Question

Let $f:R\to \left(0,1\right)$ be a continuous function. Then which of the following function $\left(s\right)$ 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. $x^9-f\left(x\right)$
• C. $x-{\int }_0^{\frac{\pi }{2}-x}f\left(t\right)\cos tdt$
• D. $e^x-{\int }_0^{x}f\left(t\right)\cos tdt$

Answer 1

Answer: (B), (C)

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

$\left(a\right)f\left(x\right)+{\int }_0^{\frac{\pi }{2}}f\left(t\right)\sin tdt>0$ for all $x\in (0,1)$

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

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

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

$\therefore f\left(x\right)\in (0,1)$

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

$\therefore x^9-f\left(x\right)=0$ for some $x\in (0,1)$

$\left(c\right)$Let $h\left(x\right)=x-{\int }_0^{\frac{\pi }{2}-x}f\left(t\right)\cos tdt$

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

and $h\left(1\right)=1-{\int }_0^{\frac{\pi }{2}-1}f\left(t\right)\cos tdt>0$

$h\left(x\right)=x-{\int }_0^{\frac{\pi }{2}-x}-f\left(t\right)\cos tdt=0$ at some $x\in (0,1)$

$\left(d\right)e^x-{\int }_0^x-f\left(t\right)\cos tdt$

$\because x\in (0,1)\Rightarrow e^x\in (1,e)$

and $0<f\left(t\right)<1$ and $0<\cos t<x$ for all $x\in (0,1)$

$\therefore 0<{\int }_0^{x}f\left(t\right)\cos tdt<1$

$\therefore e^x-{\int }_0^{x}f\left(t\right)\cos tdt\ne 0$ for any $x\in (0,1)$

Option $\left(b\right)$ and $\left(c\right)$ are correct.