Class 11th

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}\left(i\right)Thegivenexpressionis{\left(x+a\right)}^n\\ {\left(x+a\right)}^n={}_n{C}_0{x}^n{a}^0+{}_n{C}_1{x}^{n-1}a+{}_n{C}_2{x}^{n-2}{a}^2+{}_n{C}_3{x}^{n-3}{a}^3+\dots +{}_n{C}_n{a}^n\\ Sumofoddterms,\\ O={}_n{C}_0{x}^n+{}_n{C}_2{x}^{n-2}{a}^2+{}_n{C}_4{x}^{n-4}{a}^4+\dots \\ andthesumofeventerms,\\ E={}_n{C}_1{x}^{n-1}a+{}_n{C}_3{x}^{n-3}{a}^3+{}_n{C}_5{x}^{n-5}{a}^5+\dots \\ Now{\left(x+a\right)}^n=O+E\dots \left(i\right)\\ Similarly,{\left(x-a\right)}^n=O-E\dots \left(ii\right)\\ Multiplyingeqn.\left(i\right)andeqn.\left(ii\right),weget\\ {\left(x+a\right)}^n{\left(x-a\right)}^n=\left(O+E\right)\left(O-E\right)\\ {\left(x^2-a^2\right)}^n=O^2-E^2\\ Hence,O^2-E^2={\left(x^2-a^2\right)}^n\\ \left(ii\right)4OE={\left(O+E\right)}^2-{\left(O-E\right)}^2\\ ={\left[{\left(x+a\right)}^n\right]}^2-{\left[{\left(x-a\right)}^n\right]}^2\\ ={\left[x+a\right]}^{2n}-{\left[x-a\right]}^{2n}\\ Hence,4OE={\left(x+a\right)}^{2n}-{\left(x-a\right)}^{2n}\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Thegivenexpressionis{\left(+\frac{1}{}\right)}^n\\ ={\left(2^{1/3}+\frac{1}{3^{1/3}}\right)}^n\\ GeneralTerm T_{r+1}={}_n{C}_r{x}^{n-r}{y}^r\\ T_7=T_{6+1}={}_n{C}_6{\left(2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{n-6}{\left(\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}^6\\ ={}_n{C}_6{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.\left(\frac{1}{3^2}\right)={}_n{C}_6{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.{\left(3\right)}^{-2}\\ 7thtermfromtheend={\left(n-7+2\right)}^{th}termfromthebeginning\\ ={\left(n-5\right)}^{th}termfromthebeginning\\ So,T_{n-6+1}={}_n{C}_{n-6}{\left(2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{n-n+6}{\left(\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}^{n-6}\\ ={}_n{C}_{n-6}{\left(2\right)}^2.\left(\frac{1}{3^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)={}_n{C}_{n-6}{\left(2\right)}^2.{\left(3\right)}^{\raisebox{1ex}{$6-n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ Accordingtothequestion,weget\\ \frac{{}_n{C}_6{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.{\left(3\right)}^{-2}}{{}_n{C}_{n-6}{\left(2\right)}^2.{\left(3\right)}^{\raisebox{1ex}{$6-n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{1}{6}\\ \Rightarrow \frac{{}_n{C}_{n-6}{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.{\left(3\right)}^{-2}}{{}_n{C}_{n-6}{\left(2\right)}^2.{\left(3\right)}^{\raisebox{1ex}{$6-n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{1}{6}\Rightarrow {\left(2\right)}^{\frac{n-6}{3}-2}.{\left(3\right)}^{-2-\frac{6-n}{3}}=\frac{1}{6}\\ \Rightarrow {\left(2\right)}^{\frac{n-6-6}{3}}.{\left(3\right)}^{\frac{-6-6+n}{3}}=\frac{1}{6}\Rightarrow {\left(2\right)}^{\frac{n-12}{3}}.{\left(3\right)}^{\frac{n-12}{3}}={\left(6\right)}^{-1}\\ \Rightarrow {\left(6\right)}^{\frac{n-12}{3}}={\left(6\right)}^{-1}\Rightarrow \frac{n-12}{3}=-1\\ \Rightarrow n-12=-3\Rightarrow n=12-3=9\\ Hence,therequiredvalueofnis9.\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Totalnumberofmarbles=6white+5red=11marbles\\ \left(i\right)Since,wehavetodraw4marblesofanycolourfromthe11marbles\\ \therefore Requirednumberofways={}_{11}{C}_4\\ \left(ii\right)If2mustbewhiteand2mustbered,thentherequirednumberofways={}_6{C}_2*{}_5{C}_2\\ \left(iii\right)Ifallthe4marblesareofthesamecolour,thentherequirednumberofways={}_6{C}_4+{}_5{C}_4\\ Hence,therequirednumberofwaysare\\ \left(i\right){}_{11}{C}_4\left(ii\right){}_6{C}_2*{}_5{C}_2\left(iii\right){}_6{C}_4+{}_5{C}_4\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

 $\begin{array}{l}Thesideofthefirstequilateral\Delta ABC=20cm\\ Byjoiningthemidpointsofthesidesofthistriangle,wegetthesecondequilateraltriangle\\ whicheachside=\frac{20}{2}=10cm\left[\begin{array}{l}?Thelinejoiningthemid-pointsoftwosidesofatriangle\\ is\frac{1}{2}andparalleltothethirdsideofthetriangle\end{array}\right]\\ Similarly,eachsideofthethirdequilateraltriangle=\frac{10}{2}=5cm\\ \therefore Perimeteroffirsttriangle=20*3=60cm\\ Perimeterofsecondtriangle=10*3=30cm\\ andtheperimeterofthirdtriangle=5*3=15cm\\ Therefore,theserieswillbe60,30,15,\dots \\ whichisG.P.inwhicha=60,andr=\frac{30}{60}=\frac{1}{2}\\ Now,wehavetofindtheperimeterofsixthinscribedequilateraltriangle\\ \therefore a_6=a{r}^{6-1}\\ =60*{\left(\frac{1}{2}\right)}^5=60*\frac{1}{32}=\frac{15}{8}cm\\ Hence,therequiredperimeter=\frac{15}{8}cm\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat18micewereplacedequallyintwoexperimentalgroupsand\\ onecontrolgroupi.e.3groups\\ \therefore Therequirednumberofarrangements=\frac{Totalarrangements}{Equallylikelyarrangements}\\ =\frac{18!}{6!6!6!}=\frac{18!}{{\left(6!\right)}^3}\\ Hence,therequiredarrangements=\frac{18!}{{\left(6!\right)}^3}\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Since,thesumofallinterioranglesofapolygonofnsides=\left(2n-4\right)*90^0\\ \therefore Sumofinterioranglesofapolygonof3sides=\left(2*3-4\right)*90^0=180^0\\ Sumofinterioranglesofapolygonof4sides=\left(2*4-4\right)*90^0=360^0\\ Similarly,thesumofinterioranglesofapolygonofsides5,6,7,\dots are540^0,720^0,900^0,\dots \\ Therefore,theserieswillbe180^0,360^0,540^0,720^0,900^0,\dots whichisA.P.\\ Herea=180^0,d=180^0\\ Wehavetofindthesumofallinterioranglesofapolygonof21sidesi.e.,19thterm\\ a_n=a+\left(n-1\right)d\\ a_{19}=180^0+\left(19-1\right)180^0=180^0+18*180^0\\ =180^0+3240^0=3420^0\\ Hence,therequiredsumofinteriorangles=3420^0.\end{array}$