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

Answer 1

Answer: (C), (D)

The correct options are
C $x-{\int }_0^{\frac{\pi }{2}-x}f\left(t\right)\cos tdt$
D $x^9-f\left(x\right)$

Let $g\left(x\right)=f\left(x\right)+{\int }_0^{\frac{\pi }{2}}f\left(t\right)\sin tdt$

(A)

$f\left(0\right)+0\le f\left(0\right)+{\int }_0^{\frac{\pi }{2}}f\left(t\right)\sin dt\le f\left(0\right)+\frac{\pi }{2}$

$f\left(0\right)\le g\left(0\right)\le f\left(0\right)+\frac{\pi }{2}$

hence $g\left(0\right)$ is always positive

$g\left(1\right)$

$f\left(1\right)\le f\left(1\right)+{\int }_0^{\frac{\pi }{2}}f\left(t\right)\sin dt\le f\left(1\right)+\frac{\pi }{2}$

$f\left(1\right)\le g\left(1\right)\le f\left(1\right)+\frac{\pi }{2}$

Hence $g\left(1\right)$ is also positive

thus $g\left(0\right).g\left(1\right)>0$

A is incorrect

(B)

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

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

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

$g\left(0\right).g\left(1\right)\ge 0$

No root

B is incorrect

(C)

$g\left(0\right)$ $=0-{\int }_0^{\frac{\pi }{2}}f\left(t\right)\cos tdt$

$-\frac{\pi }{2}\le g\left(0\right)\le 0$

$g\left(0\right)$ is negative.

$g\left(1\right)$

$1-{\int }_0^{\frac{\pi }{2}-1}f\left(t\right)\cos t$

$1-(\frac{\pi }{2}-1)\le g\left(1\right)\le 1$

$2-\frac{\pi }{2}\le g\left(1\right)<1$

Hence $g\left(0\right)-g\left(1\right)<0$

C is correct

(D)

$g\left(0\right)=0-f\left(0\right)\in (-1,0)$

$g\left(1\right)=1-f\left(1\right)\in (0,1)$

Hence $g\left(0\right).g\left(1\right)<0$

D is correct

Hence, options C and D are the answers.