Let $f : R \to (- \infty, - 1]$ be a function defined by $f (x) = (a b + 2 a - b - 2) x^{5} - (a^{3} - 2 a + 1) x^{3} + (a^{2} - 2 a - 3) x^{2} + (a^{2} + 2 b) x - 5,$ $a, b \in R$ . If $f (x)$ is surjective then the possible value of a-b is _______

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$\frac{9}{2}$

$\frac{11}{2}$

$- \frac{5}{2}$

$- \frac{7}{2}$

answer is C, D.

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Detailed Solution

$f : R \to (- \infty, - 1] f (x) = (a b + 2 a - b - 2) x^{5} - (a^{3} - 2 a + 1) x^{3} + (a^{2} - 2 a - 3) x^{2} + (a^{2} + 2 b) x - 5 = (a - 1) (b + 2 x) x^{5} - (a^{3} - 2 a + 1) x^{3} + (a^{2} - 2 a - 3) x^{2} + (a^{2} + 2 b) x - 5$

If f(x) is a polynomial of odd degree then its range will be R.
Thus, $a = 1$
$f (x) = (1 - 2 - 3) x^{2} + (1 + 2 b) x - 5 = - 4 x^{2} + (1 + 2 b) x - 5$

Max. Value of $f (x) = - 1$

$- \frac{D}{4 a} = - 1; - \frac{{(1 + 2 b)}^{2} - 4 (- 4) (- 5)}{4 (- 4)} = - 1 {(1 + 2 b)}^{2} - 80 = - 16 {(1 + 2 b)}^{2} = 64 1 + 2 b = \pm 8 2 b = 7; 2 b = - 9 b = \frac{7}{2} o r - \frac{9}{2}$