Mathematics - Functions Question with Solution | TestHub

We are asked to solve the question by taking 25 in place of 4.

First, let's define the function $f(x)$ and $f(1-x)$ with 25 instead of 4:

$$f(x)=\frac{25^x}{25^x+5}$$

Now, let's find $f(1-x)$:

$$f(1-x)=\frac{25^{1-x}}{25^{1-x}+5}=\frac{\frac{25}{25^x}}{\frac{25}{25^x}+5}=\frac{25}{25+5\cdot 25^x}=\frac{5}{5+25^x}$$

Now, let's sum $f(x)$ and $f(1-x)$:

$$f(x)+f(1-x)=\frac{25^x}{25^x+5}+\frac{5}{5+25^x}=\frac{25^x+5}{25^x+5}=1$$

So, we have the property $f(x)+f(1-x)=1$.

We need to evaluate the sum:

$$f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+\dots +f\left(\frac{39}{40}\right)-f\left(\frac{1}{2}\right)$$

We can group the terms using the property $f(x)+f(1-x)=1$.

The sum can be rewritten as:

$$\left(f\left(\frac{1}{40}\right)+f\left(\frac{39}{40}\right)\right)+\left(f\left(\frac{2}{40}\right)+f\left(\frac{38}{40}\right)\right)+\dots +\left(f\left(\frac{19}{40}\right)+f\left(\frac{21}{40}\right)\right)+f\left(\frac{20}{40}\right)-f\left(\frac{1}{2}\right)$$

There are 19 such pairs, and each pair sums to 1.

For example, $f\left(\frac{1}{40}\right)+f\left(1-\frac{1}{40}\right)=f\left(\frac{1}{40}\right)+f\left(\frac{39}{40}\right)=1$.

So, the sum of the pairs is $19\times 1=19$.

The term $f\left(\frac{20}{40}\right)$ simplifies to $f\left(\frac{1}{2}\right)$.

So the expression becomes:

$$19+f\left(\frac{1}{2}\right)-f\left(\frac{1}{2}\right)$$

$$=19$$