Let f: R R be a twice differentiable function such that f''(x) > 0 for all x R and f'(a-1) = 0, where a is a real number. Let g(x) = f ( ^ 2 x - 2 x + a ), \; 0 Consider the following two statements: (I) g is increasing in (0, 4 ) (II) g is decreasing in ( 4 , 2 ). Then,
A. Neither (I) nor (II) is True
Given f''(x) > 0 for all x R , which means f'(x) is a strictly increasing function. We are given f'(a-1) = 0. Since f'(x) is strictly increasing, f'(x) 0 for x > a-1. The function g(x) is defined as g(x) = f( ^2 x - 2 x + a) for x (0, /2). Let u(x) = ^2 x - 2 x + a = ( x - 1)^2 + a - 1. Differentiating g(x) with respect to x: g'(x) = f'(u(x)) u'(x) = f'(( x - 1)^2 + a - 1) (2 x ^2 x - 2 ^2 x) g'(x) = f'(( x - 1)^2 + a - 1) 2 ^2 x ( x - 1). Case 1: x (0, /4). In this interval, 0 Also, ( x - 1)^2 > 0, so u(x) = ( x - 1)^2 + a - 1 > a - 1. Since f'(x) is increasing and f'(a-1) = 0, for u(x) > a - 1, we have f'(u(x)) > 0. Thus, g'(x) = (+)(+)(-) Statement (I) is False. Case 2: x ( /4, /2). In this interval, x > 1, so ( x - 1) > 0. Also, ( x - 1)^2 > 0, so u(x) = ( x - 1)^2 + a - 1 > a - 1. Thus, f'(u(x)) > 0. Then g'(x) = (+)(+)(+) > 0. So g(x) is increasing in ( /4, /2). Statement (II) is False. Therefore, neither (I) nor (II) is true.
Related: Mathematics — Application of Derivatives · All PYQ Banks