Question

Let $f\left(x\right)=\sqrt{\frac{x^2+ax+4}{x^2+bx+16}}$ is defined for all real $x$, then find the number of possible ordered pairs $(a,b)$ (where $a,b\in I$)

Answer 1

$f\left(x\right)=\sqrt{\frac{x^2+ax+4}{x^2+bx+16}}$ to be defined for all real a,

$x^2+ax+4⩾0$ & $x^2+bx+16>\mathrm{0...}\left(1\right)$

$x^2+ax+4<0$ & $x^2+bx+16<0$ (both simultaneously)...(2)

$\left(2\right)\to$ [Not Possible as Positive coefficient]

for (1) to be true Discriminant $\Rightarrow B^2-4AC=D$

for $x^2+ax+4\ge 0\to D⩽0$

$x^2+bx+16>0\to D<0$

$a^2-4\times 4\times 1⩽0$ $b^2-4\times 16\times 1<0$

$a^2⩽16$ $b^2-64<0\to b^2<64$

$aϵ[-4,4]$ $bϵ(-8,8)$

Ordered Pair, $(-4,-7)(-4,-6)(-4,-5)(-4,-4)(-4,-3)....(-4,7)$

$(-3,-7)(-3-6).......(-3,7)$

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$(3,-7)(3,-6)....(3,7)$

$(4,-7)(4,-6)(4,-5)...(4,0).....(4,7)$