Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$1 \times 2 + 2 \times 5 + 3 \times 8 + \cdots + n(3n - 1) = n^2(n + 1)$$ (b) When a particular biased coin is tossed, the probability of obtaining a head is $\frac{3}{5}$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2020 - Paper 1

The expected value for a binomial distribution is given by the formula:

$E(X) = n \cdot p,$

where $n$ is the number of trials and $p$ is the probability of success.

In this case:

• $n = 100$ (the number of tosses)
• $p = \frac{3}{5}$ (the probability of obtaining a head)

Therefore:

$E(X) = 100 \cdot \frac{3}{5} = 60.$

The variance of a binomial distribution is given by:

$Var(X) = n \cdot p \cdot (1-p).$

Substituting our values:

$Var(X) = 100 \cdot \frac{3}{5} \cdot \frac{2}{5} = 100 \cdot \frac{6}{25} = 24.$

The standard deviation is the square root of the variance:

$\sigma = \sqrt{Var(X)} = \sqrt{24} \approx 5.$

Using the normal approximation and the continuity correction, we want:

$P(54.5 < X < 65.5).$

Converting to Z-scores: $Z = \frac{X - \mu}{\sigma}$ where $\mu = 60$ and $\sigma \approx 5$.

Calculating: For $54.5$: $Z_{54.5} = \frac{54.5 - 60}{5} = -1.1$ For $65.5$: $Z_{65.5} = \frac{65.5 - 60}{5} = 1.1.$

Looking these values up in the Z-table, we find:
$P(-1.1 < Z < 1.1) \approx 0.726.$

Thus, the approximate probability that $X$ is between 55 and 65 is about $72.6\%$.

The number of different combinations of three topics from eight topics is given by:

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.$

Since there are 400 students and 56 combinations, by the pigeonhole principle, if each combination can have at most 7 students, we have: $7 \times 56 = 392.$

Therefore, at least one topic combination must have at least 8 students, since 400 students exceed 392.

We can use the product-to-sum identities to rewrite the integral:

$\cos A \sin B = \frac{1}{2} [\sin(A + B) - \sin(A - B)].$

Applying this identity:

$\cos 5x \sin 3x = \frac{1}{2} [\sin(8x) - \sin(2x)].$

Rewriting the integral:

$\int_0^{\pi} \cos 5x \sin 3x \, dx = \frac{1}{2} \left[\int_0^{\pi} \sin(8x) \, dx - \int_0^{\pi} \sin(2x) \, dx\right].$

Calculating both integrals, we find:

• For $\int \sin(num \, x) \, dx$, the integral is: $\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx).$

Thus, we have: $= \frac{1}{2} \left[-\frac{1}{8} \cos(8x) \bigg|_0^{\pi} + \frac{1}{2} \cos(2x) \bigg|_0^{\pi}\right].$

Evaluating: $= \frac{1}{2} \left[0 + 0 \right] = 0.$

We start with the given differential equation:

$\frac{dy}{dx} = -x + y.$

Rearranging gives us: $\frac{dy}{dx} - y = -x.$

This is a first-order linear differential equation. We can solve it using an integrating factor:

• The integrating factor is $e^{-x}$.

Multiplying through by the integrating factor and integrating both sides: $e^{-x} \frac{dy}{dx} - e^{-x} y = -e^{-x} x.$

Integrating: $\int e^{-x} \frac{dy}{dx} \, dx - \int e^{-x} y \, dx = -\int e^{-x} x \, dx.$

Solving yields the general solution: $y = Ce^{x} - x - 1.$

Using the initial condition (1, 0): When $x=1, y=0$, $$0 = Ce^{1} - 1 - 1.$

Thus: $C = 1 + e^{1}.$

The final equation of the curve is: $y = (1 + e^{1})e^{-x} - x - 1.$