Maths NCERT Exemplar Solutions Class 11th Chapter Two

Equation of normal   $y=mx-2m-m^3$ must pass through centre $\left(\mathrm{0,3}\right)$

$\begin{array}{rr}& m^3+2m+3=0\\ & m^2(m+1)-m(m+1)+3\cdot 1(m+1)=0\\ & (m+1)\left(m^2-m+3\right)=0\\ & m=-1\end{array}$

  $\therefore \mathrm{}$ point of contact of normal at parabola is $\left({\mathrm{a}\mathrm{m}}^2,-2\mathrm{a}\mathrm{m}\right)=\left(\mathrm{1,2}\right)$

So distance between parabola and circle is $=\sqrt[]{2}-1$

  $x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

Equation of any plane passing through the line of intersection of planes (1) and (2)

$x-cy-bz+\lambda (cx-y+az)=0$

$(1+\lambda c)x-(c+\lambda )y+(-b+a\lambda )z=0$

If (3) &  (4) is same plane.

$\frac{1+c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$

(i)

(ii) (iii)

By (i)  & (ii) $\lambda =-\frac{(a+bc)}{ac+b}$

By (ii) & (iii)

$\lambda =\frac{-(ab+c)}{1-a^2};a^2+b^2+c^2+2abc=1$

$n(A\cup B\cup C)+n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=50$

$n(\cup )=50$

$n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=n(\cup )-n(A\cup B\cup C)\Rightarrow 50-29\Rightarrow 21$

$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(C\cap A)-n(A\cap B\cap C)$

$\Rightarrow \mathrm{}10+15+20-5-6-10-5\Rightarrow 29$

$n(A\cup B\cup C{)}^{\mathrm{\text{'}}}=21$

$E\left(\frac{-a{x}_1+b{x}_2+c{x}_3}{-a+b+c},\frac{-a{y}_1+b{y}_2+c{y}_3}{-a+b+c}\right)E\left(\frac{-4*0+4*3+0*5}{-4+3+5},\frac{-4*3+3*0+5*0}{-4+3+5}\right)$

Mean $=np$ , variance $=npq$

$np-npq=1;p+q=1$

$n^2{p}^2-n^2{p}^2{q}^2=11$

We get   $q=\frac{5}{6},p=\frac{1}{6},n=36\mathrm{}$  Probability of ' 3 ' success $={}^{36}{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{33}$

$\bar{a}+\bar{b}+\bar{c}+\bar{d}=0$

Magnitude of all vectors will be same as well as angle between these vectors will also be same. $\bar{a}+\bar{b}+\bar{c}=-\bar{d}$

$\begin{array}{c}|\bar{a}{|}^2+|\bar{b}{|}^2+|\bar{c}{|}^2+2\bar{a}\cdot \bar{b}+2\bar{b}\cdot \bar{c}+2\bar{c}\cdot \bar{a}=|d{|}^2\\ \mathrm{}1+1+1+2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =1\\ 6\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-2\\ \mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-\frac{1}{3}\\ \alpha ={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\left(\frac{-1}{3}\right)=\pi -{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\frac{1}{3}\end{array}$

TSA  of cube $=6a^2$

$\frac{d}{dt}\left(6a^2\right)=4.8;12a\frac{da}{dt}=4.8;a\frac{da}{dt}=0.4;\frac{dv}{dt}=\frac{d}{dt}\left(a^3\right)\Rightarrow 3a\left(a\frac{da}{dt}\right)$

$3*15*0.4=3*15*\frac{04}{10}\Rightarrow 18$

${\mathrm{l}\mathrm{o}\mathrm{g}}_{(2x+3)}?x^2<{\mathrm{l}\mathrm{o}\mathrm{g}}_{(2x+3)}?(2x+3)$

Case-I: $0<2x+3<1$  

$x^2>2x+3$

Case-II: $2x+3>1$

$0<x^2<2x+3$  

By solving (1) & (2) $\begin{array}{rr}& \frac{-3}{2}<x<-1\&(x-3)(x+1)>0.\\ & \frac{-3}{2}<x<-1,x>3\end{array}$

Equation (4) has no solution $\frac{-3}{2}<x<-1,x<-1$

.(5) By equation (5) . $x\in \left(\frac{-3}{2},-1\right)$

We obtain solving equation (2) $x>-1\mathrm{}x>-1\mathrm{}\Rightarrow x\in (-\mathrm{1,3})$

or $(x-3)(x+1)<0\mathrm{}-1<x<3,x^2>0\Rightarrow x\ne 0$

$x\in \left(\frac{-3}{2},-1\right)\cup (-\mathrm{1,3})-\left\{0\right\}$

$\boxed{\begin{array}{lll}\text{12.(D)}& {\int }_0^{\frac{\pi }{6}}??\frac{4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \frac{2}{5}*2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta }{\left(4-4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \right)}=\frac{16}{5*8}\int \frac{{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \mathrm{c}\mathrm{o}\mathrm{s}?d\theta }{{\mathrm{c}\mathrm{o}\mathrm{s}}^3?\theta }& x^{5/2}=2\mathrm{s}\mathrm{i}\mathrm{n}?\theta \\ & & \frac{5}{2}{x}^{3/2}dx=2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta \\ & & =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??{\mathrm{t}\mathrm{a}\mathrm{n}}^2?\theta d\theta =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??\left({\mathrm{s}\mathrm{e}\mathrm{c}}^2?\theta -1\right)d\theta =\frac{2}{5}[\mathrm{t}\mathrm{a}\mathrm{n}?\theta -\theta {]}_0^{\pi /6}=\frac{2}{5}\left(\frac{1}{\sqrt[]{3}}-\frac{\pi }{6}\right)\end{array}}$

Clearly PM. PN $={\mathrm{O}\mathrm{P}}^2$ =OP2

i.e. $PM\cdot PN=16$ PM⋅PN=16