Class 12th

$\begin{array}{l}{\displaystyle \int x}sinxdx=x{\displaystyle \int sin}dx-{\displaystyle \int \frac{d}{dx}x{\displaystyle \int sinx}dxdx}\\ =x(-cosx)-{\displaystyle \int (}-cosx)dx\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, d)

Explanation- potential drop across AB is independent to R'  also when we decrease the value of R the current will  also very large. According to the relation I=e/R+r

$\begin{array}{l}\frac{1}{x(x^2+1)}=\frac{x}{x^2(x^2+1)}\\ Putting,x^2=t\Rightarrow 2xdx=dt\end{array}$

$\begin{array}{l}{\displaystyle \int \frac{dx}{x(x^2+1)}}={\displaystyle \int \frac{xdx}{x^2(x^2+1)}=\frac{1}{2}}{\displaystyle \int \frac{dt}{t(t+1)}}\\ =\frac{1}{2}{\displaystyle \int \frac{(t+1)-t}{t(t+1)}}dt\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{dt}{t}}-{\displaystyle \int \frac{dt}{t+1}}\right\}\\ =\frac{1}{2}\left[log\left|t\right|-log\left|t+1\right|\right]+\; C\\ =\frac{1}{2}log\left|x^2\right|-\frac{1}{2}log\left|x^2+1\right|+\; C\\ =\frac{2}{2}log\left|x\right|-\frac{1}{2}log\left|x^2+1\right|+c\\ =log\left|x\right|-\frac{1}{2}log\left|x^2+1\right|+c\\ So,option\left(A\right)iscorrect.\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation-Algebraic sum of the currents flowing towards any point in an electric network is zero. It also tell us about the conservation of charge.

Kindly go through the solution

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- I=AneV_d  , current is directly proportional to drift velocity.

Putting e^x = t so that e^x dx = dt=>dx = $\frac{dt}{e^x}=\frac{dt}{t}$

$\begin{array}{l}\therefore {\displaystyle \int \frac{dx}{e^x-1}={\displaystyle \int \frac{dt}{(t-1)*t}={\displaystyle \int \frac{dt}{t(t-1)}}}}\\ ={\displaystyle \int \frac{t-(t-1)}{t(t-1)}}dt\\ ={\displaystyle \int \frac{dt}{t-1}}-{\displaystyle \int \frac{dt}{t}}\\ =log\left|t-1\right|-log\left|t\right|+\; C\\ =log\left|e^n-1\right|-log{e}^x+C.\end{array}$

Putting x⁴ = tÞ 4x³dx = dt

$\begin{array}{l}\therefore {\displaystyle \int \frac{1}{x(x^4-1)}dx=\frac{1}{4}{\displaystyle \int \frac{dt}{t(t-1)}=\frac{1}{4}{\displaystyle \int \frac{t-(t-1)}{t(t-1)}}dt}}\\ =\frac{1}{4}\left\{{\displaystyle \int \frac{dt}{t-1}-{\displaystyle \int \frac{dt}{t}}}\right\}\\ =\frac{1}{4}\left\{log\left|t-1\right|-logt\right\}+\; C\end{array}$

$\begin{array}{l}=\frac{1}{4}log\left|\frac{t-1}{t}\right|+\; C\\ =\frac{1}{4}log\left|\frac{x^4-1}{x^4}\right|+\; C\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- R= $\frac{\delta l}{A}$ for greater value of R, A should be less and its possible value when connected across 1cm $*$ 1/2cm faces.

Putting x² = t such that 2xdx = dt

$\begin{array}{l}{\displaystyle \int \frac{2xdx}{(x^2+1)(x^2+3)}={\displaystyle \int \frac{dt}{(t+1)(t+3)}=\frac{1}{2}{\displaystyle \int \frac{2dt}{(t+1)(t+3)}}}}\\ =\frac{1}{2}{\displaystyle \int \frac{(t+3)-(t+1)dt}{(t+1)(t+3)}}\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{dt}{t+1}}-\frac{dt}{t+3}\right\}\\ =\frac{1}{2}\left\{log\left|t+1\right|-log\left|t+3\right|\right\}+\; c\\ =\frac{1}{2}log\left|\frac{t+1}{t+3}\right|+\; c\end{array}$