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

Hint: 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.