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This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Giventhat{3}^{400}={\left(9\right)}^{200}={\left(10-1\right)}^{200}\\ \therefore {\left(10-1\right)}^{200}={}_{200}{C}_0{\left(10\right)}^{200}+{}_{200}{C}_1{\left(10\right)}^{199}+\dots -{}_{200}{C}_{199}{\left(10\right)}^1+{}_{200}{C}_{200}{\left(1\right)}^{200}\\ =10^{200}-200\times 10^{199}+\dots -10\times 200+1\\ So,itisclearthatlasttwodigitsare01.\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$

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

Sol:

$\begin{array}{l}LetabethefirsttermanddbethecommondifferenceofthegivenA.P.\\ \therefore S_p=\frac{p}{2}\left[2a+\left(p-1\right)d\right]=q\Rightarrow 2a+\left(p-1\right)d=\frac{2q}{p}\dots \left(i\right)\\ and{S}_q=\frac{q}{2}\left[2a+\left(q-1\right)d\right]=p\Rightarrow 2a+\left(q-1\right)d=\frac{2p}{q}\dots \left(ii\right)\\ Subtractingeqn.\left(ii\right)fromeqn.\left(i\right)weget\\ \left(p-q\right)d=\frac{2q}{p}-\frac{2p}{q}\Rightarrow \left(p-q\right)d=\frac{2\left(q^2-p^2\right)}{pq}\\ \Rightarrow \left(p-q\right)d=\frac{-2}{pq}\left(p^2-q^2\right)\Rightarrow \left(p-q\right)d=\frac{-2}{pq}\left(p+q\right)\left(p-q\right)\\ \Rightarrow d=\frac{-2}{pq}\left(p+q\right)\\ Substitutingthevalueofdineqn.\left(i\right)weget\\ 2a+\left(p-1\right)\left[\frac{-2}{pq}\left(p+q\right)\right]=\frac{2q}{p}\\ \Rightarrow 2a=\frac{2q}{p}+\frac{2\left(p-1\right)\left(p+q\right)}{pq}\Rightarrow a=\frac{q}{p}+\frac{\left(p-1\right)\left(p+q\right)}{pq}\\ \Rightarrow a=\frac{q^2+p^2+pq-p-q}{pq}\\ Now{S}_{p+q}=\frac{p+q}{2}\left[2a+\left(p+q-1\right)d\right]\\ =\frac{p+q}{2}\left[\frac{2q^2+2p^2+2pq-2p-2q}{pq}+\frac{\left(p+q-1\right)\left[-2\left(p+q\right)\right]}{pq}\right]\\ =\frac{p+q}{2}\left[\frac{2q^2+2p^2+2pq-2p-2q-2p^2-2pq+2p-2pq-2q^2+2q}{pq}\right]\\ =\frac{p+q}{2}\left[\frac{-2pq}{pq}\right]=-\left(p+q\right)Henceproved.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Givenexpressionis{\left(1+x\right)}^{2n-1}\\ Numberofterms=2n-1+1=2n\left(even\right)\\ \therefore Middletermsare\frac{2n}{2}thtermand{\left(\frac{2n}{2}+1\right)}^{th}terms=nthtermsand\left(n+1\right)thterms\\ Coefficientofnthterm={}_{2n-1}{C}_{n-1}\\ andthecoefficientof\left(n+1\right)thterm={}_{2n-1}{C}_n\\ Sumofthecoefficients={}_{2n-1}{C}_{n-1}+{}_{2n-1}{C}_n\\ ={}_{2n-1+1}{C}_n={}_{2n}{C}_n\left[?{}_n{C}_r+{}_n{C}_{r-1}={}_{n+1}{C}_r\right]\\ Hence,thegivenstatementis\text{'}False\text{'}.\end{array}$

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This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Totalnumberof5-digittelephonenumberifallthedigitsarerepeated={\left(10\right)}^5\\ \left[?digitsarefrom0to9\right]\\ Ifdigitsarenotrepeated,then5-digittelephones,canbeformedin{}_{10}{P}_{5}ways\\ \therefore Requirednumberofways={\left(10\right)}^5-{}_{10}{P}_5\\ =100000-\frac{10!}{\left(10-5\right)!}=100000-\frac{10\times 9\times 8\times 7\times 6\times 5!}{5!}\\ =100000-30240=69760\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

Students willing to study BBA at the Jyothy Institute of Commerce and Management (JICM) can estimate the total expenditure by the information on fee structure. The Jyothy Institute of Commerce and Management BBA fee is INR 2.1 Lakh. The information about the Jyothy Institute of Commerce and Management BBA course fees is taken from the official sources and could be subject to change.

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Givenexpressionis{\left[{\left(2x+y^3\right)}^4\right]}^7={\left(2x+y^3\right)}^{28}\\ Here,n=28\\ \therefore Thenumberoftermsingivenexpressionis\left(n+1\right)\\ i.e.,28+1=29\\ Hence,thegivenstatementis\text{'}False\text{'}.\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Totalnumberofplayers=22\\ 2playersarealwaysincludedand4arealwaysexcludingorneverincluded=22-2-4=16\\ \therefore Requirednumberofselection={}_{16}{C}_9\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}{7}^9+9^7={\left(1+8\right)}^7-{\left(1-8\right)}^9\\ =\left[{}_7{C}_0+{}_7{C}_1.8+{}_7{C}_2{\left(8\right)}^2+{}_7{C}_3{\left(8\right)}^3+\dots +{}_7{C}_7{\left(8\right)}^7\right]-\left[\begin{array}{l}{}_9{C}_0-{}_9{C}_1.8+{}_9{C}_2{\left(8\right)}^2-{}_9{C}_3{\left(8\right)}^3\\ +\dots {}_9{C}_9{\left(8\right)}^9\end{array}\right]\\ =\left(7\times 8+9\times 8\right)+\left(21\times 8^2-36\times 8^2\right)+\dots \\ =\left(56+72\right)+\left(21-36\right)\times 8^2+\dots \\ =128+64\left(21-36\right)+\dots \\ =64\left[2+\left(21-36\right)+\dots \right]\\ whichisdivisibleby64.\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$

Yes, MTech at Indian Institute for Aeronautical Engineering and Information Technology is offered in full-time mode. The course is for a duration of two years. It is offered in two different specialisations, i.e., Space Technology and Aerospace Engineering. Interested candidates can apply offline or online through the ofiicial website of the institute.

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