Class 12th

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$  .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$      = $5-\sqrt[]{2}-\sqrt[]{3}$  

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method       

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$  

Sound level is given in dB = 10 log $\frac{{}_{}}{{}_{}}\frac{}{{}_{}}$ $\left(\frac{P_0}{P_i}\right)or10log\left(\frac{l}{l_0}\right)$

As sound level decreases 5 dB every km,

So, in 20 km sound level will decrease by 20 * 5 = 100 dB.

$\Delta \beta ={\beta }_2-{\beta }_1=10log\left(\frac{l_2}{l_1}\right)$

$-100=10log\left(\frac{l_2}{l_1}\right)$

$10^{-10}=\frac{l_2}{l_1}\Rightarrow l_2=10^{-10}{l}_1$

$P_2=10^{-10}{P}_1$

$P_2=10^{-10}*\left(0.1*10^3\right)\Rightarrow P_2=10^{-8}W=10^{-x}W$

$$ $\therefore$  x = 8

${\omega }_0=10^{5}rad/sec$

P = 16w, 120v at resonance

$P=\frac{v^2}{R}\Rightarrow 16=\frac{{\left(120\right)}^2}{R}\Rightarrow R=\frac{14400}{16}=900\Omega$

$l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$        .(A)

Put   $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$            .(i)

Using properties ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$  

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$    .(ii)

Adding (i) and (ii) we get

  $2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$         

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

Flux through the 6 sides of square (i.e. cube)

${?}_T=\frac{q_{in}}{{\epsilon }_0}=\frac{12*10^{-6}C}{{\epsilon }_0}$

$\therefore$ Flux through a square

$?=\frac{{?}_T}{6}=\frac{12*10^{-6}}{{\epsilon }_0*6}=\frac{2*10^{-6}}{{\epsilon }_0}$  

$\Rightarrow ?=225.98*10^{3}N{m}^2/C?226N{m}^2/C$

$\lambda =\frac{h}{p}or\lambda =\frac{h}{mv}$

${\lambda }_p={\lambda }_{\alpha }\left[Given\right]$

$\frac{h}{m_p{V}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}$

$\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{4m}{m}=\frac{4}{1}$ 0r 4: 1

kx + y + 2z = 1    . (i)

 3x – y – 2z = 2       . (ii)

-2x – 2y – 4z = 3   . (iii)

(ii) * 5 - (i)  $\equiv$ (iii) * 3 -> (15 – k) = -6

K = 21

${\lambda }_{Red}>{\lambda }_{violet}$

$\beta =\frac{\lambda D}{d}\Rightarrow$ Fringe width

If the source of light used in a Young's double slit experiment is changed from red to violet : consecutive fringe lines will come closer.

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

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

$\therefore 5.f\left(2\right)=144$