Class 11th

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x = 2t $A\left(2t,\frac{t^2}{3}\right)$ $\frac{{}^{}}{}$

$\frac{{}^{}}{}$  $y=\frac{t^2}{3}$ S (0, 3)

$3y={\left(\frac{x}{2}\right)}^2$ $B\left(0,?\right)$

$3k=\frac{t^2}{3}+3+?=\frac{2t^2}{3}+3?\frac{12t^2}{9?t^2}$

$\underset{t?1}{lim}3k=\frac{2}{3}+3?\frac{12}{8}=3?\frac{5}{6}=\frac{13}{6}$

Here are some of the best NCERT Class 11 Chemistry Chapter 11 exercise solutions you can download and study .

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$y^2=4\left(\frac{1}{4}\right)x$

^{$x-ty+\frac{1}{4}{t}^2=0$}

^{$\frac{1}{t}=m,m_2\Rightarrow \left(8m_1{m}_2\right)=3\sqrt[]{2}-4$}

$t_n=\frac{1}{2^n.3^{11-n}}={\left(\frac{1}{3}\right)}^{11}.{\left(\frac{3}{2}\right)}^n$

$S_{10}={\left(\frac{1}{3}\right)}^{11}.\frac{\left(\frac{3}{2}\right).\left({\left(\frac{3}{2}\right)}^{10}-1\right)}{\frac{3}{2}-1}$

$=\frac{1}{3^{10}}.\frac{3^{10}-2^{10}}{2^{10}}$

$=1{+}^{10}{C}_1.2^1{+}^{10}{C}_2.2^2+....{+}^{10}{C}_9.2^9$

k = 21 + 6m -> R = 3

$\frac{x^2}{\frac{1}{25}}+\frac{y^2}{\frac{1}{4}}=1$      

$25{\alpha }^2+4{\beta }^2=1..........(i)$

Equation of tangent to parabola y = mx + $\frac{1}{m}$  passes

though (a, b)

$\alpha m^2-\beta m+1=0$

$m_1+m_2=\frac{\beta }{\alpha },4m_1^2=\frac{1}{\alpha }$

$\therefore from(i)\&(ii)$

25(a² +  a) = 1 …………(iii)

$\therefore {\left(10\alpha +5\right)}^2+{\left(16{\beta }^2+50\right)}^2=2929$

$\therefore a=1+0+b{\left(sin-\frac{\pi }{2}\right)}^{\frac{\pi }{2}}+2b+1$           

$B\left(-\frac{3}{\sqrt[]{a}},\sqrt[]{a}\right)andc\left(-\frac{3}{\sqrt[]{a}},-\sqrt[]{a}\right)$

$\therefore Areaof\Delta ACD=\frac{1}{2}\left|\begin{array}{ccc}\hfill \frac{3}{\sqrt[]{a}}\hfill & \hfill \sqrt[]{a}\hfill & \hfill 1\hfill \\ \hfill \frac{-3}{\sqrt[]{a}}\hfill & \hfill -\sqrt[]{a}\hfill & \hfill 1\hfill \\ \hfill 3cos\theta \hfill & \hfill asin\theta \hfill & \hfill 1\hfill \end{array}\right|$  = 12 

$\Rightarrow \Delta =3\sqrt[]{a}\left|cos\theta +sin\theta \right|=12$

$\therefore \Delta max=3\sqrt[]{a}.\sqrt[]{2}=12\Rightarrow a=8$

Ortho product has intramolecular H- bonding & para product has intermolecular H- bonding. Thus it can be separated by steam distillation due to difference in B.

Let x = correct answer, y = incorrect answer

  $\therefore 3x-2y=5,x+y\le 5,x,y\in w$              

 only possible (x, y) is (3, 2)

$\therefore$ Required number of ways =