Let S be the set of all functions ƒ : [0,1] → R, which are continuous on [0,1] and differentiable on (0,1). Then for every ƒ in S, there exists a c ∈ (0,1), depending on ƒ, such that

Solution

Maths sir · 3 steps

The claim must hold for every f, so try the simplest f there is: a constant function.

1. One counterexample. Each option claims a strict inequality for every f in S. A single function in S that breaks it for every c ∈ (0,1) rules that option out.
2. Take a constant. Let f(x) = k on [0,1]. It is continuous and differentiable, so it is in S. For every c, f(c) = f(1) = k and f'(c) = 0.

   Option	Left side	Right side	Strict inequality?
   (a)	|k - k| = 0	|0| = 0	0 < 0, false
   (b)	|k + k| = 2|k|	(1 + c) × 0 = 0	2|k| < 0, false
   (c)	|k - k| = 0	(1 - c) × 0 = 0	0 < 0, false

3. So. No choice of c can rescue (a), (b) or (c) for a constant function, because the right side is always 0 and a strict inequality needs it to be larger. The answer is (d) None.

The idea to keep. When a statement must hold for every function in a class, first test the simplest members: constants, f(x) = 0, f(x) = x. One failure is enough to reject it.

Answer from the NTA final answer key. Solution by PrepWiser.