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Step-by-Step Explanation
Step 1: Identify the Possible Counterexample
Observe that each option in the question places an inequality involving
$f(c)$, $f(1)$, and $f'(c)$. A constant function is often a good test
when checking for contradictions because its derivative is zero everywhere.
So let us consider the constant function
$f(x) = K$ for some constant $K \in \mathbb{R}$.
Step 2: Compute the Derivative for the Constant Function
For $f(x) = K$, its derivative is
$f'(x) = 0$ for all $x \in (0,1)$.
Step 3: Substitute into Each Option
Option 1:
$|\,f(c) - f(1)\,| < |\,f'(c)\,|$
Since $f(c) = K$ and $f(1) = K$, the left-hand side becomes
$|\,K - K\,| = 0$ and the right-hand side is $|\,0\,| = 0$.
This gives $0 < 0$, which is false.
Option 2:
$|\,f(c) + f(1)\,| < (1 + c)\,\bigl|f'(c)\bigr|$
Here, $f(c) + f(1) = K + K = 2K$, while
$\bigl|f'(c)\bigr| = 0$.
So the inequality becomes
$|\,2K\,| < (1+c)\times 0$, i.e. $|\,2K\,| < 0$,
which can never be true for any real $K$.
Option 3:
$|\,f(c) - f(1)\,| < (1 - c)\,\bigl|f'(c)\bigr|$
As before, $f(c) - f(1) = K - K = 0$ and
$\bigl|f'(c)\bigr| = 0$. So this becomes
$0 < (1 - c)\times 0 = 0$, which is false.
Step 4: Conclude the Correct Answer
Since a simple choice of $f(x) = K$ (constant function) violates
all the given inequalities, none of the provided options
can hold for every continuous and differentiable function on $[0,1]$.
Therefore, the correct answer is
None.
