Mathematics - Application of Derivative Question with Solution | TestHub

For option (i),

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

$\therefore M\left(0\right)=0-{\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(t\right).\cos t dt<0$

Also,  $M\left(1\right)=1-{\int }_0^{\frac{\mathrm{\pi }}{2}-1}f\left(t\right).\mathrm{cost }dt>0$

$\Rightarrow M\left(0\right). M\left(1\right)<0$     (so $M\left(x\right)$ can be zero)

For option (ii),

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

Now,  $k\left(0\right)=-f\left(0\right)<0$, as $f\in \left(0, 1\right)$

Also,  $k\left(1\right)=1-f\left(1\right)>0 \left(As f\in \left(\mathrm{0,1}\right)\right)$

$\Rightarrow k\left(0\right). k\left(1\right)<0$     (so $k(x)$ can be zero)

For option (iii),

Let  $g\left(x\right)=e^x-{\int }_0^{x}f\left(t\right)\mathrm{si}\mathrm{n}tdt$

$\therefore g\text{'}\left(x\right)=e^x-\left(f\left(x\right)·\sin x\right)>0 \forall x\in \left(0, 1\right)$    (as $f\left(x\right)·\text{sin x}$ , is less than 1)

$\Rightarrow g\left(x\right)$ is strictly increasing function.

$\Rightarrow g\left(x\right)>1 \forall x\in \left(0, 1\right)$

Also,  $g\left(0\right)=1$, (so $g(x)$ is, never zero)

For option (iv),

Let, $T\left(x\right)=f\left(x\right)+{\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(t\right).\sin t dt$

$\Rightarrow T\left(x\right)>0 \forall x\in \left(0, 1\right) \left(As f\in \left(0, 1\right)\right)$     (so $T(x)$ can not be zero)