Maths NCERT Exemplar Solutions Class 12th Chapter Nine

2cos⁴ x – cos²x = $2{\left(\frac{1+cos2x}{2}\right)}^2-cos2x$

$=\frac{1+co{s}^{2}2x}{2}$

$\sqrt[]{2}{\displaystyle \int \frac{2dx}{1+co{s}^{2}2x}=\sqrt[]{2}{\displaystyle \int 2}\cdot \frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}$

$=\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}\cdot ta{n}^{-1}\left(\frac{tanx}{\sqrt[]{2}}\right)$

now IF = $e^{{\displaystyle \int Pdx}}$

$e^{\sqrt[]{2}{\displaystyle \int 2\frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}}=e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}$

Solution: $y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}={\displaystyle \int x.e^{ta{n}^{-1}\left(\sqrt[]{2}\cdot cot2x\right)}\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}dx}$

$y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}=\frac{x^2}{2}\cdot e^{\pi /2}+C$

at $x=\frac{\pi }{4},y=\frac{{\pi }^2}{32},C=0$

at $x=\frac{\pi }{3},y=\frac{{\pi }^2}{18}{e}^{ta{n}^{-1}\alpha }$

$y\cdot e^{ta{n}^{-1}}\frac{\left(tan2\frac{\pi }{3}\right)}{\sqrt[]{2}}=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

$\frac{{\pi }^2}{18}\cdot e^{ta{n}^{-1}\alpha }\cdot e^{ta{n}^{-1}}\left(-\frac{\sqrt[]{3}}{2}\right)=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

tan^-1 a + tan^-1 $\left(-\sqrt[]{3}/2\right)=\frac{\pi }{2}$

cot^-1a = tan^-1 $\left(-\frac{\sqrt[]{3}}{2}\right)$

tan^-1 $\frac{1}{\alpha }=ta{n}^{-1}\left(-\frac{\sqrt[]{3}}{2}\right)$

$\frac{1}{\alpha }=-\sqrt[]{3}/2$

2 = $\frac{2}{3}$

$a_{n+2}=2\cdot a_{n+1}-a_{n+1}$

$a_2=2a_1-a_0+1$

a₂ = 1

a₃ = 3

a₄ = 6

So for n $\ge 2,a_n=\frac{n\left(n-1\right)}{2}$

${\displaystyle \sum _{n=2}^{}\frac{n\left(n-1\right)}{2\cdot 7^n}=\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+......}$

Let S = $\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+.....$

$\frac{\frac{S}{7}=\frac{1}{7^3}+\frac{3}{7^4}+....}{S-\frac{S}{7}=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....}$

$\frac{\frac{6}{7}S\cdot \frac{1}{7}=\frac{1}{7^3}+\frac{2}{7^3}+....}{\frac{6}{7}S-\frac{6}{49}S=\frac{1}{7^2}+\frac{1}{7^3}+....}$

$\frac{36}{49}S=\frac{\frac{1}{7^2}}{1-\frac{1}{7}}$

$S=\frac{1}{7^2}*\frac{7}{6}*\frac{49}{36}$

$S=\frac{7}{216}$

 $\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}$

Put y = $V\cdot x$

differentiable worst = x

$\frac{dy}{dx}=V+x\cdot \frac{dV}{dx}$

V + $x\frac{dV}{dx}=V+\sqrt[]{V^2+16}$

$x\frac{dV}{dx}=\sqrt[]{V^2+16}$

apply variable separable method

${\displaystyle \int \frac{dV}{\sqrt[]{V^2+16}}={\displaystyle \int \frac{dx}{x}+\mathcal{l}nC}}$

$\Rightarrow \mathcal{l}n\left|V+\sqrt[]{V^2+16}\right|=\mathcal{l}nCx$

$\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}=Cx$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2}+\frac{\sqrt[]{y^2+16.4}}{2}=8.2$

y2 + 64 = (32 – y)2

y = 15

$d_1=\frac{199-100}{2}\cancel{\in }l$

$d_2=\frac{199-100}{3}=33$

$d_3=\frac{199-100}{4}\cancel{\in }l$

$d_n=\frac{199-100}{i+1}\in l$

$\Rightarrow di=33+11or9$

$\therefore$ sum of common differences = 33 + 11 + 9 = 53

 $\frac{dy}{dx}-y=2-e^{-x}$

$I.F.=e^{-{\displaystyle \int dx}}=e^{-x}$

$\therefore$ soln

$y{e}^{-x}={\displaystyle \int \left(2e^{-x}-e^{-2x}\right)dx}$

$y=-2+\frac{e^{-x}}{2}+C{e}^x$

as for x $\infty$ y finite c = 0

$\therefore y=\frac{e^{-x}}{2}-2$

$\begin{array}{l}\Rightarrow x+2y=-3\\ \Rightarrow a=-3b=\frac{-3}{2}\end{array}$

$\therefore a=4b=-3+6=3$

 $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt[]{1-x^2}},$ $I.F.e^{{\displaystyle \int \frac{xdx}{x^2-1}}}=\sqrt[]{\left|x^2-1\right|}=\sqrt[]{1-x^2}$ $\left(?x\in \left(-1,1\right)\right)$

Solution of differential equation is $y\sqrt[]{1-x^2}={\displaystyle \int \left(x^4+2x\right)dx=\frac{x^5}{5}+x^2+c}$

Curve is passing through origin, c = 0 $y=\frac{x^5+5x^2}{5\sqrt[]{1-x^2}}$

$\therefore {\displaystyle \underset{-\frac{\sqrt[]{3}}{2}}{\overset{\frac{\sqrt[]{3}}{2}}{\int }}\frac{x^5+5x^2}{5\sqrt[]{1-x^2}}dx=\frac{\pi }{3}-\frac{\sqrt[]{3}}{4}}$

 $\frac{6}{3^{12}}+10\left(\frac{1}{3^{11}}+\frac{2}{3^{10}}+\frac{2^2}{3^9}+\frac{2^3}{3^8}+....+\frac{2^{10}}{3}\right)$

$\frac{6}{3^{12}}=\frac{10}{3^{11}}\left(\frac{6^{11}-1}{6-1}\right)$

$=2^{12}.1m.n=12$

 $1+\left(1+2^{49}\right)\left(2^{49}-1\right)=2^{98}$

M = 1, n = 98

M + n = 99

Equation of circle passing through (0, 2) and (0, 2) is

$x^2+\left(y^2-4\right)+\lambda x=0,\left(\lambda \in R\right)$

Divided by x we get

$\frac{x^2+\left(y^2-4\right)}{x}+\lambda =0$

Differentiating w.r.t. x

$\frac{x\left[2x+2y.\frac{dy}{dx}\right]-\left[x^2+y^2-4\right].1}{x^2}=0$

$\Rightarrow 2xy.\frac{dy}{dx}+\left(x^2-y^2+4\right)=0$

 $\frac{dy}{dx}+\frac{1}{x^2-1}y={\left(\frac{x-1}{x+1}\right)}^{1/2}$

$\frac{dy}{dx}+py=Q$

$I.F=e^{{\displaystyle \int Pdx}}={\left(\frac{x-1}{x+}\right)}^{\frac{1}{2}}$

$x-2lo{g}_e\left|x+1\right|+C$

Curves passes through $\left(2,\frac{1}{\sqrt[]{3}}\right)$

$\Rightarrow C=2lo{g}_{e}3-\frac{5}{3}$

$atx=8,\sqrt[]{7}y\left(8\right)=19-6lo{g}_{e}3$