এখন, ${\int }_0^1\frac{{\mathrm{xe}}^{\mathrm{x}}}{{\left(1+\mathrm{x}\right)}^2}$

$={\mathrm{e}}^{\mathrm{x}}\left\{\frac{1}{\left(1+\mathrm{x}\right)}+\frac{\left(-1\right)}{{\left(1+\mathrm{x}\right)}^2}\right\}={\mathrm{e}}^{\mathrm{x}}/\left(1+\mathrm{x}\right)$

$\therefore {\int }_0^1\frac{e^{x}x}{{\left(1+x\right)}^2}={\left[\frac{e^x}{\left(1+x\right)}\right]}_0^1$

$=\left(\frac{e}{2}-\frac{e^{\circ }}{1}\right)=\frac{e}{2}-1 \left(Ans\right)$

$\left(a^6-a^5\right)-\left(a^4-a^3\right)+\left(a^3-a\right)$

$=a^5-a^6-a^4+2a^3-a \left(Ans\right)$

ax+by=0……..(i) এবং bx-ay+c=0…….(ii)

(i) নং এবং (ii) নং রেখার ছেদবিন্দুগামী রেখার সমীকরণ

ax+by+k(bx-ay+c)=0

$\Rightarrow$(a+kb)x+b(b-ka)y+kc=0……(iii)

(iii) নং রেখা x অক্ষের সমান্তরাল

$\therefore a+kb=0 \therefore k=-\frac{a}{b}$

(iii) নং হতে $\left(\mathrm{b}+\frac{\mathrm{a}}{\mathrm{b}}\mathrm{a}\right)\mathrm{y}-\frac{\mathrm{a}}{\mathrm{b}}\mathrm{c}=0$

$\Rightarrow \left({\mathrm{b}}^2+{\mathrm{a}}^2\right)\mathrm{y}-\mathrm{ac}=0 \left(\mathrm{Ans}\right)$

ধরি, পরাবৃত্তের উপর চলমান একটি বিন্দু (x, y)

সংজ্ঞানুসারে, PS=PM

$\Rightarrow \sqrt{{\left(\mathrm{x}+8\right)}^2+{\left(\mathrm{y}+2\right)}^2}=\frac{\left|2\mathrm{x}-\mathrm{y}-9\right|}{\sqrt{5}}$

$\Rightarrow \left\{{\left(\mathrm{x}+8\right)}^2+{\left(\mathrm{y}+2\right)}^2\right\}\times 5={\left(2\mathrm{x}-\mathrm{y}-9\right)}^2$

$\Rightarrow 5\left({\mathrm{x}}^2+16\mathrm{x}+64+{\mathrm{y}}^2+4\mathrm{y}+4\right)=4{\mathrm{x}}^2+{\mathrm{y}}^2+81-4\mathrm{xy}+18\mathrm{y}-36\mathrm{x}$

$\Rightarrow {\mathrm{x}}^2+4{\mathrm{y}}^2+116\mathrm{x}+2\mathrm{y}+4\mathrm{xy}+259=0 \left(\mathrm{Ans}\right)$

$\cos \mathrm{ec}\mathrm{\theta }+cot\mathrm{\theta }=\sqrt{3}$

$\Rightarrow \frac{1+cos\mathrm{\theta }}{\sin \mathrm{\theta }}=\sqrt{3}$

$\Rightarrow \frac{2co{s}^2\frac{\mathrm{\theta }}{2}}{2sin\frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}}=\sqrt{3}$

$\Rightarrow cos\frac{\mathrm{\theta }}{2}\left(\cos \frac{\mathrm{\theta }}{2}-\sqrt{3}\sin \frac{\mathrm{\theta }}{2}\right)=0$

হয়, $\cos \frac{\mathrm{\theta }}{2}-\sqrt{3}\sin \frac{\mathrm{\theta }}{2}=0$

$\Rightarrow \cos \frac{\mathrm{\theta }}{2}=\sqrt{3}$

অথবা, $cos\frac{\mathrm{\theta }}{2}=0$

$\therefore \theta =\left(2n+1\right)\pi$

$\Rightarrow tan\frac{\theta }{2}=\frac{1}{\sqrt{3}}=tan\frac{\pi }{6}$ [ $\theta$ এর এই মান শুদ্ধি পরীক্ষা দ্বারা সমীকরণ সিদ্ধ করে না]

$\therefore \frac{\theta }{2}=n\pi +\frac{\pi }{6} \therefore \theta =2n\pi +\frac{\pi }{3}$ ( যেখানে n শূন্য বা যেকোন পূর্ণ সংখ্যা) Ans.

