Question

If the range of $\frac{x^2+x+d}{x^2+2x+d}$ is $[\frac{5}{6},\frac{3}{2}]$ for all real x, then the value of 'd' is

• A. 4
• B. -4
• C. 1
• D. -1

Answer 1

Answer: (A)

4

Let $y=\frac{x^2+x+d}{x^2+2x+d}$

$\Rightarrow x^2(1-y)+x(1-2y)+d(1-y)=0$ has real roots

$\Rightarrow D=(1-2y{)}^2-4d(1-y{)}^2>0$

$\Rightarrow y^2(4d-4)+(4-8d)y+(4d-1)<\mathrm{0....}\left(i\right)$

Now given, $\frac{5}{6}\le y\le \frac{3}{2}$

$\Rightarrow (y-\frac{5}{6})(y-\frac{3}{2})\le 0$ or $y^2-\frac{7y}{3}+\frac{15}{12}\le \mathrm{0.....}\left(ii\right)$

Comparing $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow \frac{4d-1}{4d-4}=\frac{15}{12}$

$\Rightarrow d=4$