Maths

${\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^{3/4}{\left(x+2\right)}^{5/4}}={\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^2}put\frac{x-1}{x+2}}=t\Rightarrow \frac{3}{{\left(x+2\right)}^2}dx=dt}$

$\therefore \frac{1}{3}{\displaystyle \int \frac{dt}{t^{3/4}}=\frac{1}{3}.\frac{t^{1/4}}{\frac{1}{4}}+C=\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{1/4}+C}$

$LHS=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to \infty }{lim}\frac{-1}{h}ln\left(\frac{1-\frac{h}{a}}{1+\frac{h}{b}}\right)=\underset{h\to 0}{lim}\frac{ln\left(1-\frac{h}{a}\right)}{a\left(\frac{-h}{a}\right)}+\underset{h\to 0}{lim}\frac{ln\left(1+\frac{h}{b}\right)}{b\left(\frac{h}{b}\right)}=\frac{1}{a}+\frac{1}{b}.............\left(i\right)$

$RHS=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{cos2x-1}{x^2+1-1}\left(\sqrt[]{x^2+1}+1\right)=\underset{x\to 0}{lim}\frac{-2si{n}^{x}x}{x^2}*2=-4...............\left(ii\right)$

$and\underset{x\to 0}{lim}f\left(x\right)=k.............\left(iii\right)$

$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=-4=k$

From (i), (ii) and (iii) we get $\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}=-4-1=-5$

Since the line $\frac{x}{5}cos\theta +\frac{y}{12}sin\theta =1$ is the equation of tangent to the ellipse $\frac{x^2}{5^2}+\frac{y^2}{12^2}=1$ at (5 cos q, 12 sin q). Hence option (A) is correct option.

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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