Answer: (a) \(k e^{-2x}(1-2x)\); (b) \(\left(\frac12,\frac5e\right)\); (c) \(-(2x+1)e^{-2x}+C\); (d) \(1-3e^{-2}\).

Answer: (a) \(k e^{-2x}(1-2x)\); (b) \(\left(\frac12,\frac5e\right)\); (c) \(-(2x+1)e^{-2x}+C\); (d) \(1-3e^{-2}\).

(a) Use the product rule on \(y=kxe^{-2x}\):

\(\frac{dy}{dx}=k\left(e^{-2x}+x(-2e^{-2x})\right).\)

Therefore

\(\frac{dy}{dx}=ke^{-2x}(1-2x).\)

(b) For \(y=10xe^{-2x}\),

\(\frac{dy}{dx}=10e^{-2x}(1-2x).\)

At a stationary point,

\(10e^{-2x}(1-2x)=0.\)

Since \(e^{-2x}\ne0\),

\(1-2x=0,\qquad x=\frac12.\)

Then

\(y=10\left(\frac12\right)e^{-1}=\frac5e.\)

So the stationary point is

\(\left(\frac12,\frac5e\right).\)

(c) From part (a), take \(k=-2\):

\(\frac{d}{dx}(-2xe^{-2x})=-2e^{-2x}+4xe^{-2x}.\)

Thus

\(4xe^{-2x}=\frac{d}{dx}(-2xe^{-2x})+2e^{-2x}.\)

Therefore

\(\int4xe^{-2x}\,dx=-2xe^{-2x}+\int2e^{-2x}\,dx.\)

Since \(\int2e^{-2x}\,dx=-e^{-2x}\),

\(\int4xe^{-2x}\,dx=-2xe^{-2x}-e^{-2x}+C.\)

So

\(\int4xe^{-2x}\,dx=-(2x+1)e^{-2x}+C.\)

(d) Hence

So

\(-3e^{-2}-(-1)=1-3e^{-2}.\)