Option 4 : \(\frac{247}{16}\)

__Concept:__

*Remainder Theorem: *When polynomial **p(x)** is divided by another polynomial **q(x) = x - a**, then the **remainder is equal to p(a)**, where p(x) be any polynomial of degree greater than or equal to one and 'a' be any real number.

**Calculation:**

By remainder theorem, we know that f(x) when divided by g(x) = 2x - 5 gives remainder equal to f(5/2).

Now, f(x) = -x4 + 2x3 + 5x2 - 6x + 7

\(\Rightarrow f\left ( \frac{5}{2} \right ) = -\left ( \frac{5}{2} \right )^{4} \ + \ 2\left ( \frac{5}{2} \right )^{3} \ + \ 5\left ( \frac{5}{2} \right )^{2} \ - \ 6\left ( \frac{5}{2} \right ) \ + \ 7\)

\(\Rightarrow f\left ( \frac{5}{2} \right ) = - \frac{625}{16} \ + \ 2\left ( \frac{125}{8} \right ) \ + \ 5\left ( \frac{25}{4} \right ) \ - \ \frac{30}{2} \ + \ 7\)

\(\Rightarrow f\left ( \frac{5}{2} \right ) = - \frac{625}{16} \ + \ \frac{125}{4} \ + \ \frac{125}{4} \ - \ 15 \ + \ 7\)

\(\Rightarrow f\left ( \frac{5}{2} \right ) = - \frac{625}{16} \ + \ \frac{250}{4} \ - \ 8\)

\(\Rightarrow f\left ( \frac{5}{2} \right ) = \frac{-625 \ + 1000 \ - \ 128}{16}\)

\(\Rightarrow f\left ( \frac{5}{2} \right ) = \frac{247}{16}\)

**Hence, the remainder if f(x) is divided by g(x) is \(\frac{247}{16}\).**