If \( \displaystyle\frac{d}{dx}[f(x)]=5x^4-\frac{2}{x^3} \) such that \( f\left(-\frac{1}{2}\right)=0 \), then \( f(x) \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( x^5+\displaystyle\frac{1}{x^2}-\frac{127}{32} \)
Integrating the derivative term by term:
\( \displaystyle f(x)=\int\left(5x^4-2x^{-3}\right)dx \)
\( \displaystyle =x^5-2\cdot\frac{x^{-2}}{-2}+C \)
\( \displaystyle =x^5+\frac{1}{x^2}+C \)
Since \( f\left(-\frac{1}{2}\right)=0 \):
\( \displaystyle \left(-\frac{1}{2}\right)^5+\frac{1}{\left(-\frac{1}{2}\right)^2}+C=0 \)
\( \displaystyle -\frac{1}{32}+4+C=0 \)
\( \displaystyle C=\frac{1}{32}-4 \)
\( \displaystyle =-\frac{127}{32} \)
\( \displaystyle f(x)=x^5+\frac{1}{x^2}-\frac{127}{32} \)
Shortcut: Options (2) and (4) start with \( x^4 \), whose derivative \( 4x^3 \) is not \( 5x^4 \), so they are out. Between (1) and (3), \( x^5+\frac{1}{x^2} \) at \( x=-\frac{1}{2} \) is \( -\frac{1}{32}+4=\frac{127}{32} \), so the constant must be \( -\frac{127}{32} \): Option (1).