Class 12th

Percentage Modulation =   $\frac{A_m}{A_c}*100=\frac{20}{80}*100=25\mathrm{\%}$

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$ ${}^{}{}^{}{}^{}{}^{}$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$

So at least one root will lie in (2, 1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$ ${}^{}{}^{}{}^{}$

$=10\left[x^4+2x^3+3x^2+2x+1\right]$

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$

So, f (x) be purely increasing function so exactly one root of f (x) that will lie in (-2, 1). Hence |a| = 2

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$

->Represent a line X – y $\ge -5$

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$  

Hence α + β) = 48

Potential difference across 2k $\Omega$  is 5V, thus current through it,

  $i=\frac{5}{2*10^3}=25*10^{-4}A.$          

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                          9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

K = 3.3 * 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$  

3.3 * 10^-4 =   $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$

$t=\frac{2.303}{3.3}*10^4*0.22\Rightarrow t=25.58mins$  

so; the nearest integer is 26.

Coordination no. of an atom in BCC is 8.

Given curves $\frac{x^2}{9}+\frac{y^4}{4}=1.........\left(i\right)$

&     $x^2+y^2=\frac{31}{4}...........\left(ii\right)$      

Equation of any tangent to (i) be y = mx +  $\sqrt[]{9m^2+4}............\left(iii\right)$

For common tangent (iii) also should be tangent to (ii) so by condition of common tangency

$9m^2+4=\frac{31}{4}\left(1+m^2\right)$

OR 36m² + 16 = 31 + 31m²

=>m² = 3

Using $\lambda =\frac{h}{\sqrt[]{2qVm}}$  

${\lambda }_{Li}=\frac{h}{\sqrt[]{2*3e*V*8.3m}}$                 m = mass of proton

${\lambda }_p=\frac{h}{\sqrt[]{2*e*V*m}}$  

$\frac{{\lambda }_{Li}}{{\lambda }_p}=\sqrt[]{\frac{2eVm}{2*24.9eVm}}$            

= 0.2 = 2 * 10^-1

So; x = 2

The balanced equation is :

$S_8\left(s\right)+12O{H}^{-}\left(aq\right)\to 4S^{2-}\left(aq\right)+2S_2{O}_3^{2-}\left(aq\right)+6H_{2}O\left(\mathcal{l}\right)$           

Here ; a = 12