Relations and Functions

$f\left(x\right)=2x-1g:R-\left\{1\right\}\to R$        

$g\left(x\right)=\frac{x-1/2}{x-1}$           

$f\left\{g\left(x\right)\right\}=2\left(\frac{x-1/2}{x-1}\right)-1=\frac{2x-1-x+1}{x-1}=\frac{x}{x-1},x\ne 1$           

$y=\frac{x}{x-1}\Rightarrow x=\frac{y}{y-1}\therefore y\ne 1$           

One – one but not onto

$f\left(x\right)=\frac{4x^3-3x^2}{6}-2sinx+\left(2x-1\right)cosx$      

$f\text{'}\left(x\right)=2x^2-x-2cosx+2cosx-\left(2x-1\right)sinx$     

= $\left(2x-1\right)\left(x-sinx\right)\ge 0forx\ge 0$  

  $f\text{'}\left(x\right)\ge 0\forall x\ge 1/2$          

$\therefore f\left(x\right)$ is increasing in $\left[\frac{1}{2},\infty \right)$

$\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$  

 Let sin x = t, t $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$  

 Minimum value of a for which solution exist = 9

let f : R -> R

$f\text{'}\left(x\right)=\{\begin{array}{l}-55,x<-5\\ 6x^2-6x-120,-5<x<4\\ 6x^2-6x-36,x>4\end{array}$            

f’ (x) increasing in x $\in \left(-5,-4\right)\cup \left(4,\infty \right)$  

$\frac{a^{a^x}+a^{1-a^x}}{2}\ge \sqrt[]{a^{a^x}.a^{1-a^x}}$ $\left(AM\ge GM\right)$

$\Rightarrow a^{a^x}+a^{1-a^x}\ge 2\sqrt[]{a}$

f : N ->N

f (n + 1) = f (n) + f (1)

Let f (1) = a, a  $\in$ N

f (2) = f (1) + f (1) = 2a

f (3) = f (2) + f (1) = 3a

and so on

->f (m) = ma, m, a  $\in$ N

->f is one – one, Þ option (2) is true.

Suppose f (g (x) is one-one

then f (g (x₁) $\ne$ f (g (x₂) for x₁ $\ne$  x₂

->g (x₁) $\ne$ g (x₂) (as f is one-one)

-

  $f\left(x\right)=\{\begin{array}{l}-2x^2+3,-2<x<-1\\ x^2,-1\le x<0\\ \frac{\sqrt[]{3}}{2}-1,0\le x<1\\ x^2-3+\frac{1}{\sqrt[]{2}},1\le x<2\end{array}$

$f\left(-1^{-}\right)=f\left(-1^{+}\right)=f\left(-1\right)=1$                

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

Points of discontinuity

x = 0, 1

Given expression

= 2π - 5 + 6 - 2π - (12 - 4π) = 4π - 11

f (m + n) = f (m) + f (n)

put m = n = 1, f (2) = f (1) + f (1)

again put m = 2, n = 1, f (3) = f (2) + f (1)

and put m = 3, n = 3, f (3 + 3) = f (3) + f (3), 2f (3) = f (6) = 18 Þ f (3) = 9

f (3) = 3f (1)

f (1) = 3, f (2) = 6

f (2).f (3) = 54

  $f\left(x\right)=si{n}^{-1}\left(\frac{3x^2+x-1}{{\left(x-1\right)}^2}\right)+co{s}^{-1}\left(\frac{x-1}{x+1}\right)$ ${}^{}\frac{{}^{}}{{}^{}}{}^{}\frac{}{}$

Domain of sin1 $\left(\frac{3x^2+x-1}{{\left(x-1\right)}^2}\right)$ $\frac{{}^{}}{{}^{}}$ is

$-1\le \frac{3s^2+x-1}{{\left(x-1\right)}^2}\le 1$

${}^{}{}^{}$  $-x^2-1+2x\le 3x^2+x-1$ and $3x^2+x-1\le x^2+1-2x$ ${}^{}{}^{}$

Domain of the function is $\left[\frac{1}{4},\frac{1}{2}\right]\cup \left\{0\right\}$ .