Class 12th

 $\beta =\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)},\alpha \in R\underset{x\to 0}{lim}\frac{\frac{\alpha }{3}-\left(\frac{e^{3x}-1}{3x}\right)}{\alpha x\left(\frac{e^{3x}-1}{3x}\right)}$

$=\underset{x\to 0}{lim}\frac{1-\frac{\left(1+3x+\frac{9x^2}{2}+..........-1\right)}{3x}}{1+3x+\frac{9x^2}{2}+........-1}=-\frac{1}{2}\therefore \alpha +\beta =\frac{5}{2}$

$M_A={\rho }_A*\frac{4}{3}\pi R_A^3$ ${\rho }_B=4{\rho }_A$

$M_B={\rho }_B*\frac{4}{3}\pi R_B^3$ $R_B=\frac{R_A}{2}$

$\frac{M_A}{M_B}=2,\frac{R_A}{R_B}=2$ $\frac{V_{EA}}{V_{EB}}=\sqrt[]{\frac{2G_1{M}_A}{R_A}*\frac{R_B}{2G_1{M}_B}}$

$v_{EA}=v_{EB}=12kmse{c}^{-1}$

$f_a\left(x\right)=ta{n}^{-1}2x-3ax+7\Rightarrow f_a\text{'}\left(x\right)=\frac{2}{1+4x^2}-3a\Rightarrow f_a\text{'}\left(x\right)\ge 0\Rightarrow 3a\le \frac{2}{1+4x^2}$

$a_{max}=\frac{2}{3}\left(\frac{1}{1+4*\frac{{\pi }^2}{36}}\right)=\frac{6}{9+{\pi }^2}=\overline{a}\therefore f_a\left(\frac{\pi }{8}\right)=ta{n}^{-1}\frac{2\pi }{8}-3\frac{\pi }{8}\frac{6}{9+{\pi }^2}+7=8-\frac{9\pi }{4\left(9+{\pi }^2\right)}$

$\left|\overrightarrow{A}+\overrightarrow{B}\right|=2\left|\overrightarrow{A}-\overrightarrow{B}\right|$

Given A = B

Squaring equation (1) both side.

${\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2={\left(2\left|\overrightarrow{A}-\overrightarrow{B}\right|\right)}^2$

$A^2+B^2+2\overrightarrow{A}.\overrightarrow{B}=4\left(A^2+B^2-2\overrightarrow{A}.\overrightarrow{B}\right)$

2A² + 2A² cosq = 4 (2A² – 2A² cosq)

   2A² + 2A² cosq = 8A² – 8A² cosq

 10A² cosq = 6A²

cosq =   $\frac{3}{5}$

 q = cos^-1 (3/5)

Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f   $\left(\frac{1}{4}\right)f\left(\frac{1}{3}\right)<0\Rightarrow t\in \left(\frac{1}{4},\frac{1}{3}\right)$

Let P(1 – sin q, -1 + cos q) $\therefore$  slope of normal = slope of CP $-\frac{1}{2t}=\frac{cos\theta }{-sin\theta }\Rightarrow 2t$ Þ = tan q according to question $x=1-sin\theta =1-\frac{2t}{\sqrt[]{1+4t^2}}=g\left(t\right)\therefore g\left(t\right)=1-\frac{2t}{\sqrt[]{1+4t^2}}$ ,  

Þ g'(t) < 0 g(t) is decreasing function in  $t\in \left(\frac{1}{4},\frac{1}{3}\right)\Rightarrow g\left(t\right)\in \left(0.440,0.485\right)\in \left(\frac{1}{4},\frac{1}{2}\right)$

dimension of a Pv²

        $\frac{a}{b}=pv$      dimension of b = v

Kindly consider the following figure

 $\left[Co{\left(H_{2}O\right)}_6\right]C{l}_2\Rightarrow C{o}^{2+}\Rightarrow 3d^7\Rightarrow {t_{2g}}^{5}e{g}^2$

Number of unpaired electrons = 3

$\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3\Rightarrow C{r}^{3+}\Rightarrow 3d^3\Rightarrow {t_{2g}}^{3}e{g}^0$

Number of unpaired electrons = 3

$a=\frac{F}{m}=\frac{mv}{m}=\frac{1*10}{2}=5m/se{c}^2$