Class 12th

$\frac{1-x^2}{x(1-2x)}$ which is not a proper fraction

So, division is missing

$\begin{array}{l}So,\frac{1-x^2}{x(1-2x)}=\frac{1}{2}+\frac{(\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1)}{x(1-2x)}\\ =\frac{1}{2}+\frac{1}{2}\frac{(2-x)}{x(1-2x)}\\ Let\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}\end{array}$

=> 2 -x = A(1 - 2x) + B x.

   So, - 2A + B = - 1

   A = 2.

   So, B = - 1 + 2A = - 1 + 2 ´ 2 = - 1 + 4 = 3

$\begin{array}{l}\therefore {\displaystyle \int \frac{1-x^{2}dx}{x(1-2x)}}={\displaystyle \int \frac{1}{2}}dx+{\displaystyle \int \frac{1}{2}}\frac{(2-x)}{x(1-2x)}dx\\ =\frac{x}{2}+\frac{1}{2}\left\{{\displaystyle \int \frac{A}{x}}dx+{\displaystyle \int \frac{B}{1-2x}}dx\right\}\\ =\frac{x}{2}+\frac{1}{2}\{\left\{\frac{2}{x}dx+{\displaystyle \int \frac{3}{(1-2x)}}dx\right\}\\ =\frac{x}{2}+log\left|x\right|+\frac{3}{2}log\frac{|1-2x|}{(-2)}+c\\ =\frac{x}{2}+log\left|x\right|-\frac{3}{4}log|1-2x|+c\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- when all resistance are in parallel

$\frac{1}{R_p}=\frac{1}{R_1}+\dots \dots \dots \dots ..+\frac{1}{R_n}$ by multiplying this equation by R_min we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$

But there exist a term in RHS $\frac{R_{min}}{R_{min}}=1$ and other terms are positive so we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$ >1

This shows that equivalent resistance is less than maximum resistance available.

But when all resistance are in series

R_s =R₁+R₂………R_n

here must be R_max value in RHS

R_s= R₁+……R_max+….R_n

And R_s> R_max

So equivalent resistance is less than R_max

$\frac{2x}{x^2+3x+2}=\frac{2x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

2x = A(x + 2) + B(x + 1).

= (A + B) x + (2A + B).

Comparing the coefficients we get,

A + B = 2 ---(1)

2A + B = 0 ---- (2).

So, Eqⁿ (2) - (1),

2A + B - (A + B) = 0 - 2

Þ A = - 2

And B = 2 - A = 2 - (-2) = 2 + 2 = 4

B = 4

$\begin{array}{l}So,\frac{2x}{(x+1)(x+2)}=\frac{-2}{x+1}+\frac{4}{x+2}.\\ \therefore {\displaystyle \int \frac{2xdx}{(x+1)(x+2)}}=-2{\displaystyle \int \frac{dx}{x+1}}+4{\displaystyle \int \frac{dx}{x+2}}\\ =-2log|x+1|+4log|x+2|+c\\ =4log|x+2|-2log|x+1|+c\end{array}$

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation-

 total resistance = 2+8= 10 ohm

I= $\frac{6-4}{2+8}$ = 0.2A

The direction of flow of current is always from high potential to low potential

If VB $>$ VA

VB-4V-0.2 $*$ 8 = VA

V_B-V_A= 3.6V

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$\begin{array}{l}So,\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}+\frac{(-5)}{(x-2)}+\frac{4}{(x-3)}\\ Then,{\displaystyle \int \frac{3x-1dx}{(x-1)(x-2)(x-3)}}={\displaystyle \int \frac{dx}{x-1}}-5{\displaystyle \int \frac{dx}{x-2}}+4{\displaystyle \int \frac{dx}{x-3}}.\\ =log|x-1|-5log|x-2|+4log|x-3|+c\end{array}$

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – in series combination of resistors current I =e/R+r

10I= $\frac{e}{R+\frac{r}{n}}$

$\frac{1+n}{1+\frac{1}{n}}$ $\Rightarrow$ 10 = ( $\frac{1+n}{1+n}$ )n $\Rightarrow$ n=10

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