Class 12th

Kindly consider the following figure

B = (I – adjA)⁵

$?$ gof is differentiable at x = 0

 So R.H.D = L.H.D.

$\frac{d}{dx}\left(4e^x+k_2\right)=\frac{d}{dx}\left({\left(-\left|x+3\right|\right)}^2-k_1\left|x+3\right|\right)$  

⇒ 4 = 6 – k₁ Þ k₁ = 2

Now g (f (-4) + g (f (4)

=2 (2e⁴ – 1)

$\underset{x\to \frac{1}{\sqrt[]{2}}}{lim}\frac{sin\left(co{s}^{-1}x\right)-x}{1-tan\left(co{s}^{-1}x\right)}$

Let $co{s}^{-1}x=\frac{\pi }{4}+\theta$

= $\underset{\theta \to \infty }{lim}\frac{\sqrt[]{2}sin\theta }{-2tan\theta }\left(1-tan\theta \right)=-\frac{1}{\sqrt[]{2}}$

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -8\hfill & \hfill 9\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 3\\ -1\end{array}\right)$

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 15\hfill & \hfill -24\hfill & \hfill 27\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 3a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 9\\ -3\end{array}\right)$

$R_3-2R_1,R_2-5R_1$

for no solution

3a + 9 = 0 but $\frac{3}{2}-\frac{9b}{2}\ne 0$

$\Rightarrow a=-3\Rightarrow b\ne \frac{1}{3}$

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

  $\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

  $x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

CoCl₃.NH₃ + AgNO₃ $\stackrel{}{\to }AgC\mathcal{l}\downarrow \left(2mol\right)$  

$\left[Co{\left(N{H}_3\right)}_{5}C\mathcal{l}\right]C{\mathcal{l}}_2+AgN{O}_3\stackrel{}{\to }AgC\mathcal{l}\downarrow \left(2mol\right)$  

x = 5

$M{n}^{2+}\stackrel{}{\to }{t}_{2g^{111}}{e_g}^{11}$

(Number of unpaired electron = 5)

${\mu }_S=\sqrt[]{35}=5.91\cong 6$

$K=A.e^{-Ea/RT}=\left(6.5*10^2\right)e^{-26000K/T}$

$\frac{Ea}{8.314}=26000$

Ea = 216.164kJ/mol $\cong$ $$ 216

H₂ (g) + Cu²⁺ (aq) $\stackrel{}{\to }$  2H⁺ (aq) + Cu (s)

  $\therefore E_{cell}=E_{cell}^0-\frac{2.303RT}{nf}logQ$

0.31 = 0.34 - $\frac{0.06}{2}log\frac{\left[H^{+}\right]}{C{u}^{2+}}$  

  $\left[C{u}^{2+}\right]=10^{-7}\left(?\left[H^{+}\right]=10^{-3}\right)$

x = 7