Maths

$?f\left(x\right)=ln\left(x+\sqrt[]{1+x^2}\right)\therefore f\left(-x\right)=-f\left(x\right)$ .

Hence f(x) is an odd function.

$Nowg\left(t\right)={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(\frac{\pi }{4}t+f\left(x\right)\right)dx}$

Put t = 0, g(0) = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx=2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx}...............\left(i\right)$

where cos(f(x)) is an even function.

Now again put t = 1,

$\therefore g\left(1\right)=\frac{1}{\sqrt[]{2}}*g\left(0\right),from\left(i\right)$

$\therefore g\left(0\right)=\sqrt[]{2}g\left(1\right)$

$f\left(x\right)=\frac{5x+3}{6x-\alpha }=y.....................(i)$

$\Rightarrow x=\frac{\alpha y+3}{6y-5}\Rightarrow f^{-1}\left(x\right)=\frac{\alpha x+3}{6x-5}..............(ii)$

According to question, $f\left(x\right)=f^{-1}\left(x\right)\therefore$  from (i) and (ii) we get $\propto =5$

The projection of $\overrightarrow{BA}on\overrightarrow{BC}=cos\angle ABC=7\left|\frac{7^2+3^2-5^2}{2*7*3}\right|=\frac{11}{2}$

$\frac{x+1}{a}=\frac{y}{1}=\frac{z-1}{a}...........\left(i\right)$

$\frac{x+2}{a}=\frac{y}{1}=\frac{z}{\frac{3}{b}}..........\left(ii\right)$

Let A (-1, 0, 1) and B (-2, 0, 0) $$ $\therefore$ direction ratios of AB = -1, 0, -1

$$ $\therefore$  lines are coplanar

$\therefore \left|\begin{array}{ccc}\hfill a\hfill & \hfill 1\hfill & \hfill a\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill \frac{3}{b}\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right|=0\Rightarrow b=1anda\in R-\left\{0\right\}$

A = $-\frac{y}{x}+2sinx+2$  

$\therefore \frac{dy}{dx}+\frac{y}{x}=2sinx+2$ . (i)

$\Rightarrow l.F=e^{{\displaystyle \int !lnxdx}}$ = x from (i) d (xy) = 2 ${\displaystyle \int xsinxdx+2{\displaystyle \int xdx}}$

xy = 2 [-xcos x + sin x] + x² + c       . (ii)

according to question, we get c = 0 $\therefore y\left(\frac{\pi }{2}\right)=\frac{4}{\pi }+\frac{\pi }{2}$

10 =   $\frac{7+10+11+15+a+b}{6}\Rightarrow a+b=17$  . (i)

$\Rightarrow \frac{20}{3}=\frac{7^2+10^2+11^2+15^2+a^2+b^2}{6}-10^2$

a² + b² = 145       . (ii)

solving (i) and (ii) we get a = 8, b = 9 or a = 9, b = 8

|a – b|= 1

Centre of the smallest circle is A

Centre of the largest circle is B

$r_1=\left|CP+CB\right|=3\sqrt[]{2}+3$ and

$r_2=\left|CP-CB\right|=3\sqrt[]{2}-3$    

$\frac{r_1}{r_2}=\frac{3\sqrt[]{2}+3}{3\sqrt[]{2}-3}=3+2\sqrt[]{2}$

$f\left(x\right)=x^3-3x^2-\frac{3f\text{'}\text{'}\left(2\right)}{2}x+f\text{'}\text{'}\left(1\right)$ . (i)

f' (x) = 3x² – 6x -   $\frac{3f\text{'}\text{'}\left(2\right)}{2}$ . (ii)

$f\text{'}\text{'}\left(x\right)=6x-6$ . (iii)

  $\therefore f\text{'}\text{'}\left(2\right)=12-6=6$           

 Hence f (x) is local maxima at x = -1

and f (x) is local minima at x = 3

$\therefore$ from (i) local minimum value at x = 3 is f (3) = -27

Let mid point of PQ is R (h, k)

$\therefore h=\frac{\alpha +\frac{4{\alpha }^2+\alpha +1}{2}}{2}and$

$k=\frac{4{\alpha }^2+1+\frac{4{\alpha }^2+\alpha +1}{2}}{2}$   

Eliminate a from above these two, we get

2 (3x – y)² + (x – 3y) + 2 = 0

$l={\displaystyle \sum _{r=0}^{n-1}f\left(\frac{5r}{n}\right)\frac{1}{n}}\Rightarrow l={\displaystyle \underset{0}{\overset{1}{\int }}f\left(5x\right)dx=}{\displaystyle \underset{0}{\overset{1}{\int }}\left(5x+1\right)dx=\frac{7}{2}}$