$\frac{{\mathrm{x}}^2}{\mathrm{p}}+\frac{{\mathrm{y}}^2}{5^2}=1$

$\Rightarrow \frac{6^2}{\mathrm{p}}+\frac{16}{25}=1$[ $\because \left(6, 4\right)$ বিন্দুগামী]

$\Rightarrow \mathrm{p}=100 \therefore \frac{{\mathrm{x}}^2}{{10}^2}+\frac{{\mathrm{y}}^2}{5^2}=1 \mathrm{Ans}.$

$\mathrm{a}=10, \mathrm{b}=5 \therefore \mathrm{e}=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{1-{\left(\frac{5}{10}\right)}^2}=\frac{\sqrt{3}}{2} \mathrm{Ans}.$

দেওয়া আছে, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+3\mathrm{x}+1$ এবং $\mathrm{g}\left(\mathrm{x}\right)=2\mathrm{x}-3$

$\therefore \mathrm{fog}\left(\mathrm{x}\right)=\mathrm{f}\left(2\mathrm{x}-3\right)={\left(2\mathrm{x}-3\right)}^2+3\left(2\mathrm{x}-3\right)+1$

           $=4{\mathrm{x}}^2-6\mathrm{x}+1 \mathrm{Ans}.$

এবং $\mathrm{fof}\left(\mathrm{x}\right)=\mathrm{f}\left({\mathrm{x}}^2+3\mathrm{x}+1\right)$

$={\left({\mathrm{x}}^2+3\mathrm{x}+1\right)}^2+3\left({\mathrm{x}}^2+3\mathrm{x}+1\right)+1$

$={\mathrm{x}}^4+6{\mathrm{x}}^3+14{\mathrm{x}}^2+15\mathrm{x}+5 \mathrm{Ans}.$

$\sqrt[3]{\mathrm{x}+\mathrm{iy}}=\mathrm{p}+\mathrm{iq}$

$\Rightarrow \mathrm{x}+\mathrm{iy}={\mathrm{p}}^3+3\left(\mathrm{iq}\right){\mathrm{p}}^2+3\mathrm{p}{\left(\mathrm{iq}\right)}^2+{\mathrm{i}}^3{\mathrm{q}}^3$

$={\mathrm{p}}^3-3{\mathrm{pq}}^2+\mathrm{i}\left(3{\mathrm{p}}^2\mathrm{q}-{\mathrm{q}}^3\right)$

$\therefore \mathrm{x}={\mathrm{p}}^3-3{\mathrm{pq}}^2$

$\mathrm{y}=3{\mathrm{p}}^2\mathrm{q}-{\mathrm{q}}^3$

$\therefore \mathrm{R}.\mathrm{S}=\frac{\mathrm{x}}{\mathrm{p}}+\frac{\mathrm{y}}{\mathrm{q}}=\frac{{\mathrm{p}}^3-3{\mathrm{pq}}^2}{\mathrm{p}}+\frac{3{\mathrm{p}}^2\mathrm{q}-{\mathrm{p}}^3}{\mathrm{q}}$

$={\mathrm{p}}^2-3{\mathrm{q}}^2+3{\mathrm{p}}^2-{\mathrm{q}}^2=4\left({\mathrm{p}}^2-{\mathrm{q}}^2\right)=\mathrm{L}.\mathrm{S}\left(\mathrm{Proved}\right)$

$\sqrt{\frac{\mathrm{q}}{\mathrm{p}}}=-\sqrt{\frac{\mathrm{q}}{\mathrm{p}}}+\sqrt{\frac{\mathrm{q}}{\mathrm{p}}}=0$

