Question

Domian of the function is $y=\sqrt{16x-x^5}+{\log }_{1/2}(x^2-1)$

• A. $\therefore D_1\cap D_2=[-\infty ,0]\cup [1,2]$
• B. $\therefore D_1\cap D_2=[-\infty ,-2]\cup [1,2]$
• C. $\therefore D_1\cap D_2=[-\infty ,2]\cup [2,4]$
• D. $\therefore D_1\cap D_2=[-\infty ,-1]\cup [1,2]$

Answer 1

Answer: (B)

$\therefore D_1\cap D_2=[-\infty ,-2]\cup [1,2]$

To find domain of $y=\sqrt{16x-x^5}+{\log }_{\frac{1}{2}}(x^2-1)$

let domain of $\sqrt{16x-x^5}$ and ${\log }_{\frac{1}{2}}$ be $D_1$ and $D_2$ respectively.

$D_1\ge 0$ for a square root function.

$\sqrt{16x-x^5}\ge 0$

$x(16-x^4)\ge 0x(4-x^2)(4+x^2)\ge 0x(2-x)(2+x)(4+x^2)\ge 0$

$-x(x-2)(2+x)(4+x^2)\ge 0$

$x(x-2)(2+x)(4+x^2)\le 0x\le 0,x\le 2,x\le -2,x^2\le -4$

Hence $x\in [-2,2]$

for ${\log }_{b}a,a>0$

hence for $lo{g}_{1/2}(x^2-1)$

$x^2-1>0$

$(x-1)(x-2)>0.$

Hence $x\in (-\infty ,-2)(1,\infty )$

$D_1\cap D_2=[-\infty ,-2][1,2]$