${(2y+1)}^{10}={\left(1+2y\right)}^{10}; T_{r+1}={}^{10}C_r.2^r.y^r$

অতএব, $y^{r-1}$এর সহগ $={}^{10}C_{r-1}.2^{r-1}$

$C_r={}^{10}C_{r-1}{2}^{r-1}$এবং $C_{r+2={}^{10}C_{r+2-1}.2^{r+2-1}={}^{10}C_{r+1}.2^{r+1}}$

We know, if ${}^{n}C_x={}^{n}C_y$

$$\begin{gathered} \Rightarrow x+y=n \\ Then C_{r+2}=4C_r \\ \Rightarrow {}^{10}C_{r+1}{2}^{r+1}=4. {}^{10}C_{r-1}{2}^{r-1} \\ \Rightarrow {}^{10}C_{r+1}={}^{10}C_{r-1} \\ \Rightarrow {}^{10}C_{r+1}={}^{10}C_{r-1} \\ \Rightarrow r+1+r-1=10 \end{gathered}$$

অতএব, r = 5

ধরি, B প্রান্ত থেকে ym ভেতরে C এর অবস্থান। 

$17y=20\times \left(5-y\right)+5\times \left(10-y\right)$অতএব, $y=3.571m$

অতএব, C বিন্দুকে মূল বিন্দু থেকে A এর দিকে (3.571-3.125)m = 0.446m সরাতে হবে। 

লোকের বেগ V ও বাসের ত্বরণ a হলে, 

$0+60\times a=v\Rightarrow 60a=v$

$\Rightarrow a=\frac{v}{60}$[ ধরি, বাস ধরার সময় বাস ও ব্যক্তির বেগ সমান]

$60v=50+0+\frac{1}{2}\times a\times 3600$অতএব, $v=1.67 m{s}^{-1}$এবং $a = 0.0278m{s}^{-2}$

(a) Probability (4 জন ছাত্রী) $=\frac{{}^{20}C_4}{{}^{35}C_4}=\frac{57}{616}$

(b) Probability (2 জন ছাত্র ও 2 জন ছাত্রী) $=\frac{{}^{20}C_2\times {}^{15}C_2}{{}^{35}C_4}=\frac{285}{748}$

$\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x\left(1+cosx\right)}=\underset{x\to 0^{+}}{\lim }\left(\frac{\sin x}{x}\right)\frac{1}{1+cosx}=1.\frac{1}{1+cos0}=\frac{1}{1+1}=\frac{1}{2}$

$\mathrm{Let}, \mathrm{I}=\int 2^{\ln \mathrm{x}}\mathrm{dx}=2^{\ln \mathrm{x}}\int dx-\int \left({\frac{d}{dx}}^{\left(2^{\ln x}\right)}\int dx\right)dx$

$=2^{\ln x}.x-\int 2^{\ln \mathrm{x}}.ln2.\frac{1}{x}.x dx=x 2^{\ln \mathrm{x}}-ln2\int 2^{\ln \mathrm{x}}dx$

$=x 2^{\ln \mathrm{x}}-I.ln2$

$\therefore I+I ln2=x 2^{\ln \mathrm{x}}$

$\Rightarrow I\left(I+ln2\right)=x 2^{\ln \mathrm{x}}$

$\Rightarrow I=\frac{x 2^{\ln \mathrm{x}}}{1+ln2}+C$

$\overrightarrow{\mathrm{r}}=12\sqrt{\mathrm{t}}\widehat{\mathrm{i}}+{\mathrm{t}}^{\frac{3}{2}}\widehat{\mathrm{j}}$

$\Rightarrow \frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{dt}}=6{\mathrm{t}}^{-\frac{1}{2}}\widehat{\mathrm{i}}+\frac{3}{2}.{\mathrm{t}}^{\frac{1}{2}}\widehat{\mathrm{j}}$

$\therefore \mathrm{v}=\left|\frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{dt}}\right|=\sqrt{\frac{36}{\mathrm{t}}+\frac{9}{4}\mathrm{t}}$

$\therefore \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{1}{2\sqrt{\frac{36}{\mathrm{t}}+\frac{9}{4}\mathrm{t}}}\left(\frac{36}{-{\mathrm{t}}^2}+\frac{9}{4}\right)=0$

