The number of integral values of a for which the inequality 3 - |x...

Given inequality is
3 - \left|x - a\right| > x^{2} . . . \left(1\right)$</p><p>$⇒ \left|x - a\right| < 3 - x^{2}$</p><p>or~$- 3 + x^{2} < x - a < \left(3 - x^{2}\right)$</p><p>or$x^{2} - x + a - 3 < 0 \left(2\right)$</p><p>and~$x^{2} + x - \left(a + 3\right) < 0 . . . \left(3\right)$</p><p>If~$\alpha,\beta \left(\alpha< \beta\right)$be the real and unequal roots of the corresponding equation of Eq. \left(1\right)$~for all x \in \left(\alpha,\beta\right).Since Eq. \left(1\right)$~will have at least one negative solution. The corresponding equation of \left(2\right)$~and \left(3\right)$~are</p><p>$x^{2} - x + a - 3 = 0 . . . \left(4\right)

and~x^{2} + x - \left(a + 3\right) = 0 . . . \left(5\right)$<br />respectively.</p><p>Both Eqs. \left(4\right)$~and \left(5\right)~should have real and unequal roots, then

$1 - 4$\cdot$1$\cdot \left(a - 3\right) > 0 & 1 + 4 \cdot \left(a + 3\right) > 0$</p><p>i.e.,~$a - \dfrac{13}{4} . . . \left(6\right)$</p><p>Roots of Eq. \left(6\right)$~are~$\dfrac{1 \pm \sqrt{\left(13 - 4a\right)}}{2}$

One of these will be negative provided the condition below

$13 - 4a > 1$

i.e.,~a < 3 . . . \left(7\right)$</p><p>So, it has one negative root.</p><p>Combining Eqs. \left(6\right)$~and \left(7\right), we get

$-$\dfrac{13}{4}$< a < 3$

$∴$~Integral values of a are~ $- 3, - 2, - 1, 0, 1, 2 .$

Total number~$= 6 .$