If \(\displaystyle\frac{d}{dx}(f(x))=5x^4-\frac{4}{x^5}\) such that f(1) = 0. Then \(f(2)-2f\left(\displaystyle\frac{1}{2}\right)\) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
f(x) is the antiderivative of the given expression:
\( \displaystyle f(x) = \int\left(5x^4 - 4x^{-5}\right)dx \)
\( \displaystyle = x^5 + \frac{1}{x^4} + C \)
Using f(1) = 0:
\(1 + 1 + C = 0\)
\(C = -2\)
So
\( \displaystyle f(2) = 32 + \frac{1}{16} - 2 = 30 + \frac{1}{16} \)
\( \displaystyle f\left(\frac{1}{2}\right) = \frac{1}{32} + 16 - 2 = 14 + \frac{1}{32} \)
\( \displaystyle 2f\left(\frac{1}{2}\right) = 28 + \frac{1}{16} \)
Hence
\( \displaystyle f(2) - 2f\left(\frac{1}{2}\right) = \left(30 + \frac{1}{16}\right) - \left(28 + \frac{1}{16}\right) \)
\(= 2\)