Question

Let f : [0, 1] → R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1) = 0 and satisfies f "(x) – 2f '(x) + f(x) ≥ e^x, x ∈ [0, 1].

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Linked Question 1

If the function e^–x f(x) assumes its minimum in the interval [0, 1] at $x = \frac{1}{4}$ , which of the following is true?

$f' (x) < f (x), 0 < x < \frac{1}{4}$

$f' (x) < f (x), \frac{1}{4} < x < \frac{3}{4}$

$f' (x) < f (x), \frac{3}{4} < x < 1$

$f' (x) > f (x), 0 < x < \frac{1}{4}$

$\frac{d}{dx} [e^{- x} f (x)] < 0 \forall x \in (0, \frac{1}{4})$

$\Rightarrow e^{- x} [f' (x) - f (x)] < 0 \forall x \in (0, \frac{1}{4})$

⇒ f ' (x) – f (x) < 0

Linked Question 2

Which of the following is true for 0 < x < 1?

– $\infty$ < f(x) < 0

$\frac{- 1}{4} < f (x) < 1$

0 < f(x) < $\infty$

$\frac{- 1}{2} < f (x) < \frac{1}{2}$

f '' (x) – 2f ' (x) + f (x) $\geq$ e^x

e^–x (f”(x) – 2f’(x) + f(x) $\geq$ 1)

$⇒ \frac{d^{2}}{{dx}^{2}} (e^{- x} f (x)) \geq 1$

$∵$ second order derivative of e^–x f(x) is positive

⇒ graph of e^–x f(x) is concave upwards

⇒ f (x) < 0 in (0, 1)