Let $f : R \to (0, 1)$ be a continuous function. Then, which of the following function(s) does not have the value zero at some point in the interval (0, 1).

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$f (x) + \int_{0}^{x} f (t) \cos t d t$

$x - \int_{0}^{\frac{π}{2} - x} f (t) \cos t d t$

$e^{x} - \int_{0}^{x} e^{t} f (t) d t$

$x^{8} + f (x)$

answer is A, B, C.

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Detailed Solution

Consider $g_{1} (x) = e^{x} - \int_{0}^{x} e^{t} f (t) d t$
$g_{1} (0) = 1, g_{1} (1) = e - \int_{0}^{1} e^{t} f (t) d t$
$f (t) \in (0, 1)$
$e^{t} f (t) \in (0, e^{t})$
$\int_{0}^{1} e^{t} f (t) d t \in (0, (e - 1))$
Consider $g_{2} (x) = x^{8} + f (x)$
$g_{2} (0) = f (0) \in (0, 1)$
$g_{2} (1) = 1 + f (1) \in (1, 2)$
$x^{8}$ is positive for $x \in (0, 1)$
$f (x) \in (0, 1)$ for all x
$\Rightarrow x^{8} + f (x)$ is always positive
Consider $g_{3} (x) = f (x) + \int_{0}^{x} f (t) \cos t d t$
$f (x) \in (0, 1)$ for all x
$\int_{0}^{1} f (t) \cos t d t$ is positive because f(t) and cost are positive
$\Rightarrow g_{3} (x)$ is always positive for $x \in (0, 1)$
Consider $g_{4} (x) = x - \int_{0}^{\frac{π}{2} - x} f (t) \cos t d t$
$g_{4} (0) = \int_{0}^{- \frac{π}{2}} f (t) \cos t d t < 0$
$g_{4} (1) = 1 - \int_{0}^{\frac{π}{2} - 1} f (t) \cos t d t$
$\int_{0}^{\frac{π}{2} - 1} f (t) \cot d t \in \int_{0}^{\frac{π}{2} - 1} (0, 1) \cos t d t$
$\in (0, \int_{0}^{\frac{π}{2} - 1} \cos t d t)$
$\in (0, \cos 1) g_{4} (1) > 0$
$g_{4} (x)$ has a root in (0, 1)