Maths

$?a_r=e^{i\frac{2r\pi }{9}}$

$\therefore a_1,a_2,a_3,...........$ are in G.P.

$\left|\begin{array}{ccc}\hfill a_1\hfill & \hfill a_{1}r\hfill & \hfill a_1{r}^2\hfill \\ \hfill a_1{r}^3\hfill & \hfill a_1{r}^4\hfill & \hfill a_1{r}^5\hfill \\ \hfill a_1{r}^6\hfill & \hfill a_1{r}^7\hfill & \hfill a_1{r}^8\hfill \end{array}\right|={a_1}^3{r}^9\left|\begin{array}{ccc}\hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \\ \hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \\ \hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \end{array}\right|=0$

$\left(i\right)a_2{a}_6-a_4{a}_8=a_{1}r.a_1{r}^5-a_1{r}^6{a}_1{r}^7={a_1}^2{r}^3-{a_1}^2{r}^{10}\ne 0$

$\left(ii\right)a_9=a_1{r}^8\ne 0$

$\left(iii\right)a_1.a_1{r}^8-a_1{r}^2.a_1{r}^6=0$

$\left(iv\right)a_5\ne 0$

$\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi -\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{x^4}$

$=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}*\frac{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}{x^4}=4{\pi }^2\underset{x\to 0}{lim}\frac{si{n}^{4}x}{x^4}=4{\pi }^2$

Length of perpendicular from origin to xcosec a - y sec a = k cot2 a and x sin a + ycos a = k sin 2a are

$p=kcot\alpha sin\alpha cos\alpha =\frac{k}{2}.\frac{cos2\alpha }{sin2\alpha }sin2\alpha =\frac{k}{2}cos2\alpha \Rightarrow \frac{2p}{k}=cos2\alpha$

and $q=ksin2\alpha \Rightarrow \frac{q}{k}=sin2\alpha$

on solving these two we get 4p² + q² = k²

$f\left(x\right)=\left|x^2-2x-3\right|e^{\left|9x^2-12x+4\right|}$

$\Rightarrow f\left(x\right)=\{\begin{array}{l}\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x<-1\\ -\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},-1\le x<3\\ \left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x\ge 3\end{array}$

$f\text{'}\left(-1^{-}\right)\ne f\left(-1^{+}\right),$

$f\text{'}\left(3^{-}\right)\ne f\left(3^{+}\right)$

Number of non differential points is 2 at x = -1, 3.

$e^{4x}+2e^{3x}-e^x-6=0$

$e^x=t\in \left(0,\infty \right)$

$t^4+2t^3-t-6=0$

Let f (t) = t⁴ + 2t³ – t – 6

f' (t) = 4t³ + 6t² – 1f

Þf' (0) = -1, f' (+ $\infty )$ = + $\infty$

For t > 0

Þ f' (t) = 0 has only one root.

One solution of f (t) = 0 is possible

Equation of plane is 3x – 2y + 4z – 7 + $\lambda$ (x + 5y – 2z + 9) = 0. (i)

It passes through (1, 4, -3) and we get $\lambda =\frac{2}{3}$

$\therefore$ from (i) we get 11x + 4y + 8z – 3 = 0 Þ -11x – 4y – 8z + 3 = 0

$\therefore \alpha +\beta +\lambda =-11-4-8=-23$  

$\left(p\wedge \sim q\right)\to \left(p\vee q\right)$ is tautology

$*=\wedge ,?=\vee$

Let $l={\displaystyle \int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}}dx={\displaystyle \int \frac{2e^{2x}+3}{4e^{2x}+7}dx}$  

$={\displaystyle \int \frac{2e^{2x}}{4e^{2x}+7}dx+{\displaystyle \int \frac{3e^{-2x}}{4+7e^{-2x}}dx}}$               

Put $4e^{2x}+7=t4+7e^{-2x}=\lambda$  

$8e^{2x}dx=dt-14e^{-2x}dx=d\lambda$              

$=\frac{1}{4}{\displaystyle \int \frac{dt}{t}-\frac{3}{14}{\displaystyle \int \frac{d\lambda }{\lambda }=\frac{1}{4}lnt-\frac{3}{14}ln\lambda +c}}$

$u=\frac{13}{2},v=\frac{1}{2}$              

->so, u + v = 7

Since A (sec θ, 2 tanθ) & B $\left(sec?,2tan?\right)$  lies on 2x² – y² = 2 then

2sec² θ - 4tan^2_θ = 2 or sec² θ - 2 tan² θ = 1

-> $ta{n}^2\theta =0so\theta =0$

Similarly    $?=0but\theta +?=\frac{\pi }{2}$ (given) so not possible

Hence question is not correct

Given variance of boys ${\sigma }_b^2=2\&{\overline{x}}_b=12$ (average marks of boys)

& variance of girls  ${\sigma }_g^2=2\&\mu \to$ average marks of girls

$Now,{\overline{x}}_g=\mu =\frac{50*15-12*20}{30}=17$           

${\sigma }^2=\frac{20*2+30*2}{20+30}+\frac{20*30}{{\left(20+30\right)}^2}{\left(12-17\right)}^2=8$            

$So,\mu +{\sigma }^2=17+8=25$