Question

Let $f:R\to (0,1)$ be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval $(0,1)$?

• A. $e^x-{\int }_0^{x}f\left(t\right)\sin \left(t\right)dt$
• B. $f\left(x\right)+{\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(t\right)\sin \left(t\right)dt$
• C. $x-{\int }_0^{\frac{\mathrm{\pi }}{2}-x}f\left(t\right)\cos \left(t\right)dt$
• D. $x^9-f\left(x\right)$

Answer 1

Answer: (C), (D)

$x^9-f\left(x\right)$

Step 1. Explanation for correct option:

Option (C): $x-{\int }_0^{\frac{\mathrm{\pi }}{2}-x}f\left(t\right)\cos \left(t\right)dt$

Let $g\left(x\right)=x-{\int }_0^{\frac{\mathrm{\pi }}{2}-x}f\left(t\right)\cos \left(t\right)dt$

$$\begin{gathered} \Rightarrow g\left(x\right)=x-{\int }_0^{\frac{\mathrm{\pi }}{2}-x}f\left(t\right)\cos \left(t\right)dt \\ \Rightarrow g\left(0\right)=0-{\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(t\right)\cos \left(t\right)dt \end{gathered}$$

$\therefore g\left(0\right)\le 0$ $\left[\because f\left(t\right)\in (0,1)andcost\in (0,1)fort\in \left(0,\frac{\pi }{2}\right)\right]$

$$\begin{gathered} g\left(1\right)=1-{\int }_0^{\frac{\mathrm{\pi }}{2}-1}f\left(t\right)\cos \left(t\right)dt \\ \therefore g\left(1\right)>0\mathbf{\left[}\mathbf{\because }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\in }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{cost}\mathbf{\in }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{for}\mathbf{}\mathbf{t}\mathbf{\in }\mathbf{(}\mathbf{0}\mathbf{,}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{\right]} \end{gathered}$$

So $g\left(x\right)=0$ for atleast one value of $x\in \left(0,1\right)$

Hence, Option C is correct.

Option (D): $x^9-f\left(x\right)$

Let $g\left(x\right)=x^9-f\left(x\right)$

$\Rightarrow g\left(0\right)=0-f\left(0\right)$

As $f\left(x\right)\in \left(0,1\right)$ , then $g\left(x\right)<0$

$g\left(1\right)=1-f\left(1\right)$

As $f\left(x\right)\in \left(0,1\right)$ , then $g\left(x\right)>0$

So $g\left(x\right)=0$ for atleast one value of $x\in \left(0,1\right)$

Hence, Option D is correct.

Step 2. Explanation for incorrect option:

Option (A): ${\mathit{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\left(t\right)\mathit{d}\mathit{t}$

Let $g\left(x\right)=e^x-{\int }_0^{x}f\left(t\right)\sin \left(t\right)dt$

$\Rightarrow g\left(0\right)=1$

$g\left(1\right)=e^1-{\int }_0^{1}f\left(t\right)\sin \left(t\right)dt\ge e-1>0$

As $x\in \left(0,1\right)$, and $f\left(x\right)\sin \left(x\right)\in \left(0,1\right)$. Thus $e^x\in (1,e)$. This means, $e^x>1$.

$\therefore g\left(0\right)·g\left(1\right)\ge 1$

Therefore, no root.

Hence, Option A is incorrect.

Option (B): $f\left(x\right)+{\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(t\right)\sin \left(t\right)dt$

$f\left(t\right)\in \left(0,1\right)and\sin t>0$. Thus $f\left(t\right)>0$ and also $f\left(t\right)\sin \left(t\right)>0$ for $t\in \left(0,\frac{\mathrm{\pi }}{2}\right)$

$g\left(x\right)=f\left(x\right)+{\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(t\right)\sin \left(t\right)dt>0$ for $x\in \left(0,1\right)$

Hence, option B is incorrect.

Therefore, option (C) and (D) are correct.