Class 11th

a, ar, ar², ….ar⁶³

a+ar+ar² +….+ar⁶³ = 7 [a + ar² + ar⁴ +.+ar⁶²]

$\frac{a\left(1-r^{64}\right)}{\left(1-r\right)}=7\frac{a\left(1-r^{64}\right)}{\left(1-r^2\right)}$            

1 + r = 7

r = 6

12x =

 $3x=\frac{\pi }{4}$

$cos3x=\frac{1}{\sqrt[]{2}}$

$4co{s}^{3}x-3cosx=\frac{1}{\sqrt[]{2}}$

$4\sqrt[]{2}co{s}^{3}x-3\sqrt[]{2}cosx-1=0$            

$x=\frac{\pi }{12}$ is the solution of above equation.

Statement 1 is true

$f\left(x\right)=4\sqrt[]{2}{x}^3-3\sqrt[]{2}x-1$

$f\text{'}\left(x\right)=12\sqrt[]{2}{x}^2-3\sqrt[]{2}=0$

$x=\pm \frac{1}{2}$

$f\left(-\frac{1}{2}\right)=-\frac{1}{\sqrt[]{2}}+\frac{3}{\sqrt[]{2}}-1=\sqrt[]{2}-1>0$            

f(0) = – 1 < 0

one root lies in $\left(-\frac{1}{2},0\right)$ , one root is $cos\frac{\pi }{12}$  which is positive. As the coefficients are real, therefore all the roots must be real.

Statement 2 is false.

  $F_x=\frac{-dv}{dx}$           

  $\overrightarrow{F}=-3x\widehat{i}-5\widehat{j}-6\widehat{k}$          

  ${\left|\overrightarrow{F}\right|}_{x=6}=\sqrt[]{18^2+5^2+6^2}$          

$=\sqrt[]{324+25+36}$

= $\sqrt[]{385}$    

= 19.62N

%error= $\left(\frac{dV}{V}+\frac{dR}{R}\right)*100$  

$=\left(\frac{5}{200}+\frac{0.2}{20}\right)*100$                          

= 3.5

Since $a=\frac{gsin?}{1+\frac{I}{M{R}^2}}$

$a=\frac{g*\frac{\sqrt[]{3}}{2}}{1+\frac{1}{2}}=\frac{g\frac{\sqrt[]{3}}{2}}{\frac{3}{2}}$

$a=\frac{g}{\sqrt[]{3}}$

PV = nRT

->Pµn

->Ratio= $\frac{32}{2}$ =16

KE =  $\frac{1}{2}$ m W² (A²– n²)

TE = $\frac{1}{2}$ mw²A²

^{$\frac{KE}{TE}=\frac{A^2-n^2}{A^2}=\frac{1-\frac{1}{9}}{1}=\frac{8}{9}$}

  $y=xtan\theta -\frac{g{x}^2}{2v^{2}co{s}^2\theta }$

(2.5, 2.5) must lie on this

-> $1=tan\theta -\frac{g*2.5}{2v^{2}co{s}^2\theta }$

-> $\frac{25}{2v^{2}co{s}^2\theta }=tan\theta -1$

-> $v^2=\frac{25}{2}\left\{\frac{1+ta{n}^2\theta }{tan\theta -1}\right\}$   

-> $v_{min}=5\sqrt[]{\sqrt[]{2}+1}$  

[Happens when tan θ = $\sqrt[]{2}$  + 1]

Work Area

->W = $\frac{1}{2}$ * 600 * 4

= 1200J

$a_c=\frac{v^2}{r}$ , $\frac{{\left(a_c\right)}_1}{{\left(a_c\right)}_2}={\left(\frac{v_1}{v_2}\right)}^2$

$\frac{3}{4}={\left(\frac{v_1}{v_2}\right)}^2\to \frac{v_1}{v_2}=\sqrt[]{3}:2$