6.4-36\)
\(=24-36\)
\(=-12\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore x=4\) বিন্দুতে ফাংশনটির সর্বোচ্চ মান আছে।
ফাংশনটির সর্বোচ্চ মান \(=4^3-18.4^2+96.4 \) ➜Note \((1)\) হতে \(\because y=x^3-18x^2+96x\)
\(=64-288+384\)
\(=448-288\)
\(=160\)
আবার, \(x=8\) হলে,
\(\frac{d^2y}{dx^2}=6.8-36\)
\(=48-36\)
\(=12\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=8\) বিন্দুতে ফাংশনটির সর্বনিম্ন মান আছে।
ফাংশনটির সর্বনিম্ন মান \(=8^3-18.8^2+96.8 \) ➜Note \((1)\) হতে \(\because y=x^3-18x^2+96x\)
\(=512-1152+768\)
\(=1280-1152\)
\(=128\)

দেওয়া আছে,
\(f(x)=3x^2-2x-4\) একটি ফাংশন।
\((a)\)
ধরি,
\(y=f(x)=x^{\sin^{-1}{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{\sin^{-1}{x}})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{\sin^{-1}{x}}\left(\frac{\sin^{-1}{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\sin^{-1}{x})\right)\) ➜Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\sin^{-1}{x}}\left(\frac{\sin^{-1}{x}}{x}.1+\ln{x}\times{\frac{1}{\sqrt{1-x^2}}}\right)\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{dy}{dx}=x^{\sin^{-1}{x}}\left(\frac{\sin^{-1}{x}}{x}+\frac{\ln{x}}{\sqrt{1-x^2}}\right)\)
\((b)\)
ধরি,
\(y=f(x)=3x^2-2x-4 ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(3x^2-2x-4)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=3\frac{d}{dx}(x^2)-2\frac{d}{dx}(x)-\frac{d}{dx}(4)\) ➜Note\(\because \frac{d}{dx}(u-v-w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)-\frac{d}{dx}(w)\)
\(=3.2x-2.1-0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=6x-2\)
\(\therefore \frac{dy}{dx}=6x-2\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 6x-2=0\) ➜Note\(\because \frac{dy}{dx}=6x-2\)
\(\Rightarrow 6x-2=0\)
\(\Rightarrow 6x=2\)
\(\Rightarrow x=\frac{2}{6}\)
\(\therefore x=\frac{1}{3}\)
\(x=\frac{1}{3}\) বিন্দুটি সকল বাস্তব সংখ্যাকে \(\frac{1}{3}>x\) ও \(x>\frac{1}{3}\) ব্যবধিতে বিভক্ত করে।
\(\frac{1}{3}>x\) ব্যবধিতে \(0>\frac{dy}{dx}\)
\(\therefore \frac{1}{3}>x\) ব্যবধিতে ফাংশনটি হ্রাস পায়।
আবার,
\(x>\frac{1}{3}\) ব্যবধিতে \(\frac{dy}{dx}>0\)
\(\therefore x>\frac{1}{3}\) ব্যবধিতে ফাংশনটি বৃদ্ধি পায়।
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(6x-2)\) ➜Note\(\because \frac{dy}{dx}=6x-2\)
\(=6.1-0\) ➜Note\(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=6\)
এখন, \(x=\frac{1}{3}\) হলে,
\(\frac{d^2y}{dx^2}=6\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=\frac{1}{3}\) বিন্দুতে ফাংশনটির লঘুমান আছে।
ফাংশনটির লঘুমান \(=3\left(\frac{1}{3}\right)^2-2\left(\frac{1}{3}\right)-4\) ➜Note \((1)\) হতে \(\because y=3x^2-2x-4\)
\(=3\left(\frac{1}{9}\right)-\frac{2}{3}-4\)
\(=\frac{1-2-12}{3}\)
\(=\frac{1-14}{3}\)
\(=\frac{-13}{3}\)
\(=-\frac{13}{3}\)
\((c)\)
ধরি,
\(y=f(x)=3x^2-2x-4 ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(3x^2-2x-4)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=3\frac{d}{dx}(x^2)-2\frac{d}{dx}(x)-\frac{d}{dx}(4)\) ➜Note\(\because \frac{d}{dx}(u-v-w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)-\frac{d}{dx}(w)\)
\(=3.2x-2.1-0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=6x-2\)
