Maths

Let $?\frac{a}{r}$ , a, ar are in G.P. $\therefore \frac{a}{r},$ 2a, ar are in A.P.

$2a-\frac{a}{r}=ar-2a\Rightarrow 2-\frac{1}{r}=r-2\Rightarrow r^2-4r+1=0\Rightarrow r=2+\sqrt[]{3}$

$A/q,a{r}^2=3r^2\Rightarrow a=3$

$\begin{array}{l}\therefore d=ar-2a=3\sqrt[]{3}\\ \therefore r^2-d=7+\sqrt[]{3}\end{array}$

Length of latus rectum = 4 (distance between vertex and focus) = 4 (S – R)

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

$\therefore a_1,a_2,a_3,...........$ 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

$*=\wedge ,?=\vee$