Class 12th

  $f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=x^5+64$

$\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}\left(\frac{f\left(1\right)}{0}\right)f\left(1\right)=0$

Let f(x) = x⁵ + ax⁴ + bx³ + cx² + dx + e

f(1) = 0 Þ 1 + a + b + c + d + e = 0

  $\Rightarrow x^5+\left(a+5\right)x^4+\left(b+4a+20\right)x^3+\left(c+3b+12a\right)x^2+\left(d+2c+6b\right)x+e+d+2c=x^5+64$

Þ e + d + 2c = 64

b + 4a + 20 = 0                 c – 1 – a – b = 64

c + 3b + 12a = 0

d + 2c + 6b = 0                 a + b – c = -65

1 + a + b + c + d + e = 0   13a + 4b = -65

b + 4c = -20 * 4

-3a = 15 a = -5

$\begin{array}{l}f\left(x\right)=x^5-5x^4+60x^2-120x+64\\ f\text{'}\left(x\right)=5x^4-20x^3+12x-120\\ f\text{'}\left(1\right)=5-20+120-120=-15\end{array}$

${\overrightarrow{b}}_1*{\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -a\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|$ = $-a\widehat{i}-\widehat{j}+\left(a-1\right)\widehat{k}$

${\overrightarrow{a}}_1-{\overrightarrow{a}}_2=-\widehat{i}+\widehat{j}+\widehat{k}$

Shortest distance = $\left|\frac{\left({\overrightarrow{a}}_1-{\overrightarrow{a}}_2\right).\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}\right|$

$=\frac{2\left(a-1\right)}{\sqrt[]{a^2+1+{\left(a+1\right)}^2}}=\sqrt[]{\frac{2}{3}}=\frac{4{\left(a-1\right)}^2}{a^2+1+{\left(a-1\right)}^2}=\frac{2}{3}$

$12{\left(a-1\right)}^2=2\left(a^2+1\right)+2{\left(a-1\right)}^2$

$10{\left(a-1\right)}^2=2\left(a^2+1\right)$

$5a^2-10a+5=a^2+1\Rightarrow 4a^2-10a+4=0$

$2a^2-5a+2=0$

$\left(2a-1\right)\left(a-2\right)=0$

$\therefore a=\frac{1}{2},2$

$R_2={\displaystyle \underset{0}{\overset{1}{\int }}\left(x-x^3\right)dx={\left(\frac{x^2}{2}-\frac{x^4}{4}\right)}_0^1}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

$R_1={\displaystyle \underset{0}{\overset{\frac{1}{4}}{\int }}\left(\sqrt[]{x}-2x\right)dx={\left(\frac{2x^{3/2}}{3}-x^2\right)}_0^{\frac{1}{4}}}=\frac{1}{12}-\frac{1}{16}$

$=\frac{4-3}{48}=\frac{1}{48}$

$R_1+R_2={{\displaystyle \underset{0}{\overset{1}{\int }}\left(\sqrt[]{x}-x^3\right)dx=\left(\frac{2x^{3/2}}{3}-\frac{x^4}{4}\right)}}_0^1=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$

$\frac{R_1+R_2}{R_1}=1+\frac{R_2}{R_1}=\frac{\frac{5}{12}}{\frac{1}{48}}=20$ $\therefore \frac{R_2}{R_1}=19$

 

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.