Class 11th

$\alpha =\underset{x\to \pi /4}{lim}\frac{ta{n}^{3}x-tanx}{cos\left(x+\pi /4\right)}$

a = -4

$\beta =\underset{x\to 0}{lim}{\left(cosx\right)}^{cotx}$      

$\beta =e^0=1$

Equation whose roots are a and b

x² + 3x – 4 = 0

a = 1, b = 3

$32^{ta{n}^{2}x}+32^{se{c}^{2}x}=81$

$32^{ta{n}^{2}x}+32^{1+ta{n}^{2}x}=81$

$33.32^{ta{n}^{2}x}=81$

$32^{ta{n}^{2}x}=\frac{81}{33}$   

$forx\in \left[0,\pi /4\right]ta{n}^{2}x\in \left[0,1\right]$

One solution

Given expression

= 2π - 5 + 6 - 2π - (12 - 4π) = 4π - 11

f (m + n) = f (m) + f (n)

put m = n = 1, f (2) = f (1) + f (1)

again put m = 2, n = 1, f (3) = f (2) + f (1)

and put m = 3, n = 3, f (3 + 3) = f (3) + f (3), 2f (3) = f (6) = 18 Þ f (3) = 9

f (3) = 3f (1)

f (1) = 3, f (2) = 6

f (2).f (3) = 54

  $f\left(x\right)=si{n}^{-1}\left(\frac{3x^2+x-1}{{\left(x-1\right)}^2}\right)+co{s}^{-1}\left(\frac{x-1}{x+1}\right)$ ${}^{}\frac{{}^{}}{{}^{}}{}^{}\frac{}{}$

Domain of sin1 $\left(\frac{3x^2+x-1}{{\left(x-1\right)}^2}\right)$ $\frac{{}^{}}{{}^{}}$ is

$-1\le \frac{3s^2+x-1}{{\left(x-1\right)}^2}\le 1$

${}^{}{}^{}$  $-x^2-1+2x\le 3x^2+x-1$ and $3x^2+x-1\le x^2+1-2x$ ${}^{}{}^{}$

Domain of the function is $\left[\frac{1}{4},\frac{1}{2}\right]\cup \left\{0\right\}$ .

$2cosx\left(4sin\left(\frac{\pi }{4}+x\right)sin\left(\frac{\pi }{4}-x\right)-1\right)=1$  

->2cos x (2 cos 2x – 1) = 1

$\Rightarrow cos3x=\frac{1}{2},For0\le x\le \pi ,x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}$                

Lithium salts are extensively hydrated due to high hydration enthalpy of Li⁺

$L{i}^{+}>N{a}^{+}>K^{+}>R{b}^{+}>C{s}^{+}$ (order of polarizing power)

Moles of sodium = $\frac{8}{23}mol$  

Number of atoms = $\frac{8}{23}*6.02*10^{23}$  

$\approx 2*10^{23}$                                            

$B_2^{+}\to \sigma 1s^{2}s*1s^2\sigma 2s^2\sigma *2s^2\pi 2p^1$  

Here, number of unpaired electrons, n = 1

Spin only moment ;    $\mu =\sqrt[]{n\left(n+2\right)}B.M$

= 173 * 10^-2 B.M

$H_2\underset{hv}{\overset{high\text{?}\text{?}T}{?}}2H$

Zn + 2HCl -> ZnCl₂ + H₂

$NaO{H}_{\left(aq\right)}+Zn?N{a}_{2}Zn{O}_2+H_2$        

$H_2\stackrel{2000K}{?}2H$    

It is nearly 0.081%.