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JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 15)

Let S be the set of all functions ƒ : [0,1] $$ \to $$ R, which are continuous on [0,1] and differentiable on (0,1). Then for every ƒ in S, there exists a c $$ \in $$ (0,1), depending on ƒ, such that
$$\left| {f(c) - f(1)} \right| < \left| {f'(c)} \right|$$
$$\left| {f(c) + f(1)} \right| < \left( {1 + c} \right)\left| {f'(c)} \right|$$
$$\left| {f(c) - f(1)} \right| < \left( {1 - c} \right)\left| {f'(c)} \right|$$
None

Explanation

If we consider the case where f(x) is a constant function, then its derivative f'(x) is equal to 0 for all x in the interval (0,1).

Therefore, if we substitute this into the expressions provided in Options A, B and C, we would have :

Option A : |f(c) - f(1)| < |f'(c)| would become |constant - constant| < |0|, which is 0 < 0. This is not true.

Option B : |f(c) + f(1)| < (1 + c)|f'(c)| would become |constant + constant| < (1 + c)$$ \times $$0, which is a positive number < 0. This is not true.

Option C : |f(c) - f(1)| < (1 - c)|f'(c)| would become |constant - constant| < (1 - c)$$ \times $$0, which is 0 < 0. This is not true.

Hence, for the case where f(x) is a constant function, none of the options A, B and C are correct.

So, the correct answer would be Option D : None.

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