Let the set of all values of p, for which f(x) = (p2 - 6p + 8 ) ( 2 2x - 2 2x ) + 2(2 - p)x + 7 does not have any critical point, be the interval (a, b). Then 16ab is equal to .
Answer
252
Solution
2 2x - 2 2x = - 4x, so f(x) = -(p2 - 6p + 8) 4x + 2(2 - p)x + 7
f'(x) = 4(p - 2)(p - 4) 4x - 2(p - 2) = 2(p - 2) [2(p - 4) 4x - 1 ]
f is differentiable everywhere, so it has no critical point only if f'(x) 0 for every x.
p = 2 makes f'(x) = 0 for all x, so p 2 and we need 2(p - 4) 4x 1 for all x.
Since 2(p - 4) 4x takes every value in [-2|p - 4|, 2|p - 4| ], we need 2|p - 4| < 1
72 < p < 92, so 16ab = 16 72 92 = 252
Answer: 252