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 ( ²x - 2 x + a ), ; 0 Consider the following two statements: (I) g is increasing in (0, 4 ) (II) g is de

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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 ( ²x - 2 x + a ), ; 0 Consider the following two statements: (I) g is increasing in (0, 4 ) (II) g is decreasing in ( 4 , 2 ) . Then,

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Correct answer: 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)

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 ( ²x - 2 x + a ), ; 0 Consider the following two statements: (I) g is increasing in (0, 4 ) (II) g is decreasing in ( 4 , 2 ) . Then,

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