Class 12th

$f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

      $\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$  

$d=4\sqrt[]{3}$ cm (Lateral shift)

By Snell's law

${\mu }_{air}sin60°={\mu }_{g}sin\theta$

θ = 30

$\Rightarrow 1*\frac{\sqrt[]{3}}{2}=\sqrt[]{3}sin\theta$  

$sin\theta =\frac{1}{2}$  

t = 12 cm

Since a is a odd natural number then $\left|{\displaystyle \underset{1}{\overset{3}{\int }}{y}^{a}dy}\right|=\frac{364}{3}\Rightarrow \left|{\left(\frac{y^{a+1}}{a+1}\right)}_1^3\right|=\frac{364}{3}\Rightarrow \frac{3^{a+1}}{a+1}=\frac{364}{3}$  

 Þ a = 5

| (A + I) (adj A + I)| = 4 Þ |A adj A + A + Adj A + I| = 4 Þ | (A)I + A + adj A + I|= 4|A| = -1

Þ |A + adj A| = 4

$A=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow adjA=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill -c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow \left|\begin{array}{cc}\hfill \left(a+d\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \left(a+d\right)\hfill \end{array}\right|=4\Rightarrow a+d=\pm 2$  

$\Delta =\left|\begin{array}{ccc}\hfill 8\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \lambda \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|=12-3\lambda$  

So for  $\lambda$ = 4, it is having infinitely many solutions. ${\Delta }_x=\left|\begin{array}{ccc}\hfill -2\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \mu \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|$  = -6 - $3\mu =0\Rightarrow -6-3\mu =0$  

For  $\mu =-2$ distance of $\left(4,-2,-\frac{1}{2}\right)$ from 8x + y + 4z + 2= 0 $\left|\frac{32-2-2+2}{\sqrt[]{64+1+16}}\right|=\frac{10}{3}$  units

 f (3x)- f (x) = x

Replace  $x\to \frac{x}{3}\Rightarrow f\left(x\right)-f\left(\frac{x}{3}\right)=\frac{x}{3}$  

Again replace  $x\to \frac{x}{3}\Rightarrow f\left(\frac{x}{3}\right)-f\left(\frac{x}{3^2}\right)-f\left(\frac{x}{3^2}\right)=\frac{x}{3^2}$

$\Rightarrow f\left(3x\right)-f\left(0\right)=\frac{3x}{2}puttingx=\frac{8}{3}\Rightarrow f\left(8\right)-f\left(0\right)=4\therefore f\left(0\right)=3$  

Also putting x =  $\frac{14}{3}$ in f (3x) – 3 = $\frac{3x}{2}\Rightarrow$ F (14) – 3 = 7 Þ f (14) = 10

$\left[Cu{\left(en\right)}_2{\left(SCN\right)}_2\right]$

More stable isomers = 3 (trans isomers)

t_1/2 = 0.301 min

              t = 2 min

              $K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{2.303*0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{Co}{Ct}=10^2=100$  

              Ans. 100

Ka for C₃H₇COOH = 2 * 10^-5

              $pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$  

              =5 – 0.3 = 4.7

              pH of 0.2 (M) solution =

              $pH=\frac{pKa-\mathcal{l}ogC}{2}$  

              $=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

              $pH=27*10^{-1}$     

          $\therefore$    Ans 27