Let f: R → (0, 1) , be a continuous function then, which of the...

For option (i),

Let $M\left(x\right) = x - {∫}_{0}^{\dfrac{\pi}{2} - x}f\left(t\right) \cdot costdt$

$∴ M\left(0\right) = 0 - {∫}_{0}^{\dfrac{\pi}{2}}f\left(t\right) . costdt < 0$

Also, $M\left(1\right) = 1 - {∫}_{0}^{\dfrac{\pi}{2} - 1}f\left(t\right) . costdt > 0$

$⇒ M\left(0\right) . M\left(1\right) < 0$ (so $M ( x )$ 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, $k\left(1\right) = 1 - f\left(1\right) > 0\left(Asf \in \left(0,1\right)\right)$

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

For option (iii),

Let $g\left(x\right) = {e}^{x} - {∫}_{0}^{x}f\left(t\right)sintdt$

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

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

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

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

For option (iv),

Let, $T\left(x\right) = f\left(x\right) + {∫}_{0}^{\dfrac{\pi}{2}}f\left(t\right) . sintdt$

$⇒ T\left(x\right) > 0 \forall x \in \left(0, 1\right)\left(Asf \in \left(0, 1\right)\right)$ (so $T ( x )$ can not be zero)