Class 12th

  ${E^o}_{Cell}={E^o}_{A{g}^{+}/Ag}-{E^o}_{Z{n}^{+2}/Zn}=0.8+0.76=1.56V$

Anode :  $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$    

Cathode :  2Ag⁺(aq) + 2e^- -> 2Ag(s)

Zn(s) + 2Ag⁺(aq) -> Zn²⁺ (aq) + 2Ag(s)

  $E_{cell}={E^o}_{Cell}-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}=1.56-\frac{0.0591}{2}log\left(\frac{0.1}{10^{-4}}\right)$          

x = 147

Cis (optically active)                          Trans (optically inactive)

d & l form (2)                                                    (1)

So, total 3 isomerism.

$f\left(x\right)=2co{s}^{-1}x+4co{t}^{-1}x-3x^2-2x+10\forall x\in \left[-1,1\right]$

$\Rightarrow f\text{'}\left(x\right)=-\frac{2}{\sqrt[]{1-x^2}}-\frac{4}{1+x^2}-6x-2<0\forall x\in \left[-1,1\right]$

So, f (x) is decreasing function and range of f (x) is

$\left[f\left(1\right),f\left(-1\right)\right],$ which is $\left[\pi +5,5\pi +9\right]$

Now 4a – b = 4 (p + 5) - (5p + 9) = 11 - “π”

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

  $f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$

$?8x^2?2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

  $9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

  $f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

  $9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

Fraction of molecules having enough energy to form product  =   $e^{-Ea/RT}$

 Fraction of molecules having enough energy to form product =   $e^{-\frac{80.9*10^3}{8.314*700}}$

$=e^{-13.8}\cong e^{-14}$

So, x = 14

$\Delta T_f=K_f*m*i$  

0.93 = 1.86 * 1 * i

i =  $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  

$i=1+\left(\frac{1}{n}-1\right)$

$\therefore n=2$            

$Mn{O}_4^{-}+S_2{O}_3^{2-}\stackrel{O{H}^{-}}{\to }Mn{O}_2+S{O}_4^{2-}$

O.S of S in    $S{O}_4^{2-}=+6$

Effective no. of octahedral voids in a lattice = n

Effective no. of lattice point in a lattice = n

Ratio = n/n = 1

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$  

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$         

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$  

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (-2, 2)