Application of DerivativeseasyFree

Possible Values of f(2) When f(1) = 1, f(3) = 4 and f'(x) at least 1 | JEE

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Question
Let ff be a differentiable function for all x∈Rx \in \mathbb{R}, where f(1)=1f(1) = 1 and f(3)=4f(3) = 4 and f′(x)≥1f'(x) \ge 1 for all x∈Rx \in \mathbb{R}. Then f(2)f(2) can be equal to
A11
B22correct
C33correct
D44
Solution
Step 1: Apply Lagrange's mean value theorem on the left interval [1,2][1,2]. There is c1∈(1,2)c_1 \in (1,2) with
f′(c1)=f(2)−f(1)2−1=f(2)−1.f'(c_1) = \frac{f(2)-f(1)}{2-1} = f(2) - 1 .
 Step 2: Use f′(c1)≥1f'(c_1) \ge 1.
f(2)−1 ≥ 1  ⟹  f(2) ≥ 2.f(2) - 1 \ \ge\ 1 \implies f(2) \ \ge\ 2 .
 Step 3: Apply the theorem again on the right interval [2,3][2,3]. There is c2∈(2,3)c_2 \in (2,3) with
f′(c2)=f(3)−f(2)3−2=4−f(2).f'(c_2) = \frac{f(3)-f(2)}{3-2} = 4 - f(2) .
 Step 4: Use f′(c2)≥1f'(c_2) \ge 1.
4−f(2) ≥ 1  ⟹  f(2) ≤ 3.4 - f(2) \ \ge\ 1 \implies f(2) \ \le\ 3 .
 Step 5: Combine the two bounds.
2 ≤ f(2) ≤ 3.2 \ \le\ f(2) \ \le\ 3 .
 Step 6: Compare with the options. Only 22 and 33 lie in that range; 11 is too small and 44 too large. Step 7: Confirm both ends are genuinely reachable. Taking ff piecewise linear through (1,1), (2,2), (3,4)(1,1),\ (2,2),\ (3,4) gives slopes 11 and 22, both at least 11, with f(2)=2f(2) = 2; through (1,1), (2,3), (3,4)(1,1),\ (2,3),\ (3,4) gives slopes 22 and 11, with f(2)=3f(2) = 3. (Either can be smoothed into a differentiable function keeping f′≥1f' \ge 1.) Answer: (2) and (3).
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