Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (b), (d)

Explanation- if we place a charge then we must experience some forces but if there would be no charge so field is continuous

$\begin{array}{l}\begin{array}{l}Let{x}^3=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}3x^{2}dx=dt\hfill \end{array}\\ I=\int \frac{3x^2}{x^6+1}dx=\int \frac{dt}{t^2+1}\\ \begin{array}{ll}= ta{n}^{–1}t+C\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= ta{n}^{–1}\left(x^3\right) +C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Let{e}^{2}x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}{e}^{2}x+ e^{x}1dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}{e}^x\left(x+1\right)dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{e^x(1+x)}{co{s}^2\left(e^{2}x\right)}}dx={\displaystyle \int \frac{dt}{co{s}^{2}t}}\\ \begin{array}{l}={\displaystyle \int se{c}^2}t dt = tan \left(e^x,x\right) + C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\therefore The correct answer is \left(B\right).\hfill \end{array}\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (c), (d)

Explanation- It is only possible when charges must be outside the surface or field line entering or leaving the surface are equal.

$\begin{array}{l}={\displaystyle \int \frac{si{n}^{2}x-co{s}^{2}x}{si{n}^{2}co{s}^{2}x}}dx\\ \begin{array}{l}={\displaystyle \int (se{c}^{2}x-cose{c}^{2}x)}dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= tanx+ cotx+ C.\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Therefore, the correct answer is \left(A\right).\hfill \end{array}\end{array}$

$\begin{array}{l}\frac{1}{cos(x-a)cos(x-b)}=\frac{1}{sin(a-b)}*\left[\frac{sin(a-b)}{cos(x-a)cos(x-b)}\right]\\ =\frac{1}{sin(a-b)}\left[\frac{sin\{(x-b)-(x-a)\}}{cos(x-a)cos(x-b)}\right]\end{array}$

$\begin{array}{l}=\frac{1}{sin(a-b)}[sin(x-b)cos(x-a)-cos(x-b)\cdot sin(x-a)cos(x-a)cos(x-b)\\ =\frac{1}{sin(a-b)}[tan(x-b)-tan(x-a)]\\ =\frac{1}{sin(a-b)}{\displaystyle \int tan}(x-b)-tan\left(x-a\right)]dx\\ =\frac{1}{sin(a-b)}[-log|cos(x-b)|+log|cos(x-a)?]\\ =\frac{1}{sin(a-b)}\left[log\left|\frac{cos(x-a)}{cos(x-b)}\right|\right]+\; C\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (a)

Explanation- field lines are always perpendicular to the surface.

$\begin{array}{l}I={\displaystyle \int si{n^{-}}^1}\left(cosx\right)dx\\ ={\displaystyle \int si{n}^{-1}}\left(sin\left\{\frac{\pi }{2}-x\right\}\right)dx\\ ={\displaystyle \int \{\frac{\pi }{2}-x\}dx}\\ =\frac{\pi }{2}x-\frac{x^2}{2}+\; C\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (a)

Explanation- when a point positive charge brought near an isolated conducting plane, some negative charge develops on the surface of the plane towards the charge and an equal positive charge develops on opposite side of the plane. By process called induction.

$\begin{array}{l}=\frac{cos2x}{co{s}^{2}x+si{n}^{2}x+2sinxcosx}\\ =\frac{cos2x}{1+sin2x}\end{array}$

$\begin{array}{l}={\displaystyle \int \frac{cos2x}{{(cosx+sinx)}^2}}dx\\ ={\displaystyle \int \frac{cos2x}{1+sin2x}}dx\\ \begin{array}{l}Put 1 + sin 2x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}2 cos 2x dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{cos2x}{{(cosx+sinx)}^2}}dx\\ =\frac{1}{2}{\displaystyle \int \frac{1}{t}}dt=\frac{1}{2}log\left|t\right|+\; C\\ =\frac{1}{2}log\left|1+sin2x\right|+\; C\\ -\frac{1}{2}log\left|{(cosx+sinx)}^2\right|+\; C\\ =log\left|cosx+sinx\right|+C\end{array}$