Class 11th

y = ax² + bx + c

a + b + c = 2

$\therefore \frac{dy}{dx}=2ax+b\Rightarrow {\frac{dy}{dx}|}_{\left(0,0\right)}=b=1$    

It passes through (0, 0)

0 = 0 + 0 + c -> c = 0

a = 1

$\therefore a=1,b=1,c=0$           

$tan\left(\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}\right)$

$Put\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}=\theta$

$sin4\theta =\frac{\sqrt[]{63}}{8}\therefore cos4\theta =\frac{1}{8}$

$co{s}^2\theta =\frac{7}{8}\Rightarrow se{c}^2\theta =\frac{8}{7}$

$ta{n}^2\theta =\frac{8}{7}-1=\frac{1}{7}$

$tan\theta =\frac{1}{\sqrt[]{7}}$

y = x² + 4

x² = y – 4

y = 4x – 1

$PQ=\frac{\left|4*\frac{1}{2}t-\frac{1}{4}{t}^2\right|-4-1}{\sqrt[]{17}}$           

$p=\frac{\left|t^2-8t-2\right|}{4\sqrt[]{17}}$          

$\frac{dp}{dt}=\frac{1}{4\sqrt[]{17}}$           (2t – 8) for maximum / minimum, $\frac{dp}{dt}=0\Rightarrow t=4$  

$also\frac{d^{2}p}{d{t}^2}>0att=4$           

Hence the closest point becomes at t = 4 is (2, 8)

Kindly consider the following figure

Linear (180°)

Li -> Crimson Red

Na -> Yellow

Rb -> Red violet

Cs -> Blue

VIBGYOR

Wavelength

Red violet > Crimson Red >   Yellow   >   Blue

      $\downarrow$                           $\downarrow$             $\downarrow$             $\downarrow$           

780 nm             670.8 nm     589.2 nm    455.5 nm

According to Bohr's theory,

$K.E.=13.6\frac{z^2}{n^2}$ frequency $\left(\upsilon \right)$ of revolution of electron $\propto \frac{z^2}{n^3}$

  $v=2.18*10^6\frac{z}{n}$          

$F=\frac{k{q}_1{q}_2}{r^2}=\frac{Kz{e}^2}{r^2}\left(r\propto \frac{n^2}{z}\right)$       

$F\propto \frac{z^3}{n^4}$         

$\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1+\lambda \left(\overrightarrow{r}.\left(\widehat{i}-2\widehat{j}\right)+2\right)=0$

$\therefore point\left(1,0,2\right)=\widehat{i}+2\widehat{k}$

$\therefore \overrightarrow{r}\left(\frac{\widehat{i}}{3}+\frac{7\widehat{j}}{3}+\widehat{k}\right)-\frac{7}{3}=0$

$\overrightarrow{r}.\left(\widehat{i}+7\widehat{j}+3\widehat{k}\right)=7$

${=}^{n+1}{C}_2+2{\displaystyle \sum _{r=1}^{n-1}\frac{\left(r+1\right)!}{\left(r-1\right)!2!}}{=}^{n+1}{C}_2+{\displaystyle \sum _{r=1}^{n-1}\left(r+1\right)r}$

  =   $\frac{\left(n+1\right)!}{2!\left(n+1\right)!}+\frac{\left(n-1\right)n\left(2\left(n-1\right)+1\right)}{6}+\frac{\left(n-1\right)n}{2}=\frac{n\left(n+1\right)}{2}+\frac{n\left(n-1\right)\left(2n-1\right)}{6}+\frac{n\left(n-1\right)}{2}$

=   $\frac{n\left(6n+2n^2-3n+1\right)}{6}=\frac{\left(2n^2+3n+1\right)}{6}=\frac{n\left(2n+1\right)\left(n+1\right)}{6}$

$x+\sqrt[]{3}y=2\sqrt[]{3},andm=-\frac{1}{\sqrt[]{3}},C=2$          not possible

 (2) , $\left(1\right)\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{1}{2}}=1\Rightarrow c=\sqrt[]{a^2{m}^2-b^2}=\sqrt[]{\frac{9}{2}*\frac{1}{3}-\frac{1}{2}}=1,$ not possible

(3) $c=a\sqrt[]{1+m^2}=\sqrt[]{7}*2=2\sqrt[]{7},$ not possible

(4) $\frac{x^2}{9}+\frac{y^2}{1}=1,C=\sqrt[]{9m^2+1}=\sqrt[]{9*\frac{1}{3}+1}=2$