Ncert Solutions Maths class 11th

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 $=\sqrt[]{2}-1$

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

  $co{s}^{3}3x+co{s}^{3}5x={\left(2cos4xcosx\right)}^3$  

$={\left(cos5x+cos3x\right)}^3$            

$co{s}^{3}3x+co{s}^{3}5x$            

$\Rightarrow cosx.cos3x.cos4x.cos5x=0$            

$\therefore x=\left(2n+1\right)\frac{\pi }{2},\left(2n+1\right)\frac{\pi }{6},\left(2n+1\right)\frac{\pi }{8},\left(2n+1\right)\frac{\pi }{10}$            

$\Rightarrow$ Smallest +ve values of x is $\frac{\pi }{10}$  i.e. 18°.

$?arg\left(z\right)=\pi ifz=x+iy$

$\therefore y=0andx<0$

$\therefore z=-3+i.0\Rightarrow \left|z\right|=3$

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

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

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

Note that D^r < 0, hence given inequality (1) is true only if $N^r\ge 0$  

  i.e.   $\left(x-8\right)\left(x-2\right)\le 0and{2}^{x-3}>31$

i.e.    $2\le x\le 8and{2}^{x-3}\ge 31$

only x = 8 satisfies both the inequality.

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

i.e. $PM\cdot PN=16$ PM⋅PN=16

$\Delta APC,AC=AP=12$  

$\Delta ABP,tan60°=\frac{12}{AB}$           

$AB=\frac{12}{\sqrt[]{3}}=4\sqrt[]{3}$            

$Area=4\sqrt[]{3}.4\sqrt[]{6}=48\sqrt[]{2}$            

$S_1=1+2+3\dots \dots \dots .n$

$S_2=1+3+5\dots \dots \dots .n$

$S_3=1+4+7\dots \dots \dots .$

$\begin{array}{ll}\left(S_1+S_3\right)=\lambda S_2& S_1=\frac{n}{2}[2a+(n-1\left)1\right]\\ \frac{n}{2}[2a+(n-1\left)1\right]+\frac{n}{2}[2a+(n-1\left)3\right]& S_2=\frac{n}{2}[2a+(n-1\left)2\right]\\ =\lambda \frac{n}{2}[2a+(n-1\left)2\right]& S_3=\frac{n}{2}[2a+(n-1\left)3\right]\end{array}$

$\lambda =2$