Ncert Solutions Maths class 11th

$\frac{24}{\pi }{\displaystyle \underset{0}{\overset{\sqrt[]{2}}{\int }}\frac{\left(2-x^2\right)}{\left(x^2+2\right)\sqrt[]{4+x^4}}dx}$

$\frac{24}{\pi }{\displaystyle \underset{0}{\overset{\sqrt[]{2}}{\int }}\frac{x^2\left(\frac{2}{x^2}-1\right)dx}{x\left(x+\frac{2}{x}\right)*x\sqrt[]{\frac{4}{x^2}+x^2}}}$

$x+\frac{2}{x}=t$

$dt=\left(1-\frac{2}{x^2}\right)dx$

$=-\frac{12}{\pi }\left[\frac{\pi }{4}-\frac{2\pi }{2*2}\right]=-\frac{12}{\pi }\left[-\frac{\pi }{4}\right]=3$

  $LetG.P.be{a}_1=a,a_2=ar,a_3=a{r}^2$ ${}_{}{}_{}{}_{}{}^{}$ , ….

$?3a_2+a_3=2a_4$

^{$\Rightarrow 3ar+a{r}^2=2a{r}^3$}

$\Rightarrow 2a{r}^2-r-3=0$

$\therefore a_2+a_4+2a_5=a\left(r+r^3+2r^4\right)$

$\frac{}{}$ $=\frac{8}{3}\left(\frac{3}{2}+\frac{27}{8}+\frac{81}{8}\right)$  = 40

Kindly consider the following figure

36 = 2 * 2 * 3 * 3

Number should be odd multiple of 2 and does not having factor 3 and 9

Odd multiple of 2 are

102, 106, 110, 114….998 (225 no.)

No. of multiplies of 3 are

102, 114, 126 ….990 (75 no.)

Which are also included multiple of 9

Hence,

Required = 225 – 75 = 150

$?X=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore X^2=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore Y=\alpha l+\beta X+\gamma X^2=\left[\begin{array}{ccc}\hfill \alpha \hfill & \hfill \beta \hfill & \hfill \gamma \hfill \\ \hfill 0\hfill & \hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \alpha \hfill \end{array}\right]$

$\begin{array}{l}\therefore \alpha =5,\beta =10,y=15\\ \therefore {\left(\alpha -\beta +y\right)}^2=100\end{array}$

$z^2+z+1=0\Rightarrow z=w,w^2$

$\left|{\displaystyle \sum _{n=1}^{15}{\left(z^n+{\left(-1\right)}^n\frac{1}{z^n}\right)}^2}\right|=\left|{\displaystyle \sum _{n=1}^{15}\left(z^{2n}+\frac{1}{z^{2n}}+2{\left(-1\right)}^n\right)}\right|=\left|{\displaystyle \sum _{n=1}^{15}{w}^{2n}+\frac{1}{w^{2n}}+2{\left(-1\right)}^n}\right|$

$=\left|\frac{w^2\left(1-1\right)}{1-w^2}+\frac{\frac{1}{w^2}\left(1-1\right)}{1-\frac{1}{w^2}}-2\right|=2$

$co{s}^{-1}\left(\frac{3}{10}cos\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}sin\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)\right)$

$=co{s}^{-1}\left(\frac{3}{10}.\frac{3}{5}+\frac{2}{5}.\frac{4}{5}\right)$

$=co{s}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

$16sin20°.sin40°.sin80°$

$=4sin60°\left\{?4sin\theta .sin\left(60°-\theta \right).sin\left(60°+\theta \right)=sin3\theta \right\}=2\sqrt[]{3}$

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

Here, I.F.

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

$=e^{\left(x^2-2x\right)e^x}$

$$ $\therefore$  Solution of the differential equation is

$y.e^{\left(x^2-2x\right)e^x}=\left(x^2-2x-1\right)e^{\left(x^2-2x\right)e^x}$

$$ $\therefore$  y(0) = 0

$\therefore c=1$

$\therefore y\left(2\right)=-1+1=0$

  $x\frac{dy}{dx}+2y=x{e}^x$ $\frac{}{}{}^{}$

$\frac{dy}{dx}+\frac{2y}{x}=e^x$

$\frac{dz}{dx}=e^x.2\left(x-1\right)+e^x{\left(x-1\right)}^2=0$

$e^x\left(x-1\right)\left(2+x-1\right)=0$

$x=-1,1$

x = 1 local maxima. Then maximum value is

$z\left(-1\right)=\frac{4}{e}-e$