Let us consider f(x), a constant function, as constant functions are continuous and differentiable everywhere.
Then, f(1)=f(c)&f'(c)=0
So, 1−cf(1)−f(c)=f′(c)=0.
JEE Main 2020 — Mathematics Calculus
Let S, be the set of all functions f:[0,1]→R, which are continuous on [0,1], and differentiable on (0,1). Then for every f in S, there exists c∈(0,1), depending on f, such that.
Held on 8 Jan 2020 · Verified 6 Jul 2026.
∣f(c)−f(1)∣<(1−c)∣f′(c)∣
1−cf(1)−f(c)=f′(c)
∣f(c)+f(1)∣<(1+c)∣f′(c)∣
∣f(c)−f(1)∣<∣f′(c)∣
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