Question

If $f\left(x\right)=\frac{x^2+4x+3}{x^2+x+1}\in [\frac{a-b\sqrt{c}}{d},\frac{a+b\sqrt{c}}{d}]$ (where $a,b,d$ are integers and $c$ prime, also $a$ and $d$ are coprime), then which of the following are prime ?

• A. $a+c$
• B. $a+d$
• C. $b+d$
• D. $c+d$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a+c$
B $a+d$
C $b+d$

Let $y=f\left(x\right)=\frac{x^2+4x+3}{x^2+x+1}\Rightarrow x^2(y-1)+x(y-4)+y-3=0$
.
For $x$ to be real, discriminant of the above equation should be greater than or equal to zero.
${(y-4)}^2-4(y-1)(y-3)\ge 0\Rightarrow -3y^2+8y+4\ge 0\Rightarrow 3y^2-8y-4\le 0\Rightarrow (y-\frac{4-2\sqrt{7}}{3})(y-\frac{4+2\sqrt{7}}{3})\le 0\Rightarrow y\in [\frac{4-2\sqrt{7}}{3},\frac{4+2\sqrt{7}}{3}]$
.
Comparing with the range given in the question, we get $a=4,b=2,c=7,d=3.$
Hence, $a+c=11,a+d=7,b+d=5$, which are prime.