Question

Domain of the explicit form of the function y represented implicitly by the equation $(1+x)\cos y-x^2=0$ is -

• A. $(-1,1]$
• B. $[-1,\frac{1-\sqrt{5}}{2}]$
• C. $[\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}]$
• D. $[0,\frac{1+\sqrt{5}}{2}]$

Answer 1

Answer: (C)

$[\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}]$

Let $y={\cos }^{-1}\frac{x^2}{1+x}$
Now $-1\le \frac{x^2}{1+x}\le 1$
$\frac{x^2}{1+x}\le 1$
$\Rightarrow x^2-x-1\le 0$ or $x\in [\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}]$
The other inequality gives $\frac{x^2}{1+x}\ge -1$
Or, $x^2+x+1\ge 0$ or $x\in (-\infty ,\frac{-1-\sqrt{5}}{2}]\cup [\frac{-1+\sqrt{5}}{2},\infty )$
$\Rightarrow x\in [\frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}]$