$\Rightarrow \frac{36}{-{\mathrm{t}}^2}+\frac{9}{4}=0$

$\Rightarrow \frac{36}{t^2}=\frac{9}{4}$

$\Rightarrow t^2=16$

$\Rightarrow t=\pm 4 sec$

$\because$সময় সর্বদা অঋণাত্মক $\therefore 4 sec$

এখানে, $t=4 sec$ গতি সর্ব নিম্ন হবে।

$V_{min}=\sqrt{\frac{36}{4}+\frac{9}{4}.4}=\sqrt{18}=3\sqrt{2}$

$=4.24 m{s}^{-1}$

${\tan }^{-1}\left(x+\alpha \right)-{\tan }^{-1}\left(x\right)=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{x+\alpha -x}{1+\left(x+\alpha \right)x}\right)=\frac{\pi }{4}$

$\Rightarrow \frac{\alpha }{1+x\left(x+\alpha \right)}=tan\frac{\pi }{4}$

$\Rightarrow \frac{\alpha }{1+x\left(x+\alpha \right)x}=1$

$\Rightarrow 1+x\left(x+\alpha \right)=\alpha$

$\Rightarrow 1+x^2+\alpha x=\alpha$

$\Rightarrow x^2+\alpha x+1-\alpha =0$

$\Rightarrow x=\frac{-\alpha \pm \sqrt{{\alpha }^2-4\left(1-\alpha \right)}}{2}$

$=\frac{-\alpha \pm \sqrt{{\alpha }^2+4\alpha -4}}{2}$

$=\frac{-\alpha \pm \sqrt{{\left(\alpha +2\right)}^2-8}}{2}$

$\alpha$ সমান হলে, ${\left(\alpha +2\right)}^2-8=0$

$\Rightarrow {\left(\alpha +2\right)}^2=8$

$\alpha +2=2\sqrt{2}$

$\alpha =2\sqrt{2}-2=2\left(\sqrt{2}-2\right)$

তবে সম্ভাব্য মান

$x=\frac{-\alpha +\sqrt{{\alpha }^2+4\alpha -4}}{2}$

$\because x>0 \alpha >0$

$\because P\left(\overline{A}\right)=0.4$

$\Rightarrow P\left(\overline{A}\right)=1-0.4=0.6$

আবার, $P\left(A\cap B\right)=0.2$

$P\left(A\cap \overline{B}\right)=P\left(A\right)-P\left(A\cap B\right)=0.6-0.2=0.4$

$L.S.=\frac{2}{1+z}=\frac{2}{1+cos\theta +isin\theta }=\frac{2\left(1+cos\theta -isin\theta \right)}{\left\{{\left(1+cos\theta \right)}^2+si{n}^2\theta \right\}}$

$=\frac{2\left(1+cos\theta -isin\theta \right)}{1+2cos\theta +co{s}^2\theta +si{n}^2\theta }=\frac{2\left(1+cos\theta -isin\theta \right)}{1+1+2cos\theta }$

$=\frac{2\left(1+cos\theta -isin\theta \right)}{2+2cos\theta }=\frac{1+cos\theta -isin\theta }{1+cos\theta }$

$=1-i\frac{\sin \theta }{1+cos\theta }=1-i\frac{2sin\frac{\theta }{2}\cos \frac{\theta }{2}}{2co{s}^2\frac{\theta }{2}}$

$=1-i\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=1-i tan\frac{\theta }{2}=R.S \left[showed\right]$

সর্বোচ্চতা, $H=\frac{u^2}{2g}$  এর অর্ধেক দূরত্ব, $\frac{H}{2}=\frac{u^2}{4g}$

যদি এই অর্ধেক দূরত্ব যেতে  সময় লাগে তবে

$\frac{H}{2}=ut-\frac{1}{2}g{t}^2\Rightarrow \frac{u^2}{4g}=ut-\frac{1}{2}g{t}^2$

$\Rightarrow u^2=4gut-2g^2{t}^2\Rightarrow 2g^2{t}^2-4gut+4^2=0$

$\therefore t=\frac{4gu\pm \sqrt{16g^2{u}^2-8g^2{u}^2}}{4g^2}=\frac{4gu}{4g^2}\pm \frac{\sqrt{8g^2{u}^2}}{4g^2}$

