Class 12th

Let f(x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$                   Non zero finite

So, d = e = f = 0

f(x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$          Non zero finite

f'(x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$  

f'(1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 .(i)

f'(-1) = 0

-6 + 5a – 4b + 3 = 0 .(ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$  

(A) Source of microwave frequency   => Magnetron

(B) Source of infrared frequency                =>Vibration of atoms and molecules

(C) Source of Gamma rays                           => Radioactive decay of nucleus

(D) Source of x-rays                                      => inner shell electrons.

$1+{\alpha }^2=1\Rightarrow \alpha =0$

$\&{\alpha }^2+{\beta }^2=1\Rightarrow {\beta }^2=\pm 1$

$\&\alpha -\alpha \beta =0\Rightarrow \alpha \left(1-\beta \right)=0\Rightarrow \alpha =0or\beta =1$

= 0 & = 1 or = 0 & = 1

4 + 4 = 1

$A=\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill y\hfill & \hfill z\hfill & \hfill x\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill y\hfill \end{array}\right],\left|A\right|=3xyz-\left(x^3+y^3+z^3\right)=-\left(x+y+z\right)\left[{\left(x+y+z\right)}^2-3\left(xy+yz+zx\right)\right]$

A² = l

A. A' = l    (as A = A')

$\therefore x^2+y^2+z^2=1andxy+yz+zx=0$         

$x^3+y^3+z^3=3*2+1*\left(1-0\right)=7$             

V = 1.24 * 10⁶ volt

${\lambda }_{min}=?$

${\lambda }_{min}=\frac{hc}{eV}=\frac{1242nm}{e\left(1.24*10^6\right)}=\frac{1000*10^{-9}}{10^6}\Rightarrow {\lambda }_{min}=10^{-3}nm$

$\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j},\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k}$      

-> $\overrightarrow{r}*\overrightarrow{a}=\overrightarrow{c}*\overrightarrow{a}$

$\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{a}$

Now, $0=\overrightarrow{b}.\overrightarrow{c}+\lambda \overrightarrow{a}.\overrightarrow{b}as\overrightarrow{r}.\overrightarrow{b}=0$  

$\lambda =\frac{-\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}=2$        

$\therefore \overrightarrow{r}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{c}+2a^2=12$           

$\left[\begin{array}{cc}\hfill 0\hfill & \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \end{array}\right],I_2+A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right],I_2-A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right]$           

  $\left(I_2+A\right){\left(I_2-A\right)}^{-1}=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill b\hfill & \hfill a\hfill \end{array}\right]$          

  $a^2+b^2=\left|\left(I_2+A\right){\left(I_2-A\right)}^{-1}\right|=se{c}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.*co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1$          

$\therefore 13\left(a^2+b^2\right)=13*1=13$

Critical point of function are x = $\frac{-1}{2},$ 1 and -2 but x = -2 is making zero .

$\therefore nondifferentiableatx=\frac{-1}{2},1$      

$\Delta E=hv\to \left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

(1) n₁ = 3, n₂ = 2

    $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{9-4}{36}=\frac{5}{36}$         

(2) n₁ = 4, n₂ = 3

$\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{16-9}{144}=\frac{7}{144}$        

(3) n₁ = 2, n₂ = 1

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

(4) n₁ = 5, n₂ = 4

$\left(\frac{1}{4^2}-\frac{1}{5^2}\right)=\frac{25-16}{400}=\frac{9}{400}$