Class 12th

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

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

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.

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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

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

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-The potential drop along the wires of potentiometer should be greater than emfs of cells.
In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V.