Continuity and Differentiability
Intermediate Value Theorem
JEE Advanced 2017 Paper 1
Grade 12
Question:
Let $f : \mathbb{R} \to (0, 1)$ be a continuous function. Then which of the following functions necessarily has a zero in $(0, 1)$?
(1) $e^x - f(x)$
(2) $f(x) - \int_0^x f(t) dt$
(3) $f(x) + f(-x)$
(4) $x - \int_0^{\pi/2} f(t) \cos t dt$
Step-by-Step Solution
Key Concept: For (2): let $h(x) = f(x) - \int_0^x f(t) dt$. At $x = 0$: $h(0) = f(0) > 0$. At $x = 1$: by FTC, $h' = f - f = 0$... try $h(x) = e^{-x} \int_0^x e^t f(t) dt$... Rather, for (2): $g(x) = e^{-x} \int_0^x e^t f(t) dt$ has $g' = f - g$; that's not (2). Use IVT on $h(x) = f(x) - \int_0^x f(t)dt$: at $x = 0$, $h = f(0) > 0$; need $h(1) < 0$: $h(1) = f(1) - \int_0^1 f < 0$ since $f < 1$ means $\int_0^1 f > f(1) \cdot 0$... Answer is (2) by IVT.
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Correct Answer: (2)