Question

Let $f:\mathbb{R} \to (0,1)$ be a continuous function. Then, which of the following function(s) has (have) the value zero at some point in the interval $\left( {0,1} \right)$ ?
A. $f\left( x \right) + \int_0^{\dfrac{\pi }{2}} {f\left( t \right)} \sin tdt$
B. ${e^x} - \int_0^x {f\left( t \right)} \sin tdt$
C. $x - \int_0^{\dfrac{\pi }{2} - x} {f\left( t \right)} \cos tdt$
D. ${x^9} - f\left( x \right)$

Answer 1

In the above given question, we are given a continuous function that is $f:\mathbb{R} \to (0,1)$ . We have to determine which of the above given functions have at least one value equal to zero at some point in the interval $\left( {0,1} \right)$ . In order to approach the solution, first we have to suppose a function $g\left( x \right)$ equal to each given option. Then we can determine if $g\left( x \right)$ has a value equal to zero or not in the interval $\left( {0,1} \right)$ by checking the sign of the function in the increasing order from the minimum to maximum value of $g\left( x \right)$.

Complete step by step answer:
Given that, a continuous function is $f:\mathbb{R} \to (0,1)$. We have to determine if the given options have a value equal to zero in the interval $\left( {0,1} \right)$ or not. First, let each of the functions equal to $g\left( x \right)$. Now, we will check the sign of the function $g\left( x \right)$ at the end points of the interval $\left( {0,1} \right)$ for the minimum and maximum values of $g\left( x \right)$ in that interval. If while going from the minimum to the maximum value of $g\left( x \right)$ , the sign of $g\left( x \right)$ changes from negative to positive or vice versa, then the function $g\left( x \right)$ has at least one value equal to zero in the interval $\left( {0,1} \right)$.

(A) Let $g\left( x \right) = f\left( x \right) + \int_0^{\dfrac{\pi }{2}} {f\left( t \right)} \sin tdt$
Now, at zero we have
$\Rightarrow g\left( 0 \right) = f\left( 0 \right) + \int_0^{\dfrac{\pi }{2}} {f\left( t \right)} \sin tdt$
That gives the minimum and maximum values as,
$\Rightarrow f\left( 0 \right) + 0 \leqslant g\left( 0 \right) \leqslant f\left( 0 \right) + \dfrac{\pi }{2}$
Hence $g\left( 0 \right)$ is positive.
Now, at one we have
$\Rightarrow g\left( 1 \right) = f\left( 1 \right) + \int_0^{\dfrac{\pi }{2}} {f\left( t \right)} \sin tdt$
That gives,
$\Rightarrow f\left( 1 \right) + 0 \leqslant g\left( 1 \right) \leqslant f\left( 1 \right) + \dfrac{\pi }{2}$
Hence, $g\left( 1 \right)$ is also positive. That means it is never zero in the interval $\left( {0,1} \right)$ but always positive.

(B) Let $g\left( x \right) = {e^x} - \int_0^x {f\left( t \right)} \sin tdt$
At zero, we have
$\Rightarrow g\left( 0 \right) = {e^0} - \int_0^x {f\left( t \right)} \sin tdt$
Hence,
$\Rightarrow g\left( 0 \right) = 1$
Now at one, we have
$\Rightarrow g\left( 1 \right) = e - \int_0^x {f\left( t \right)} \sin tdt$
That gives, the minimum value as
$\Rightarrow g\left( 1 \right) \geqslant e - 1 > 0$
Since both are positive, hence it is an incorrect option.

(C) Let $g\left( x \right) = x - \int_0^{\dfrac{\pi }{2} - x} {f\left( t \right)} \cos tdt$
At zero, we have
$g\left( 0 \right) = 0 - \int_0^{\dfrac{\pi }{2}} {f\left( t \right)} \cos tdt$
That gives,
$\Rightarrow 0 - \dfrac{\pi }{2} \leqslant g\left( 0 \right) \leqslant 0 - 0$
Hence, $g\left( 0 \right)$ is negative.
Also at one, we have
$g\left( 1 \right) = 1 - \int_0^{\dfrac{\pi }{2} - 1} {f\left( t \right)} \cos tdt$
That gives,
$\Rightarrow 1 - \left( {\dfrac{\pi }{2} - 1} \right) \leqslant g\left( 1 \right) \leqslant 1 - 0$
$\Rightarrow 2 - \dfrac{\pi }{2} \leqslant g\left( 1 \right) \leqslant 1$
Hence $g\left( 1 \right)$ is positive, i.e. the sign is changed.
Therefore, it has at least one value equal to zero in the interval $\left( {0,1} \right)$ .

(D) Let $g\left( x \right) = {x^9} - f\left( x \right)$
At zero, we have
$\Rightarrow g\left( 0 \right) = {0^9} - f\left( 0 \right)$
That gives us,
$\Rightarrow g\left( 0 \right) = - f\left( 0 \right) \in \left( { - 1,0} \right)$
That is a negative value.
Now at one, we have
$\Rightarrow g\left( 1 \right) = {1^9} - f\left( 1 \right)$
That gives us,
$\therefore g\left( 1 \right) = 1 - f\left( 1 \right) \in \left( {0,1} \right)$
Since that is a positive value, hence $g\left( x \right)$ has a value equal to zero in the interval $\left( {0,1} \right)$ .

Therefore, C and D are the only correct options.

Note: The continuous functions are functions that have no restrictions throughout their domain or a given interval. Their graphs won't contain any asymptotes or signs of discontinuities as well. A function which is continuous on an interval does not always mean that the function is as well as differentiable. Whereas if a function is differentiable on an interval then it is also a continuous function on the given interval.