Class 12th

$\begin{array}{l}\frac{6x+7}{x^2-9x+20}\\ Let6x+7=A\frac{d}{dx}\left(x^2–9x+20\right)+B\\ \Rightarrow 6x+7=A\left(2x–9\right)+B\\ Equatingthecoefficientofxandconstantterm,wehave\\ \begin{array}{l}2A = 6 and –9A+B= 7\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow A=3 \Rightarrow B= 34\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\therefore 6x+ 7 = 3\left(2x–9\right) + 34\hfill \end{array}\end{array}$

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

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{\rho l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{\rho l}{A}$ l²= 1.7 $*$ 10^{-8 $*\frac{10}{\pi *{10}^{-6}}*81$} j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000}*$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000}*$ 100= 0.32%

$\begin{array}{l}Let5x–2=A\frac{d}{dx}\left(1+2x+3x^2\right)+B\\ \Rightarrow 5x–2=A\left(2+6x\right)+B\\ =2A+A6x+B \end{array}$

$\begin{array}{l}Equatingthecoefficientofxandconstanttermonbothsides,wehave\\ 5=6A; and 2A+B=–2 \\ \Rightarrow A= \frac{5}{6}\\ \Rightarrow B=–2–2A \\ B=\frac{-11}{3} \\ Therefore,\\ 5x-2=\frac{5}{6}(2+6x)+\left(\frac{-11}{3}\right)\\ \therefore I=\int \frac{5x-2}{1+2x+3x^2}dx\\ =\int \frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.(2+6x)+\left(\raisebox{1ex}{$-11$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}{1+2x+3x^2}dx\\ =\frac{5}{6}\int \frac{(2+6x)}{1+2x+3x^2}dx-\frac{11}{3}\int \frac{1}{1+2x+3x^2}dx\\ Let,I_1=\int \frac{2+6x}{1+2x+3x^2}dxand{I}_2=\int \frac{1}{1+2x+3x^2}dx\\ \therefore I=\int \frac{5x-2}{1+2x+3x^2}dx\\ =\frac{5}{6}{I}_1-\frac{11}{3}{I}_2-----(1)\\ \begin{array}{l}Let 1 + 2x+ 3x^2=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow \left(2+6x\right)dx=dt\hfill \end{array}\\ I_1=\int \frac{dt}{t}=log\left|t\right|\\ =log\left|1+2x+3x^2\right| \_\_\_\_\left(2\right)\\ I_2=\int \frac{1}{1+2x+3x^2}dx\\ Now,1+2x+3x^2=1+3\left(x^2+\frac{2}{3}x\right)\\ Therefore,1+3\left(x^2+\frac{2}{3}x\right)=1+3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}7–6x–x^2=7–\left(x^2+6x+9–9\right)\end{array}$

$\begin{array}{l}=7–\left(x^2+6x+9\right)+9\end{array}$

$\begin{array}{l}=16–\left(x^2+6x+9\right) \end{array}$

$\begin{array}{l}=16–{\left(x+3\right)}^2\end{array}$

$\begin{array}{l}=\left(4^2\right)–{\left(x+3\right)}^2\end{array}$

$\begin{array}{l}I=\int \frac{1}{9x^2+6x+5}dx=\int \frac{1}{{(3x+1)}^2+2^2}\\ \begin{array}{l}Let \left(3x+1\right) =t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow 3dx=dt\hfill \end{array}\\ I=\int \frac{1}{{(3x+1)}^2+2^2}dx\\ =\frac{1}{3}\int \frac{1}{t^2+2^2}dt\\ =\frac{1}{3}\left[\frac{1}{2}ta{n}^{-1}\left(\frac{t}{2}\right)\right]+C\\ =\frac{1}{6}ta{n}^{-1}\left(\frac{t}{2}\right)+C=\frac{1}{6}ta{n}^{-1}\left(\frac{3x+1)}{2}\right)+C\end{array}$