$\therefore \mathrm{L}.\mathrm{S}=\mathrm{R}.\mathrm{S} \left(\mathrm{Proved}\right)$

মোট বর্ণ = 11 টি ।

E=3টি, N=3টি, G=2টি, I=2টি এবং R=1 টি

$\therefore$ মোট বিন্যাস সংখ্যা$=\frac{11!}{3!3!2!2!}$

3 টি E একত্রে এমন বিন্যাস$=\frac{9!}{3!2!2!}$  ভাবে 

3টি E কে প্রথমে রেখে,

বাকি 8টি বর্ণকে সাজানো যায়$=\frac{8!}{3!2!2!}$ উপায়ে। Ans.

r তম পদ, ${\mathrm{U}}_{\mathrm{r}}=1^2+2^2+3^2+.....+{\mathrm{r}}^2$

$=\frac{\mathrm{r}\left(\mathrm{r}+1\right)\left(2\mathrm{r}+1\right)}{6}=\frac{\left({\mathrm{r}}^2+\mathrm{r}\right)\left(2\mathrm{r}+1\right)}{6}=\frac{2{\mathrm{r}}^3+{\mathrm{r}}^2+2{\mathrm{r}}^2+\mathrm{r}}{6}$

$=\frac{2{\mathrm{r}}^3+3{\mathrm{r}}^2+\mathrm{r}}{6} \therefore \mathrm{Sn}=\sum _{\mathrm{r}=1}^{\mathrm{r}}{\mathrm{U}}_{\mathrm{r}}=\frac{1}{6}\left[\sum _{\mathrm{r}=1}^{\mathrm{n}}2{\mathrm{r}}^3+\sum _{\mathrm{r}=1}^{\mathrm{n}}3{\mathrm{r}}^2+\sum _{\mathrm{r}=1}^{\mathrm{n}}\mathrm{r}\right]$

$=\frac{1}{6}\left\{2{\left(\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right)}^2+3\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}$

$=\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{12}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{12}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{12}$

$=\frac{\mathrm{n}{\left(\mathrm{n}+1\right)}^2\left(\mathrm{n}+2\right)}{12} \left(\mathrm{Ans}.\right)$

$2{\mathrm{x}}^2+2{\mathrm{y}}^2-3\mathrm{x}-4\mathrm{y}+1=0$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\mathrm{x}-2\mathrm{y}+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=0.....\left(\mathrm{i}\right)$

আবার, $16{\mathrm{x}}^2+16{\mathrm{y}}^2-32\mathrm{x}-1=0$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-\frac{1}{16}=0.....\left(\mathrm{ii}\right)$

(i) নং বৃত্তের কেন্দ্র = $\left(\frac{3}{4}, 1\right)$ (ii) নং বৃত্তের কেন্দ্র =(1,0)

 (i) নং বৃত্তের কেন্দ্র = $\left(\frac{3}{4}, 1\right)$ (ii) নং সমীকরণে বসিয়ে পাই।

${\left(\frac{3}{4}\right)}^2+1-2\times \frac{3}{4}-\frac{1}{16}=0$

$\therefore$(i) নং বৃত্তের কেন্দ্র (ii) নং বৃত্তের পরিধির উপর অবস্থিত।

আবার, (ii) নং বৃত্তের কেন্দ্র (1,0)(i) নং সমীকরণে বসিয়ে পাই-

2+0-3-0+1=0

$\therefore$(ii) নংবৃত্তের কেন্দ্র (i) নং বৃত্তের পরিধির উপর অবস্থিত। Showed.

