Class 12th

Let f(x) = $\frac{x^2-9}{x-5}$  

  $f\text{'}\left(x\right)=\frac{2x\left(x-5\right)-\left(x^2-9\right)}{{\left(x-5\right)}^2}$

$=\frac{x^2-10x+9}{{\left(x-5\right)}^2}=\frac{\left(x-1\right)\left(x-9\right)}{{\left(x-5\right)}^2}$

$\alpha =f\left(1\right)=2,\beta =\left\{f\left(0\right),f\left(2\right)\right\}=\frac{5}{3}$

$\therefore {\displaystyle \underset{-1}{\overset{3}{\int }}max}\left\{\frac{x^2-9}{x-5},x\right\}dx+{\left[\frac{x^2}{2}\right]}_{9/5}^3$           

a₁ = 18, a₂ = 16

a₁ + a₂ = 34

  $f\left(\theta \right)=sin\theta +{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(sin\theta +tcos\theta \right)f\left(t\right)dt}$

$=sin\theta +sin\theta {\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt+cos\theta {\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}}$

  $=\left(1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt}\right)sin\theta +\left({\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}\right)cos\theta$              

f(q) = a sin q + b cos q

  $\therefore a=1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt=1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(asint+bcost\right)dt}}$

$\therefore a=2b+1..........(i)$

$b={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(asint+bcost\right)dt}$

 Solving (i) & (ii) we get a =   $-\frac{1}{3},b=-\frac{2}{3}$

    $\therefore \left|{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\theta \right)d\theta }\right|=1$           

$y=2\left|x^2-\frac{3}{2}x-\frac{7}{2}\right|$

$=2\left|{\left(x-\frac{3}{4}\right)}^2-\frac{65}{16}\right|$

  $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}=r_1$

$A\left(3r_1+7,1-r_1,-2+r_1\right)$

and   $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}=r_2$  

  $B\left(2r_2,7+3r_2,r_2\right)$

$A/q,\frac{3r_1-2r_2+7}{1}=\frac{3r_2+r_1+6}{4}=\frac{r_1-r_2-2}{2}$             

  $\Rightarrow r_1=-5,r_2=-3$

$\therefore A\left(-8,6,-7\right)andB\left(-6,-2,-3\right)$

AB² = 84

  $f\left(x\right)=\{\begin{array}{l}\left|2x^2-3x-7\right|,x\le -1\\ \left[4x^2-1\right],-1<x<1\\ \left|x+1\right|+\left|x-2\right|,x\ge 1\end{array}$              

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x=\pm \frac{1}{2},\pm \frac{1}{\sqrt[]{2}},\pm \frac{\sqrt[]{3}}{2}$          

$E_{M{n}^{+3}/M{n}^{+2}}^o=1.51V$

the strongest oxidizing agent have the highest reduction potential. So Mn³⁺ is the strongest oxidizing agent.

$\left(p\Delta q\right)\Rightarrow \left(p\Delta \sim q\right)\vee \left(\sim p\Delta q\right)$

Case I

When  $\Delta$  is same as $\vee .$  

Then  $\left(p\Delta \sim q\right)\vee \left(-p\Delta q\right)$ becomes

$\left(p\vee \sim q\right)\vee \left(\sim p\vee q\right)$ which is always true, so x becomes tautology.

Case II

When  $\Delta$  is same as $\vee$  

Then  $\left(p\wedge q\right)\Rightarrow \left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$  becomes $p\wedge q$ is T, then $\left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$ is false, so x cannot be tautology.

Case III

When  $\Delta$  is same as $\Rightarrow$  

Then  $\left(p\Rightarrow \sim q\right)\vee \left(-p\Rightarrow q\right)$ is same as $\left(\sim p\vee \sim q\right)\vee \left(p\vee q\right)$ which is true, so x becomes tautology.

Case IV

When  $\Delta$ is same as $\iff$  

Then   $\left(p\iff q\right)\Rightarrow \left(p\iff \sim q\right)\vee \left(\sim p\iff q\right)$

$p\iff q$ is true when p and q have same truth values $p\iff \sim qand\sim p\iff q$  both are false. Hence x cannot be tautology.

  $f\left(g\left(x\right)\right)=x\forall x\in R$

$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)$      

For y = g (x)

x = y³ + y – 5

$\frac{dx}{dy}=3y^2+1\Rightarrow g\text{'}\left(63\right)=\frac{1}{3\left(16\right)+1}=\frac{1}{49}$

$g\text{'}\left(63\right)=\left(\frac{dy}{dx}\right)$

$\left(\begin{array}{l}x=63\\ y=4\end{array}\right)$                

Cerium exists in two oxidation states (+3) and (+4)

$\begin{array}{l}C{e}^{+4}+e^{-}\to C{e}^{+3}{E}^0=1.61V\\ C{e}^{+3}+3e^{-}\to Ce{E}^0=-2.336V\end{array}$

It exist as Ce+4 and acts like a strong oxidizing agent by gaining electrons

$A\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)$

$A\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}-1\\ 0\\ 1\end{array}\right)$

$A\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)$

$\Rightarrow A\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

$AB=C\Rightarrow A=C{B}^{-1}$

$\left|A-2l\right|=\left(-4\right)\left(-1\right)+3\left(-1\right)+1\left(-1\right)$

4 – 3 – 1 = 0

$\mathcal{l}\left(-1,0,1\right)+m\left(-1,1,0\right)=\left(-4,3,1\right)\Rightarrow -\mathcal{l}-m=-4$

$m=3,\mathcal{l}=1$