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. $x^9-f\left(x\right)$
• B. $x-\underset{0}{\overset{\frac{\pi }{2}-x}{\int }}f\left(t\right)\cos tdt$
• C. $e^x-\underset{0}{\overset{x}{\int }}f\left(t\right)\sin tdt$
• D. $f\left(x\right)+\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)\sin tdt$

Answer 1

Answer: (A), (B)

The correct options are
A $x^9-f\left(x\right)$
B $x-\underset{0}{\overset{\frac{\pi }{2}-x}{\int }}f\left(t\right)\cos tdt$

A)
As range of $f\left(x\right)$ is $(0,1)$,
Let $g\left(x\right)=x^9-f\left(x\right)$
$\therefore g\left(x\right)=0\Rightarrow x=\left(f\right(x){)}^{\frac{1}{9}}$
$x$ will lie $(0,1)$.

B)
Let $g\left(x\right)=x-\underset{0}{\overset{\frac{\pi }{2}-x}{\int }}f\left(t\right)\cos tdt$
$g\left(0\right)=-\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)\cos tdt\Rightarrow g\left(0\right)<0$
Now,
$g\left(1\right)=1-\underset{0}{\overset{\frac{\pi }{2}-1}{\int }}f\left(t\right)\cos tdt$
The minimum value of $g\left(1\right)$ takes place when $\underset{0}{\overset{\frac{\pi }{2}-1}{\int }}f\left(t\right)\cos tdt$ is maximum
Max.$\left(\underset{0}{\overset{\frac{\pi }{2}-1}{\int }}f\right(t\left)\cos tdt\right)=\frac{\pi }{2}-1$
Min.$\left(g\right(1\left)\right)=2-\frac{\pi }{2}>0$
$\therefore g\left(1\right)>0$
As $g\left(0\right)\cdot g\left(1\right)<0$ there is atleast one value of $x\in (0,1)$ where $g\left(x\right)$ will be zero.

C)
Let $g\left(x\right)=e^x-\underset{0}{\overset{x}{\int }}f\left(t\right)\sin tdt$
$g^{\prime }\left(x\right)=e^x-f\left(x\right)\sin x$
When $x\in (0,1)$ then $e^x\in (1,e)$ and $\left(f\right(x\left)\sin x\right)\in (0,1)$
$\therefore g^{\prime }\left(x\right)>0\forall x\in (0,1)$
So $g\left(x\right)$ monotonically increasing function.
Now, $g\left(0\right)=1$
So $g\left(x\right)>1\forall x\in (0,1)$

D)
It will always remain positive in $(0,1)$

Hence option A and B are correct.