$\overrightarrow{\mathrm{A}}=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-3\widehat{\mathrm{k}}$ এবং $\overrightarrow{\mathrm{B}}=3\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}$

$\therefore \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=4\widehat{\mathrm{i}}+\widehat{\mathrm{j}}-\widehat{\mathrm{k}}$ এবং $\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}=-2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}}$

এখন, $\left(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\right)\left(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\right)=\left(4\widehat{\mathrm{i}}+\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\right)\left(-2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-5\widehat{\mathrm{k}}\right)$

$=-8+3+5=0 \mathrm{showed}.$

$\mathrm{AE}=\frac{4}{3+4}\times 10=\frac{40}{7}\mathrm{ft}$

$\mathrm{BE}=10-\frac{40}{7}=\frac{30}{7}\mathrm{ft}$

এখন, $\mathrm{P}\times \mathrm{CE}=\mathrm{P}.\mathrm{DE}$

$\Rightarrow \frac{40}{7}-\mathrm{x}=\frac{-40}{7}+\mathrm{x}+5$

$\Rightarrow \frac{45}{7}=2\mathrm{x}$

$\therefore \mathrm{x}=\frac{45}{14}\mathrm{ft} \mathrm{Ans}.$

$\mathrm{BD}=5-\frac{45}{14}=\frac{25}{14}\mathrm{ft} \mathrm{Ans}.$

দেওয়া আছে, $\cos \mathrm{A}=sin\mathrm{B}-cos\mathrm{C}$

$\Rightarrow \cos \mathrm{A}+cos\mathrm{C}=sin\mathrm{B}$

$\Rightarrow 2cos\left(\frac{\mathrm{A}+\mathrm{C}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}\right)=2sin\left(\frac{\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{B}}{2}\right)$

$\Rightarrow 2cos\left(\frac{\mathrm{\pi }-\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}\right)=2sin\left(\frac{\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{B}}{2}\right)$

$\Rightarrow 2sin\left(\frac{\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}\right)=2sin\left(\frac{\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{B}}{2}\right)$

$\Rightarrow \frac{\mathrm{A}-\mathrm{C}}{2}=\frac{\mathrm{B}}{2}\Rightarrow A=B+C\Rightarrow 2A=\pi \Rightarrow A=\frac{\pi }{2}$

$\therefore △ABC$ একটি সমকোণী ত্রিভুজ। ‍Showed.

${\sin }^{-1}\mathrm{x}+si{n}^{-1}\mathrm{y}=\frac{\mathrm{\pi }}{2}$

$\Rightarrow \mathrm{x}=sin\left(\frac{\mathrm{\pi }}{2}-si{n}^{-1}\mathrm{y}\right)=cos\left({\sin }^{-1}\mathrm{y}\right)$

$=cos\left({\cos }^{-1}\sqrt{1-{\mathrm{y}}^2}\right)$

$\Rightarrow \mathrm{x}=\sqrt{1-{\mathrm{y}}^2} \therefore {\mathrm{x}}^2+{\mathrm{y}}^2=1 \left(\mathrm{Proved}\right)$

$\mathrm{H}=\frac{{\mathrm{u}}^2 si{n}^2\mathrm{\theta }}{2\mathrm{g}}$ এবং $\mathrm{R}=\frac{{\mathrm{u}}^2 si{n}^2\mathrm{\theta }}{\mathrm{g}}$

$\therefore \frac{H}{R}=\frac{{\sin }^2\theta }{2sin 2\theta }=\frac{{\sin }^2\theta }{4sin\theta cos\theta }=\frac{1}{4}\tan \theta$

$\Rightarrow \frac{1}{4\sqrt{3}}=\frac{1}{4}\tan \theta \Rightarrow tan\theta =\frac{1}{\sqrt{3}}$

$\therefore \theta ={30}^{\circ } Ans$

২য় সন্তানের জন্মদিন ১ম সন্তানের জন্মদিনে না হওয়ার সম্ভাবতা

$=\frac{365-1}{365}=\frac{364}{365}$

উল্লেখিত ২দিন ব্যতিত অন্য কোন দিনে ৩য় সন্তানের জন্মদিন হওয়ার সম্ভাবনা

$=\frac{365-2}{365}=\frac{363}{365}$     বিকল্প:$\frac{365p_3}{{\left(365\right)}^3}$

$\therefore$নির্ণেয় সম্ভাবনা=$\frac{364}{365}\times \frac{363}{365} Ans.$