A curve passes through the point (0, 1), and the slope of its tangent at any point (x, y) is \(3x^2(y + 1)\). The y-coordinate of the point of the curve with x = 1 is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2e − 1
The slope of the tangent is \(\frac{dy}{dx}\), so the curve satisfies
\( \displaystyle \frac{dy}{dx} = 3x^2(y + 1) \)
The variables separate:
\( \displaystyle \frac{dy}{y + 1} = 3x^2\,dx \)
Integrating both sides (y + 1 is positive at the given point):
\(\log(y + 1) = x^3 + c\)
Putting x = 0, y = 1:
\(c = \log 2\)
\(y + 1 = 2e^{x^3}\)
At x = 1:
\(y + 1 = 2e\)
\(y = 2e - 1\)