Ncert Solutions Maths class 11th

 $f\left(x\right)=2co{s}^{-1}x+4co{t}^{-1}x-3x^2-2x+10\forall x\in \left[-1,1\right]$

$\Rightarrow f\text{'}\left(x\right)=-\frac{2}{\sqrt[]{1-x^2}}-\frac{4}{1+x^2}-6x-2<0\forall x\in \left[-1,1\right]$

So, f (x) is decreasing function and range of f (x) is

$\left[f\left(1\right),f\left(-1\right)\right],$ which is $\left[\pi +5,5\pi +9\right]$

Now 4a – b = 4 ( + 5) (5 + 9) = 11 - π

Pascal's Triangle is a triangular array of binomial coefficients where each number is the sum of the two numbers directly above it, and its rows provide the coefficients for binomial expansions according to the Binomial Theorem. This le man language for defition of the passcl triangle.

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

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

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

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

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

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

$?3a_2+a_3=2a_4$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$16sin20°.sin40°.sin80°$

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