\(\therefore\) স্পর্শকের ঢাল \(=6x-2\)
আবার,
\(x+10y-7=0\) রেখার ঢাল \(=-\frac{1}{10}\) ➜Note\(\because ax+by+c=0\) রেখার ঢাল \(=-\frac{a}{b}\)
শর্তমতে,
\((6x-2)\times{-\frac{1}{10}}=-1\)
\(\Rightarrow \frac{6x-2}{10}=1\)
\(\Rightarrow 6x-2=10\)
\(\Rightarrow 6x=10+2\)
\(\Rightarrow 6x=12\)
\(\therefore x=2\)
আবার,
\((1)\) হতে,
\(y=3.2^2-2.2-4\)
\(=3.4-4-4\)
\(=12-8\)
\(=4\)
\(\therefore (2, 4)\) বিন্দুতে স্পর্শক \(x+10y-7=0\) রেখার উপর লম্ব হবে।
\((2, 4)\) বিন্দুতে স্পর্শকের ঢাল \(=6.2-2\) ➜Note স্পর্শকের ঢাল \(\because \frac{dy}{dx}=6x-2\)
\(=12-2\)
\(=10\)
\(\therefore (2, 4)\) বিন্দুতে স্পর্শকের সমীকরণ \(y-4=10(x-2)\) ➜Note \((x_{1}, y_{1})\) বিন্দুগামী এবং \(m\) ঢালবিশিষ্ঠ সরলরেখার সমীকরণ \(\because y-y_{1}=m(x-x_{1})\)
\(\Rightarrow y-4=10x-20\)
\(\Rightarrow 10x-20=y-4\)
\(\Rightarrow 10x-20-y+4=0\)
\(\Rightarrow 10x-y-16=0\)

দেওয়া আছে,
দৃশ্যকল্প-১: \(\tan{\theta}=\frac{a+bx}{a-bx}\)এবং দৃশ্যকল্প-২: \(f(x)=\cos{x}\)
\((a)\)
দৃশ্যকল্প-২: \(f(x)=\cos{x}\)
ধরি,
\(f(x)=\cos{x} …..(1)\)
\((1)\) হতে,
\(f\left(\frac{\pi}{2}-x\right)=\cos{\left(\frac{\pi}{2}-x\right)}\)
\(=\sin{x}\) ➜Note \(\because \) প্রথম চৌকোণে অবস্থিত।
এখন,
\[\lim_{x \rightarrow \frac{\pi}{2}}\frac{1-f\left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)^2}\]
\[=\lim_{x \rightarrow \frac{\pi}{2}}\frac{1-\sin{x}}{\left(\frac{\pi}{2}-x\right)^2}\] ➜Note \(\because f\left(\frac{\pi}{2}-x\right)=\sin{x}\)
আবার,
ধরি,
\(x=\frac{\pi}{x}+h\)
যখন, \(x\rightarrow \frac{\pi}{x}\)
তখন, \(h\rightarrow 0\)
\[=\lim_{h \rightarrow 0}\frac{1-\sin{\left(\frac{\pi}{x}+h\right)}}{\left(\frac{\pi}{2}-\frac{\pi}{x}-h\right)^2}\]
\[=\lim_{h \rightarrow 0}\frac{1-\cos{h}}{(-h)^2}\]
\[=\lim_{h \rightarrow 0}\frac{2\sin^2{\frac{h}{2}}}{h^2}\] ➜Note \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\[=2\left(\lim_{h \rightarrow 0}\frac{\sin{\frac{h}{2}}}{\frac{h}{2}}\right)^2\times{\frac{1}{4}}\]
\[=2.1^2\times{\frac{1}{4}}\] ➜Note \[\because \lim_{\theta \rightarrow 0}\frac{\sin{\theta}}{\theta}=1\]
\[=2\times{\frac{1}{4}}\] ➜Note \(\because \lim_{\theta \rightarrow 0}\frac{\sin{\theta}}{\theta}=1\)
\[=\frac{1}{2}\]
\((b)\)
দৃশ্যকল্প-১ হতে,
দেওয়া আছে,
\(\tan{\theta}=\frac{a+bx}{a-bx}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{a+bx}{a-bx}\right)}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{1+\frac{bx}{a}}{1-\frac{bx}{a}}\right)}\) ➜Note লব ও হরের সহিত \(a\) ভাগ করে।
ধরি,
\(\tan{\alpha}=\frac{bx}{a}\Rightarrow \alpha=\tan^{-1}\left(\frac{bx}{a}\right)\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{1+\tan{\alpha}}{1-\tan{\alpha}}\right)}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{\tan{\frac{\pi}{4}}+\tan{\alpha}}{1-\tan{\frac{\pi}{4}}\tan{\alpha}}\right)}\) ➜Note \(\because \tan{\frac{\pi}{4}}=1\)
\(\Rightarrow \theta=\tan^{-1}{\tan{\left(\frac{\pi}{4}+\alpha\right)}}\) ➜Note \(\because \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
\(\Rightarrow \theta=\frac{\pi}{4}+\alpha\)
\(\Rightarrow \theta=\frac{\pi}{4}+\tan^{-1}\left(\frac{bx}{a}\right)\)
