1

If $a,b$ are the roots of the equation $$(\arcsin x+\arctan x)p^2+25p\pi+2(\arccos x+\operatorname{arccot} x)=0$$

Then minimum of $(a+1)(b+1)$

Try: $$a+b=\frac{25\pi}{\arcsin x+\arctan x}$$ and $$ab=\frac{2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}$$

So $$ab+a+b+1=\frac{25\pi+2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}$$

Could some help me to solve it, thanks

Community
  • 1
DXT
  • 11,241

2 Answers2

2

Using Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$, we need to maximize $$f(x)=\arcsin x+\arctan x$$ for $f(x)>0$ which is increasing in $[-1,1]$

Now use $x\le1$

lab bhattacharjee
  • 274,582
  • Maximum value of $\displaystyle \arcsin x+\arctan x =\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$ at $x=1$ because $f(x)$ is strictly increasing function. – DXT Jan 30 '18 at 04:01
1

$x\neq0$, otherwise our equation has unique root.

Also, $$\left(25\pi\right)^2-8(\arcsin{x}+\operatorname{arctan}x)(\arccos{x}+\operatorname{arccot}x)\geq$$ $$\geq\left(25\pi\right)^2-8\left(\frac{\arcsin{x}+\operatorname{arctan}x+\arccos{x}+\operatorname{arccot}x}{2}\right)^2=625\pi^2-2\pi^2>0,$$ which says that our equation has two real roots for all $x\neq0$ and $-1\leq x\leq 1.$

Now, $$(a+1)(b+1)=\frac{-25\pi+2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}+2-1=$$ $$=\frac{-25\pi+2(\arcsin{x}+\arccos x+\operatorname{arctan} x+\operatorname{arccot} x)}{\arcsin x+\arctan x}-1=$$ $$=\frac{-25\pi+2\left(\frac{\pi}{2}+\frac{\pi}{2}\right)}{\arcsin x+\arctan x}-1\rightarrow-\infty$$ for $x\rightarrow0^+$, which says that the minimum does not exist.

Michael Rozenberg
  • 194,933