Question

Solve the following inequalities.
$C_{x-1}^{x+1}<21,xϵN$

Answer 1

Solve $C_{x-1}^{x+1}<21,xϵN$

Sol${}^n$ $C_{x-1}^{x+1}<21$

$\Rightarrow \frac{(x+1)!}{(x-1)!(x+1-(x-1\left)\right)!}<21$ by ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$\frac{(x+1)\left(x\right)(x-1)!}{(x-1)!(x+1-x+1)!}<21$

$\Rightarrow \frac{x(x+1)}{2!}<21$ and $2!=2\times 1=2$

$\Rightarrow x(x+1)<21\times 2$

$\Rightarrow x(x+1)<42\Rightarrow x(x+1)-42<0$

Let $y=x(x+1)-42=0$

$y=x^2+x-42=0$

then value of $x=\frac{-1\pm \sqrt{1-4\times (-42)\times 1}}{2}$

$=\frac{-1\pm \sqrt{1+168}}{2}$

$=\frac{-1\pm \sqrt{169}}{2}=\frac{-1\pm 13}{2}$

So $x=\frac{-1+13}{2}=\frac{+12}{2}=6$ and $x=\frac{-1-13}{2}=\frac{-14}{2}=-7$

So value of x lien between $-7$ and $6$ [$\because$ we need $x(x+1)<42]$

So value of $xϵ(-7,6)\to$ bracket will be open because

we need $x(x+1)<42$

and at $x=-7,6$ value of

$x(x+1)=42$