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

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

Ahmedabad Institute of Business Management admissions are currently open for all courses. Candidates are selected for courses based on entrance exam/Gujarat Common Admission Services. Aspirants can apply online. Interested candidates meeting the eligibility criteria can fill out the form. 

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

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

Ahmedabad Institute of Business Management admissions 2025 are open online. Till now, no schedule has been released by the institute. Since the admission is open, candidates can apply for BBA and MBA course. To get admission to the AIBM Ahmedabad, candidates must meet the eligibility criteria set by the college. The application process at AIBM Ahmedabad is conducted online.

The University of Rajasthna will release the RULET 2026 counselling schedule on the official website. The counselling for the eligible candidates will be conducted as per the schedule to be released by the officials. Candidates will be required to carry the admission as well as counselling fee along with them to book their seat on the counselling day. They must carry the required academic as well as other documents (both original and photocopy).

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

Sol:

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

The companies belonging to private sector had the largest share of 65.4% during the IIT Bhilai placements 2024. Check out the table for more company type information that visited the campus during placements:

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

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