Class 11th

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

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

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

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

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

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

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

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

  $?p+q=3$ $$ ….(i)

and $p^4+q^4$ ${}^{}{}^{}$ = 369 ….(ii)

${\left\{{\left(p+q\right)}^2-2pq\right\}}^2-2p^2{q}^2=369$

or ${\left(9-2pq\right)}^2-2{\left(pq\right)}^2=369$ ${}^{}{}^{}$

or ${\left(pq\right)}^2-18pq-144=0$ ${}^{}$

$\therefore pq=-6or24$

But pq 24 is not possible

$\therefore pq=-6$

$Hence,{\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}={\left(\frac{pq}{p+q}\right)}^2={\left(-2\right)}^2=4$

$f\left(x-y\right)=2^{x}f\left(y\right)+4^{y}f\left(x\right)$ ……. (i)

$$ $f\left(y+x\right)=2^{y}f\left(x\right)+4^{x}f\left(y\right)$  ……. (ii)

(i) – (ii)

$\therefore \frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=\frac{k\left(16\mathcal{l}n2-256\mathcal{l}n4\right)}{k\left(4\mathcal{l}n2-16\mathcal{l}n4\right)}$

$=\frac{\left(16-512\right)\mathcal{l}n2}{\left(4-32\right)\mathcal{l}n2}=\frac{496}{28}$

$14\frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=248$

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

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

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

$16sin20°.sin40°.sin80°$

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

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

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

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

$\left|x^2-9\right|=3$

$\Rightarrow x=\pm 2\sqrt[]{3},\pm \sqrt[]{6}$

Required area = A

$\frac{A}{2}={\displaystyle \underset{0}{\overset{\sqrt[]{6}}{\int }}\left(9-x^2-3\right)dx+{\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{9+y}-\sqrt[]{9-y}\right)dy}}$

$A=16\sqrt[]{6}+32\sqrt[]{3}-72=8\left[2\sqrt[]{6}+4\sqrt[]{3}-9\right]$

Note : No option in the question paper is correct.