Question

If $f:R\to [a,6]$ defined by $f\left(x\right)=\frac{x^2-2x+d}{x^2+3x+d}$ is an onto function, then

• A. $a$ is equal to $\frac{2}{7}$
• B. $d$ is equal to 4
• C. $a$ is equal to 1
• D. $(d-7a)$ is equal to -3

Answer 1

Answer: (A), (B)

The correct options are
A $a$ is equal to $\frac{2}{7}$
B $d$ is equal to 4

As maximum value of $\frac{x^2-2x+d}{x^2+3x+d}$ is 6.
So $\frac{x^2-2x+d}{x^2+3x+d}=6$ has equal routs.
$\Rightarrow 5x^2+20x+5d=0\Rightarrow 16-4d=0\Rightarrow d=4$
So $f\left(x\right)=\frac{x^2-2x+4}{x^2+3x+4}$
min value of $f\left(x\right)$ is $\frac{2}{7}.$
So $a=\frac{2}{7}$