Find $$\frac{d}{dx}(2\sin^{-1}(5x)).$$ Use the binomial theorem to find the term independent of $x$ in the expansion of $$\left(2x-\frac{1}{x^2}\right)^{12}.$$ (i) Differentiate $$e^{x}(\cos x - 3\sin x).$$ (ii) Hence, or otherwise, find $$\int e^{3x}\sin x\,dx.$$ A salad, which is initially at a temperature of 25°C, is placed in a refrigerator that has a constant temperature of 3°C - HSC - SSCE Mathematics Extension 1 - Question 2 - 2005 - Paper 1

The binomial theorem states that:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.$

Here, let $a = 2x$ and $b = -\frac{1}{x^2}$. We need to find the term where the total power of $x$ is zero:

• The general term is given by: $T_k = \binom{12}{k} (2x)^{12-k} \left(-\frac{1}{x^2}\right)^k = \binom{12}{k} 2^{12-k} (-1)^k x^{12-k - 2k} = \binom{12}{k} 2^{12-k} (-1)^k x^{12 - 3k}.$
• Set $12 - 3k = 0$ which gives us $k = 4$.

Therefore, the independent term is:

$\binom{12}{4} 2^{8} (-1)^4 = \frac{12!}{4!8!} \cdot 256 = 495 \cdot 256 = 126720.$

To differentiate this product, we use the product rule:

If $u = e^{x}$ and $v = \cos x - 3\sin x$, then:

$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.$

Calculating:

• For $u$: $\frac{du}{dx} = e^{x}.$
• For $v$: $\frac{dv}{dx} = -\sin x - 3\cos x.$

Thus, the derivative is:

$e^{x}(-\sin x - 3\cos x) + (\cos x - 3\sin x)e^{x} = e^{x}(-\sin x - 3\cos x + \cos x - 3\sin x) = e^{x}(-4\sin x - 2\cos x).$

To find this integral, we use integration by parts or find a suitable formula.

• Using integration by parts, let $I = \int e^{3x}\sin x \,dx.$
  Using integration by parts, we can express:

$I = \frac{1}{10} e^{3x}(-\cos x) + \frac{3}{10} \int e^{3x}\sin x \,dx.$

Solving this gives:

$I = \frac{1}{10} e^{3x}(-\cos x) + \frac{3}{10} I.$

Thus, we can solve for I:

$\frac{7}{10} I = -\frac{1}{10} e^{3x} \cos x \\ I = -\frac{1}{7} e^{3x} \cos x.$

To prove that $T = 3 + Ae^{-kt}$ satisfies the differential equation $\frac{dT}{dt} = -k(T - 3)$:

• Differentiate: $\frac{dT}{dt} = -kAe^{-kt}$.
• Substitute $T$ into the differential equation: $-k(3 + Ae^{-kt} - 3) = -k(Ae^{-kt}).$ Thus, both sides are equal, confirming that $T = 3 + Ae^{-kt}$ is a valid solution.

From our equation $T = 3 + Ae^{-kt}$, we can determine $A$ using the initial condition. At $t=10$, $T(10) = 11,$ so:

• Plugging in: $11 = 3 + Ae^{-10k}$
  Thus, $A = 8 + e^{10k}$.
• Next, to find $T(15)$, substitute $t=15$: $T(15) = 3 + A e^{-15k}$. Using our earlier value for $A$ gives: $T(15) = 3 + (8 + e^{10k}) e^{-15k} = 3 + 8e^{-15k} + e^{-5k}.$
  We simplify this depending on the value of $k$.