Q: Find all the possible values of the parameter ‘a ‘ so that the function,
f (x) = x3 – 3 (7 – a) x2 – 3 (9- a2) x + 2 , has a negative point of local minimum.
Sol. f(x) = x3 –3(7-a) x2 –3(9-a2)x + 2
f ‘(x) = 3x2 –6(7-a) x – 3(9-a2)
For distinct real roots D > 0
36(7-a)2 + 4 ×3×3 (9 – a2) > 0
⇒ 49 + a2 –14a + 9 –a2 > 0
14a < 58 ⇒ a < 29/7 For local minima f ”(x) = 6x – 6(7 – a) > 0
⇒ x – 7 + a > 0
7 – a < x as x must be -ve
⇒ 7-a < 0 ⇒ a > 7
Thus, by contradiction i.e. for real roots a < and for negative point of local minimum a > 7. No possible value of a.