Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{ax}}^2+2\mathrm{x}+1}{2{\mathrm{x}}^2-2\mathrm{x}+1}$. If f : R → [– 1, 2] is onto, then the values of a are

We have

$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{ax}}^2+2\mathrm{x}+1}{2{\mathrm{x}}^2-2\mathrm{x}+1}$

which is defined $\mathrm{\forall } \mathrm{x}\in \mathrm{R}$, since

$2{\mathrm{x}}^2-2\mathrm{x}+1=2{\left(\mathrm{x}-\frac{1}{2}\right)}^2+\frac{1}{2}\ne 0$ for any real x.

Now, if f: R → [–1, 2] is onto, then

$-1\le \frac{{\mathrm{ax}}^2+2\mathrm{x}+1}{2{\mathrm{x}}^2-2\mathrm{x}+1}\le 2, \mathrm{\forall }\mathrm{x}\in \mathrm{R}$

Solving the left-hand inequality, we have

$(\mathrm{a}+2){\mathrm{x}}^2+2\ge 0$

which is true $\mathrm{\forall }\mathrm{x}\in \mathrm{R}$ if $\mathrm{a}\ge -2$          (1)

Solving the right-hand inequality, we have

$(\mathrm{a}-2){\mathrm{x}}^2+6\mathrm{x}-1\le 0$

which is true for all $\mathrm{x}\in \mathrm{R}$ if coefficient of x² < 0 and D ≤ 0

i.e., a < 2 and 36 + 4 (a – 2) ≤ 0

i.e., a < 2 and a ≤ –7

i.e., a ≤ – 7           (2)

Hence, from (1) and (2) the permissible values of a are given by

$(-\mathrm{\infty }, -7]\cup [-2, \mathrm{\infty })$