Let $f(x) = 3x^2 + x$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2001 - Paper 1

To solve this integral, we can use substitution:

Let (u = 1 + e^x), then (du = e^x dx). The integral becomes:

$\int \frac{e^x}{1+e^x} dx = \int \frac{du}{u} = \ln|u| + C = \ln(1 + e^x) + C.$

To evaluate this integral, we can use the identity (\cos^2 x = \frac{1 + \cos(2x)}{2}):

$\int 3\cos^2 x dx = 3 \int \frac{1 + \cos(2x)}{2} dx = \frac{3}{2}(x + \frac{1}{2}\sin(2x)) + C.$

Evaluating from 0 to 2:

1. At $x = 2$:
   $\frac{3}{2}(2 + \frac{1}{2}\sin(4)) = 3 + \frac{3}{4}\sin(4)$

2. At $x = 0$:
   $\frac{3}{2}(0 + \frac{1}{2}\sin(0)) = 0.$

So, $\int_0^2 3\cos^2 x dx = 3 + \frac{3}{4}\sin(4).$

The word ALgebraic has 9 letters, among which the letters A and I are repeated. To find the number of arrangements:

1. Total arrangements = ( \frac{9!}{2!} = \frac{362880}{2} = 181440. )

In this case, the vowels A, A, I can occupy 4 positions. To find arrangements, we:

1. Choose 4 positions for the vowels:
   • The arrangement of vowels = ( \frac{4!}{2!} = 12. )
2. The remaining 5 consonants can be arranged in the remaining 5 positions:
   • The arrangement of consonants = (5! = 120. )
3. Total arrangements = ( 12 \times 120 = 1440. )

We can rewrite the expression as:

$\left(x - \frac{1}{x}\right)^9.$ To find the term independent of $x$, we will look for the term where the powers of $x$ equal 0.

Using the binomial theorem, the general term is:

$T_k = \binom{9}{k} x^{9-k} \left(-\frac{1}{x}\right)^k = \binom{9}{k} (-1)^k x^{9 - 2k}.$

Set the exponent of $x$ to 0:

$9 - 2k = 0 \Rightarrow k = \frac{9}{2},$ which is not an integer.

Thus, we find the nearest integers: k = 4 leads to: $T_4 = \binom{9}{4} (-1)^4 = 126.$