A curve passes through the point (0, 2), and the slope of the tangent…

The correct answer is Option (2) → \(4e - 4\)

The slope of the tangent is \(\frac{dy}{dx}\), so

    \( \displaystyle \frac{dy}{dx} = y + 2x \)

    \( \displaystyle \frac{dy}{dx} - y = 2x \)

This is a linear equation with \(P = -1\), so the integrating factor is \(e^{-x}\):

    \( \displaystyle y\,e^{-x} = \int 2x\,e^{-x}\,dx \)

Integrating by parts:

    \( \displaystyle \int 2x\,e^{-x}\,dx = -2x\,e^{-x} - 2e^{-x} + C \)

So

    \( y = -2x - 2 + Ce^{x} \)

The curve passes through (0, 2):

    \( 2 = -2 + C \Rightarrow C = 4 \)

So \(y = 4e^x - 2x - 2\), and at \(x = 1\):

    \( y = 4e - 2 - 2 = 4e - 4 \)