Inverse Trigonometric Functions

$3sinx+4cosx=k+1,cos\alpha =\frac{3}{5},sin\alpha =\frac{4}{5}$

$5sin\left(x+\alpha \right)=k+1,$

$\therefore -5\le k+1\le 5\Rightarrow -6\le k\le 4$

$\therefore$ total number of integral values of k is 11.

 $lo{g}_4\left(x-1\right)=lo{g}_2\left(x-3\right)$

$\Rightarrow \frac{1}{2}lo{g}_2\left(x-1\right)=lo{g}_2\left(x-3\right)$

$\Rightarrow lo{g}_2\left(x-1\right)=lo{g}_2{\left(x-3\right)}^2$

$\Rightarrow x-1=x^2-6x+9$

$\Rightarrow x=2,5$

also x – 1 > 0 and x – 3 > 0

x > 1 & x > 3

Hence, x = 5 possible. Only one solution.

$\frac{log\left(\left(2x+5\right)\left(x+1\right)\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-4=0\Rightarrow \frac{log\left(2x+5\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-3=0$

$\Rightarrow x\in \left(-1,0\right)\cup \left(0,\infty \right)$

$\frac{log\left(2x+5\right)}{log\left(x+1\right)}=1\Rightarrow x=-4$ (not possible) and $\frac{log\left(2x+5\right)}{log\left(x+1\right)}=2\Rightarrow x=2$

Hence only one solution is possible.

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

$\frac{dy}{dx}=\frac{2^{x}y+2^y.2^x}{2^x+2^{x+y}lo{g}_{e}2}$

${\displaystyle \int \frac{1+2^{y}lo{g}_{e}2}{y+2^y}dy={\displaystyle \int dx}}$        

=> ln|y + 2^y| = x + c

y (0) = 0

ln |y + 2^y| = x

y = 1

x = ln 3

$x\in \left(1,2\right)$

$x^2+9y^2-4x+3=0,x,y\in R.........\left(i\right)$

$x^2-4x+9y^2+3=0$        

Since $x\in R,D\ge 0i.e.16-4\left(9y^2+3\right)\ge 0$

or $9y^2-1\le 0$

$\Rightarrow y\in \left[-\frac{1}{3},\frac{1}{3}\right]$

$\left(i\right)\Rightarrow 9y^2=-x^2+4x-3$

Since L.H.S. $\ge 0soR.H.S.\ge 0i.e.-x^2+4x-3\ge 0$

or $x^2-4x+3\le 0$

$\left(x-3\right)\left(x-1\right)\le 0$

$\Rightarrow x\in \left[1,3\right]$

f(x) = tan^-1 (sin x + cos x)

$asx\in \left[0,\frac{\pi }{2}\right]$              so  1 $\le sinx+cosx\le \sqrt[]{2}$  

so $f\left(x\right)\in \left[ta{n}^{-1}1,ta{n}^{-1}\sqrt[]{2}\right]$  

->M = tan^-1   $\sqrt[]{2}\&m=ta{n}^{-1}1=\frac{\pi }{4}$

Now, tan (M – m) = tan $\left(ta{n}^{-1}\sqrt[]{2}-\frac{\pi }{4}\right)$

$=\frac{\sqrt[]{2}-1}{1+\sqrt[]{2}.1}=\frac{\sqrt[]{2}-1}{\sqrt[]{2}+1}*\frac{\sqrt[]{2}-1}{\sqrt[]{2}-1}=3-2\sqrt[]{2}$  

$\Rightarrow \frac{1+x}{x}\ge 1and\frac{1+x}{x}\le -1$

$\Rightarrow \frac{1}{x}\ge 0\frac{1+2x}{x}\le 0$

Domain $\left[-\frac{1}{2},0\right)\cup \left(0,\infty \right)$

$=\left[-\frac{1}{2},\infty \right)-\left\{0\right\}$  

use $si{n}^{-1}x=co{s}^{-1}\sqrt[]{1-x^2}$  

$ta{n}^{-1}\sqrt[]{1-x^2}=co{t}^{-1}\frac{1}{\sqrt[]{1-x^2}}$               

$si{n}^{-1}\frac{x}{\sqrt[]{1-x^2}}=co{s}^{-1}\frac{\sqrt[]{1-2x^2}}{\sqrt[]{1-x^2}}$               

Sum of roots ->b = -1 + 2   $\frac{\left(k^2-1\right)}{k^2-2}$

Product of roots -> -5 = -2  $\left(\frac{k^2-1}{k^2-2}\right)$  

b = 4, k² = $\frac{1}{3}$  

cos^-1 x = 2 sin^-1 x = cos^-1 2x

$co{s}^{-1}x-2\left(\frac{\pi }{2}-co{s}^{-1}x\right)=co{s}^{-1}2x$          

$cos\left(3co{s}^{-1}x\right)=cos\left(\pi +co{s}^{-1}2x\right)$    

$4x^3-3x=-2x$        

$4x^3=x\Rightarrow x=0\pm \frac{1}{2}$      

All satisfy the original equation

Sum =  $-\frac{1}{2}+0+\frac{1}{2}=0$