\(\Rightarrow \frac{d\theta}{dx}=\frac{d}{dx}\left(\frac{\pi}{4}\right)+\frac{d}{dx}\{\tan^{-1}\left(\frac{bx}{a}\right)\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=0+\frac{1}{1+\left(\frac{bx}{a}\right)^2}\frac{d}{dx}\left(\frac{bx}{a}\right)\) ➜Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\)
\(=\frac{1}{1+\frac{b^2x^2}{a^2}}.\frac{b}{a}\)
\(=\frac{1}{\frac{a^2+b^2x^2}{a^2}}.\frac{b}{a}\)
\(=\frac{a^2}{a^2+b^2x^2}.\frac{b}{a}\)
\(=\frac{ab}{a^2+b^2x^2}\)
\((c)\)
ধরি,
\(y=f(2\sin^{-1}{x}) ….(1)\)
দৃশ্যকল্প-২ হতে,
\(f(x)=\cos{x}\)
\(\Rightarrow f(2\sin^{-1}{x})=\cos{(2\sin^{-1}{x})}\)
\(\therefore y=\cos{(2\sin^{-1}{x})}\) ➜Note \((1)\) হতে, \(\because y=f(2\sin^{-1}{x})\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(2\sin^{-1}{x})}\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(2\sin^{-1}{x})}\frac{d}{dx}(2\sin^{-1}{x})\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=-\sin{(2\sin^{-1}{x})}.2\frac{1}{\sqrt{1-x^2}}\) ➜Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow y_{1}\sqrt{1-x^2}=-2\sin{(2\sin^{-1}{x})}\)
\(\Rightarrow y^2_{1}(1-x^2)=4\sin^2{(2\sin^{-1}{x})}\)| Note উভয় পার্শে বর্গ করে।
\(\Rightarrow y^2_{1}(1-x^2)=4\{1-\cos^2{(2\sin^{-1}{x})}\}\)| Note \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow y^2_{1}(1-x^2)=4(1-y^2)\)| Note \(\because y=\cos{(2\sin^{-1}{x})}\)
\(\Rightarrow y^2_{1}(1-x^2)=4-4y^2\)
\(\Rightarrow \frac{d}{dx}\{y^2_{1}(1-x^2)\}=\frac{d}{dx}(4)-4\frac{d}{dx}(y^2)\) ➜Note আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y^2_{1}\frac{d}{dx}(1-x^2)+(1-x^2)\frac{d}{dx}(y^2_{1})=0-4.2yy_{1}\) ➜Note\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow y^2_{1}(-2x)+(1-x^2)2y_{1}y_{2}=-8yy_{1}\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}\}=-8yy_{1}\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}=-\frac{8yy_{1}}{2y_{1}}\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}=-4y\)
\(\therefore (1-x^2)y_{2}-xy_{1}+4y=0\)
(Showed)

ধরি,
\(y=f(x)=x^4-4x^3+4x^2+5 ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^4-4x^3+4x^2+5)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^4)-4\frac{d}{dx}(x^3)+4\frac{d}{dx}(x^2)+\frac{d}{dx}(5)\) ➜Note\(\because \frac{d}{dx}(u-v+w+..)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)+..\)
\(=4x^3-4.3x^2+4.2x+0\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=4x^3-12x^2+8x\)
\(=4x(x^2-3x+2)\)
\(=4x\{x^2-2x-x+2\}\)
\(=4x\{x(x-2)-1(x-2)\}\)
\(=4x(x-2)(x-1)\)
\(\therefore \frac{dy}{dx}=4x(x-2)(x-1)\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 4x(x-2)(x-1)=0\) ➜Note\(\because \frac{dy}{dx}=4x(x-2)(x-1)\)
\(\Rightarrow x(x-2)(x-1)=0\)
\(\Rightarrow x=0, x-2=0, x-1=0\)
\(\Rightarrow x=0, x=2, x=1\)
\(\therefore x=0, 2, 1\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(4x^3-12x^2+8x)\) ➜Note\(\because \frac{dy}{dx}=4x(x-2)(x-1)=4x^3-12x^2+8x\)
\(=4\frac{d}{dx}(x^3)-12\frac{d}{dx}(x^2)+8\frac{d}{dx}(x)\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(=4.3x^2-12.2x+8.1\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(x)=1\)
