Let f: R rightarrow (0,1) be a continuous fun | vedclass

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(A) For option $[A]$,let $g(x) = e^x - \int_0^x f(t) \sin t \, dt$. Since $f(t) \in (0,1)$ and $\sin t \in [0,1]$ for $t \in (0,1)$,the integral $\int

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$B, D$

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(A) For option $[A]$,let $g(x) = e^x - \int_0^x f(t) \sin t \, dt$. Since $f(t) \in (0,1)$ and $\sin t \in [0,1]$ for $t \in (0,1)$,the integral $\int_0^x f(t) \sin t \, dt  e^x - x$. Since $e^x - x > 0$ for all $x$,$g(x)$ cannot be zero.For option $[B]$,let $h(x) = x^9 - f(x)$. Since $f(x) \in (0,1)$,$h(0) = 0^9 - f(0) = -f(0)  0$. By the Intermediate Value Theorem $(IVT)$,there exists $c \in (0,1)$ such that $h(c) = 0$.For option $[C]$,let $k(x) = f(x) + \int_0^{\pi/2} f(t) \sin t \, dt$. Since $f(x) > 0$ and the integral is positive,$k(x) > 0$ for all $x \in (0,1)$. Thus,it cannot be zero.For option $[D]$,let $m(x) = x - \int_0^{\pi/2 - x} f(t) \cos t \, dt$. Then $m(0) = 0 - \int_0^{\pi/2} f(t) \cos t \, dt  0$. By $IVT$,there exists $c \in (0,1)$ such that $m(c) = 0$.

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

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 $(0,1)$?

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