Maths

Given ${\left(2x^2+3x+4\right)}^{10}={\sum }_{r=0}^{20}?a_r{x}^r$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{rr}& \frac{2^{10}{\left(2x^2+3x+4\right)}^{10}}{x^{20}}=\sum _{r=0}^{20}??\frac{a_r{2}^r}{x^r}\\ & \Rightarrow 2^{10}\sum _{r=0}^{20}??a_r{x}^r=\sum _{r=0}^{20}??a_r{2}^r{x}^{(20-r)}\left(\text{from}\right(i\left)\right)\end{array}$

now, comparing coefficient of $x^7$ from both sides

(take $r=7$ in L.H.S. and $r=13$ in R.H.S.)

$2^{10}{a}_7=a_{13}{2}^{13}\Rightarrow \frac{a_7}{a_{13}}=2^3=8$

$x\frac{dy}{dx}-y=x^2(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x),x>0$

$\frac{dy}{dx}-\frac{y}{x}=x(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x)\Rightarrow \frac{dy}{dx}+Py=Q$

So, I.F. $=e^{\int -\frac{1}{x}dx}\frac{1}{\left|x\right|}=\frac{1}{x}(x>0)$

Thus, $\frac{y}{x}=\int \frac{1}{x}\left(x\right(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x\left)\right)dx$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}+\mathrm{C}$

$?\mathrm{y}\left(\pi \right)=\pi \Rightarrow \mathrm{C}=1$

So, $y=x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+x\Rightarrow (y{)}_{\pi /2}=\frac{{\pi }^2}{4}+\frac{\pi }{2}$

Also, $\frac{dy}{dx}=x^2\mathrm{c}\mathrm{o}\mathrm{s}?x+2x\mathrm{s}\mathrm{i}\mathrm{n}?x+1$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+4x\mathrm{c}\mathrm{o}\mathrm{s}?x+2\mathrm{s}\mathrm{i}\mathrm{n}?x$

$\mathrm{m}\mathrm{a}\mathrm{x}\left\{n\right(A),n(B\left)\right\}\le n(A\cup B)\le n\left(U\right)$

$\Rightarrow 76\le 76+63-x\le 100$

$\Rightarrow -63\le -x\le -39$

$\Rightarrow 63\ge x\ge 39$

$f\left(x\right)={\int }_1^3?\frac{\sqrt[]{x}dx}{(1+x{)}^2}={\int }_1^{\sqrt[]{3}}?\frac{t.2tdt}{{\left(1+t^2\right)}^2}\mathrm{}$ (put $\sqrt[]{x}=t$ )

$={\left(-\frac{\mathrm{t}}{1+{\mathrm{t}}^2}\right)}_1^{\sqrt[]{3}}+{\left({\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\mathrm{t}\right)}_1^{\sqrt[]{3}}\mathrm{}$ [Applying by parts]

$\begin{array}{rr}& =-\left(\frac{\sqrt[]{3}}{4}-\frac{1}{2}\right)+\frac{\pi }{3}-\frac{\pi }{4}\\ & =\frac{1}{2}-\frac{\sqrt[]{3}}{4}+\frac{\pi }{12}\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat{\sigma }_C=5\\ WeknowthatC=\frac{5}{9}\left(F-32\right)\Rightarrow F=\frac{9C}{5}+32\\ \therefore {\sigma }_F=\frac{9}{5}{\sigma }_C=\frac{9}{5}*5=9\\ \therefore {\sigma }_F^2={\left(9\right)}^2=81\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Here,wehaveC{V}_1=50,C{V}_2=60\\ \overline{x_1}=30and\overline{x_2}=25\\ \therefore C{V}_1=\frac{{\sigma }_1}{\overline{x_1}}*100\Rightarrow 50=\frac{{\sigma }_1}{30}*100\Rightarrow {\sigma }_1=\frac{50*30}{100}=15\\ andC{V}_1=\frac{{\sigma }_2}{\overline{x_2}}*100\Rightarrow 60=\frac{{\sigma }_2}{25}*100\Rightarrow {\sigma }_2=\frac{60*25}{100}=15\\ \therefore Difference{\sigma }_1-{\sigma }_2=15-15=0\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

 $[x{]}^2+2[x+2]-7=0$

$\begin{array}{rr}& \Rightarrow [x{]}^2+2[x]+4-7=0\\ & \Rightarrow \left[x\right]=1,-3\\ & \Rightarrow x\in \left[\mathrm{1,2}\right)\cup [-3,-2)\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatVariance\left({\sigma }^2\right)=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2\\ =\frac{18000}{60}-{\left(\frac{960}{60}\right)}^2=300-256=44\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}First10positiveintegersare1,2,3,4,5,6,7,8,9,10\\ onmultiplyingeachnumberby-1,weget\\ -1,-2,-3,-4,-5,-6,-7,-8,-9,-10\\ onadding1toeachofthenumber,weget\\ 0,-1,-2,-3,-4,-5,-6,-7,-8,-9\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=0-1-2-3-4-5-6-7-8-9\\ =-45\\ and{\displaystyle \sum _{}^{}{x}_i^2}=0^2+{\left(-1\right)}^2+{\left(-2\right)}^2+{\left(-3\right)}^2+{\left(-4\right)}^2+\dots +{\left(-9\right)}^2\\ =\frac{9*10*19}{6}=285\left[?{\displaystyle \sum _{}^{}{n}^2=}\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\\ \therefore S.D=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2}=\sqrt[]{\frac{285}{10}-{\left(\frac{-45}{10}\right)}^2}\\ =\sqrt[]{\frac{285}{10}-\frac{2025}{100}}=\sqrt[]{\frac{2850-2025}{100}}=\sqrt[]{8.25}\\ \therefore Variance={\left(SD\right)}^2={\left(\sqrt[]{8.25}\right)}^2=8.25\\ Hence,thecorrectoptionis\left(a\right).\end{array}$