\(=12x^2-24x+8\)
\(\therefore \frac{d^2y}{dx^2}=12x^2-24x+8\)
এখন, \(x=0\) হলে,
\(\frac{d^2y}{dx^2}=12.0^2-24.0+8\)
\(=12.0-24.0+8\)
\(=0-0+8\)
\(=8\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=0\) বিন্দুতে ফাংশনটির লঘিষ্ঠমান আছে।
ফাংশনটির লঘিষ্ঠমান \(=0^4-4.0^3+4.0^2+5\) ➜Note \((1)\) হতে \(\because y=x^4-4x^3+4x^2+5\)
\(=0-4.0+4.0+5\)
\(=0-0+0+5\)
\(=5\)
আবার, \(x=2\) হলে,
\(\frac{d^2y}{dx^2}=12.2^2-24.2+8\)
\(=12.4-48+8\)
\(=48-48+8\)
\(=8\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=2\) বিন্দুতে ফাংশনটির লঘিষ্ঠমান আছে।
ফাংশনটির লঘিষ্ঠমান \(=2^4-4.2^3+4.2^2+5\) ➜Note \((1)\) হতে \(\because y=x^4-4x^3+4x^2+5\)
\(=16-4.8+4.4+5\)
\(=16-32+16+5\)
\(=32-32+5\)
\(=5\)
আবার, \(x=1\) হলে,
\(\frac{d^2y}{dx^2}=12.1^2-24.1+8\)
\(=12.1-24+8\)
\(=12-24+8\)
\(=20-24\)
\(=-4\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore x=1\) বিন্দুতে ফাংশনটির গরিষ্ঠ মান আছে।
ফাংশনটির গরিষ্ঠ মান \(=1^4-4.1^3+4.1^2+5\) ➜Note \((1)\) হতে \(\because y=x^4-4x^3+4x^2+5\)
\(=1-4.1+4.1+5\)
\(=1-4+4+5\)
\(=6\)

ধরি,
\(y=x^3+\frac{1}{x^3} ……(1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(x^3+\frac{1}{x^3}\right)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)+\frac{d}{dx}\left(\frac{1}{x^3}\right)\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)+\frac{d}{dx}(x^{-3})\)
\(=3x^2-3x^{-3-1}\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=3x^2-3x^{-4}\)
\(=3(x^2-x^{-4})\)
\(\therefore \frac{dy}{dx}=3(x^2-x^{-4})\)
ফাংশনটির গুরুমান ও লঘুমানের জন্য \(\frac{dy}{dx}=0\)
\(\Rightarrow 3(x^2-x^{-4})=0\) ➜Note\(\because \frac{dy}{dx}=3(x^2-x^{-4})\)
\(\Rightarrow x^2-x^{-4}=0\)
\(\Rightarrow x^2-\frac{1}{x^4}=0\)
\(\Rightarrow \frac{x^6-1}{x^4}=0\)
\(\Rightarrow x^6-1=0\)
\(\Rightarrow (x^3-1)(x^3+1)=0\)
\(\Rightarrow (x-1)(x^2+x+1)(x+1)(x^2-x+1)=0\)
\(\Rightarrow x-1=0, x+1=0\) ➜Note\(x\) এর বাস্তব মানের জন্য
\(\Rightarrow x=1, x=-1\)
\(\therefore x=1, -1\)
আবার,
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\)
\(=\frac{d}{dx}(x^2-x^{-4})\) ➜Note\(\because \frac{dy}{dx}=x^2-x^{-4}\)
\(=\frac{d}{dx}(x^2)-\frac{d}{dx}(x^{-4})\) ➜Note\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(=2x+4x^{-4-1}\) ➜Note\(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=2x+4x^{-5}\)
\(\therefore \frac{d^2y}{dx^2}=2x+4x^{-5}\)
এখন, \(x=1\) হলে,
\(\frac{d^2y}{dx^2}=2.1+4.1^{-5}\)
\(=2+4.1\)
\(=2+4\)
\(=6\)
\(\therefore \frac{d^2y}{dx^2}>0\)
\(\therefore x=1\) বিন্দুতে ফাংশনটির ক্ষুদ্রতম মান আছে।
ফাংশনটির ক্ষুদ্রতম মান \(=1^3+\frac{1}{1^3}\) ➜Note \((1)\) হতে \(\because y=x^3+\frac{1}{x^3}\)
\(=1+\frac{1}{1}\)
\(=1+1\)
\(=2\)
আবার, \(x=-1\) হলে,
\(\frac{d^2y}{dx^2}=2.(-1)+4.(-1)^{-5}\)
\(=-2+4.(-1)\)
\(=-2-4\)
\(=-6\)
\(\therefore 0>\frac{d^2y}{dx^2}\)
\(\therefore x=-1\) বিন্দুতে ফাংশনটির বৃহত্তম মান আছে।
ফাংশনটির বৃহত্তম মান \(=(-1)^3+\frac{1}{(-1)^3}\) ➜Note \((1)\) হতে \(\because y=x^3+\frac{1}{x^3}\)
\(=-1+\frac{1}{-1}\)
\(=-1-1\)
\(=-2\)
\(\therefore \) ফাংশনটির বৃহত্তম মান ক্ষুদ্রতম মান অপেক্ষা ক্ষুদ্রতর।
(Showed)