$=\frac{u}{g}\pm \frac{2\sqrt{2}gu}{4g^2}=\frac{u}{g}\pm \frac{u}{\sqrt{2}g}$

উত্থান কালে সময় কম লাগবে তাই, $t=\frac{u}{g}-\frac{u}{\sqrt{2}g}=\frac{u}{g}\left(1-\frac{1}{\sqrt{2}}\right)$

$=\frac{u}{g}.\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}.\frac{u}{g}$

ধরি, $C\left(0,b\right)$

BC এর ঢাল$=\frac{b-0}{0+a}=\frac{b}{a}$

CA এর ঢাল$=\frac{b-0}{0-2a}=-\frac{b}{2a}$

$\because BC\perp CA$

ঢালদ্বয়ের গুণফল=-1

$\Rightarrow \frac{b}{a}\times -\frac{b}{2a}=-1$

$\Rightarrow b^2=2a^2$

$\Rightarrow b=\pm \sqrt{2}a$

$\therefore c\equiv \left(0, \sqrt{2}a\right)$ এবং $\left(0, -\sqrt{2}a\right)$

ABC বৃত্তের ব্যাস AB কারণ $\angle C={90}^{\circ }$

$\therefore △ABC$ বৃত্তের সমীকরণ, $\left(x+a\right)\left(x-2a\right)+\left(y-0\right)\left(y-0\right)=0$

$\Rightarrow x^2-2ax+ax-2a^2+y^2=0$

$\Rightarrow x^2+y^2-ax-2a^2=0$

$\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\left(1+\frac{\mathrm{x}}{1!}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+....\right)-1}{\mathrm{x}}$

$=\underset{\mathrm{x}\to 0}{\lim }\frac{x\left(1+\frac{x}{2!}+\frac{x^2}{3!}+...\right)}{x} =\underset{\mathrm{x}\to 0}{\lim }\frac{1+\frac{x}{2!}+\frac{x^2}{3!}+....}{1}$

$=1 \therefore L.H.S=R.H.S \left(Proved\right)$

$\underset{x\to 0}{\lim }\frac{3sinx-sin3x}{x^3}$

$=\underset{x\to 0}{\lim }\frac{4si{n}^{3}x}{x^3}$

$=\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^3\times 4$

$=4\times 1=4 \left(Ans\right)$

$\frac{d}{dx}\left(x^{x^x}\right)=\frac{d}{dx}\left(e^{ln{x}^{x^x}}\right)=e^{{\ln }^{x^{x^x}}}\frac{d}{dx}\left(x^x \ln x\right)$

$=x^{x^x}\left\{x^x\frac{1}{x}+lnx\frac{d}{dx}\left(x^x\right)\right\}$

$=x^{x^x}\left\{x^x\frac{1}{x}+lnx.x^x\frac{d}{dx}\left(x lnx\right)\right\}$

$=x^{x^x}\left\{x^x\frac{1}{x}+lnx.x^x\left(x\frac{1}{x} lnx\right)\right\}$

$=x^{x^x} x^x\left\{\frac{1}{x}+lnx\left(1+lnx\right)\right\} \left(Ans\right)$

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

$\Rightarrow {\mathrm{y}}_1=2\left({\cos }^{-1}\mathrm{x}\right)\frac{-1}{\sqrt{1-{\mathrm{x}}^2}}$

$\Rightarrow {\mathrm{y}}_1^2\left(1-{\mathrm{x}}^2\right)=4{\left({\cos }^{-1}\mathrm{x}\right)}^2$

$\Rightarrow {\mathrm{y}}_1^2\left(1-{\mathrm{x}}^2\right)=4\mathrm{y}$

$\therefore {\mathrm{y}}_1^2\left(-2x\right)+\left(1-x^2\right)2{\mathrm{y}}_1{\mathrm{y}}_2=4{\mathrm{y}}_1$

$\Rightarrow \left(1-x^2\right){\mathrm{y}}_2-{\mathrm{xy}}_1=2 \left(\mathrm{Proved}\right)$

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

$=\int se{c}^2\mathrm{x}\left(1-sin\mathrm{x}\right)\mathrm{dx}$

$=\int se{c}^2\mathrm{x} \mathrm{dx}-\int tan\mathrm{x} sec\mathrm{x} \mathrm{dx}$

$=tan\mathrm{x}-sec\mathrm{x}+\mathrm{C} \left(\mathrm{Ans}\right)$