Definite Integration Question 46
Question 46
- Let $f: R \rightarrow(0,1)$ be a continuous function. Then, which of the following function(s) has (have) the value zero at some point in the interval $(0,1)$ ?
(2017 Adv.) (a) $e^{x}-\int_{0}^{x} f(t) \sin t d t$ (b) $f(x)+\int_{0}^{\frac{\pi}{2}} f(t) \sin t d t$ (c) $x-\int_{0} f(t) \cos t d t$ (d) $x^{9}-f(x)$
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Solution:
- $\because e^{x} \in(1, e)$ in $(0,1)$ and $\int_{0}^{x} f(t) \sin t d t \in(0,1)$ in $(0,1)$
$\therefore e^{x}-\int_{0}^{x} f(t) \sin t d t$ cannot be zero.
So, option (a) is incorrect. (b) $f(x)+\int_{0}^{\frac{\pi}{2}} f(t) \sin t d t$ always positive
$\therefore$ Option (b) is incorrect.
(c) Let $h(x)=x-\int_{0}^{2} f(t) \cos t d t$,
$$ \begin{aligned} & h(0)=-\int_{0}^{\frac{\pi}{2}} f(t) \cos t d t<0 \ & h(1)=1-\int_{0}^{\frac{\pi}{2}-1} f(t) \cos t d t>0 \end{aligned} $$
$\therefore$ Option (c) is correct.
(d) Let $g(x)=x^{9}-f(x)$
$g(0)=-f(0)<0$
$g(1)=1-f(1)>0$
$\therefore$ Option (d) is correct.
