Class 12th

From option let it be isosceles where AB = AC then

$x=\sqrt[]{r^2-{\left(h-r\right)}^2}$            

=  $\sqrt[]{r^2-h^2-r^2+2rh}$

$x=\sqrt[]{2hr-h^2}........\left(i\right)$

 Now ar $\left(\Delta ABC\right)=\Delta =\frac{1}{2}BC*AL$

$\Delta =\frac{1}{2}*2\sqrt[]{2hr-a^2}*h$        

then  $x=\sqrt[]{2*\frac{3r}{2}*r-\frac{9r^2}{4}}=\frac{\sqrt[]{3}}{2}rfrom\left(i\right)$

$\Rightarrow BC=\sqrt[]{3}r$    

So $AB=\sqrt[]{h^2+x^2}=\sqrt[]{\frac{9r^2}{4}+\frac{3r^2}{4}}=\sqrt[]{3}r$ .

Hence $\Delta$ be equilateral having each side of length $\sqrt[]{3}r.$

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20*\frac{15^2}{2}+8.\frac{15^3}{3}=11250C$

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

$A_1+A_2={\displaystyle {\int }_0^{\pi /2}cosxdx}$

$={\left(Sinx\right)}_0^{\pi /2}=1$
$A_1={\displaystyle {\int }_0^{\pi /4}\left(cosx-sinx\right)dx={\left(sinx+cosx\right)}_0^{\pi /4}}$

$=\frac{2}{\sqrt[]{2}}-1=\sqrt[]{2}-1$

$So{A}_2=1-\left(\sqrt[]{2}-1\right)=2-\sqrt[]{2}=\sqrt[]{2}\left(\sqrt[]{2}-1\right)$

$Now\frac{A_1}{A_2}=\frac{1}{\sqrt[]{2}}$             

$K_{2}C{r}_2{O}_7+H_{2}S{O}_4+\underset{GasX}{S{O}_2}\to \underset{\begin{array}{l}Green\\ Y\end{array}}{C{r}_2{\left(S{O}_4\right)}_3}+K_{2}S{O}_4+H_{2}O$

When connected in series, equivalent capacitance,

$C_1=\frac{C*C}{C+C}=\frac{C}{2}$            

When connected in parallel, equivalent capacitance

C₂ = C + C = 2C

$\frac{C_1}{C_2}=\frac{C/2}{2C}=\frac{1}{4}$            

Given

$\left(x\right)=\{\begin{array}{l}2sin\left(\frac{-\pi x}{2}\right),x<-1\\ \left|a{x}^2+x+b\right|,-1\le x\le 1\\ sin\pi x,x>1\end{array}$

If f (x) is continuous for all $x\in R$ then it should be continuous at x = 1 & x = -1

At x = -1, L.H.L = R.H.L. Þ 2 = |a + b - 1|

=>a + b – 3 = 0  OR  a + b + 1 = 0 . (i)

=>a + b + 1 = 0 . (ii)

           (i) & (ii), a + b =-1

Given f(X) = ${\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{\left(1+t\right)}dt............\left(i\right)}$  

So  $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^{1/x}\frac{\lambda lo{g}_{e}t}{1+t}dt}............\left(ii\right)$

put $t=\frac{1}{z}thendt=-\frac{1}{z^2}dz$

$f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{t\left(1+t\right)}}dt............\left(iii\right)$    

(i) + (iii), f(x) + $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\left(\frac{lo{g}_{e}t}{1+t}+\frac{lo{g}_{e}t}{t\left(1+t\right)}\right)dt}$

$={\left[\frac{{\left(lo{g}_{e}t\right)}^2}{2}\right]}_1^x=\frac{{\left(lo{g}_{x}x\right)}^2}{2}$  

Hence f(e) + $f\left(\frac{1}{e}\right)=\frac{1}{2}$

$i=\frac{6-4}{2+8}=0.2A$

$V_x+4+0.2*8=V_Y$

$\Rightarrow V_Y-V_X=5.6V$

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

=> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21*\frac{34}{7}=102$