Class 12th

The light wave contains two lights of different frequencies, so

  $E_1=h{\upsilon }_1=\frac{4.14*10^{-15}*6*10^{15}}{2\pi }=3.96eV,and$

  $E_2=h{\upsilon }_2=\frac{4.14*10^{-15}*9*10^{15}}{2\pi }=5.92eV$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5*10^{10}}{2.0*10^{10}}=\frac{3}{4}$

According to question, we can write

  $\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$  

option (C) is incorrect, there will be minima.

According to question, we can write

    $F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$                    

For maxima of force   $\frac{dF}{dx}=0,so$

$x=\frac{d}{2\sqrt[]{2}}$

$x^4-2x^3+2x-1={\left(x-1\right)}^2\left(x^2-1\right)$

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

=2

Kindly go through the solution

B = (I – adjA)⁵