Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$ for all positive integers $n$. The diagram shows a conical drinking cup of height 12 cm and radius 4 cm. The cup is being filled with water at the rate of $3 \, \text{cm}^3$ per second. The height of water at time $t$ seconds is $h$ cm and the radius of the water's surface is $r$ cm. (i) Show that $r = \frac{1}{3}h$. (ii) Find the rate at which the height is increasing when the height of water is 9 cm. (Consider the function $f(x) = 2 \sin^{-1} \sqrt{x} - \sin^{-1}(2x - 1)$ for $0 \leq x \leq 1$. (i) Show that $f'(x) = 0$ for $0 < x < 1$. (ii) Sketch the graph of $y = f(x)$.

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Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$ for all positive integers $n$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2002 - Paper 1

Question 5

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Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$ for all positive integers $... show full transcript

Worked Solution & Example Answer:Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$ for all positive integers $n$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2002 - Paper 1

Step 1

Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$

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Answer

To use mathematical induction, we first prove the base case:

Base Case ( $n = 1$ ):
- Left-hand side (LHS): $2 \times 1! = 2$
- Right-hand side (RHS): $1(1 + 1)! = 2$
- Thus, LHS = RHS for $n = 1$ .
Induction Hypothesis:
- Assume true for $n = k$ :
$2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (k^2 + 1)k! = k(k + 1)!$
Induction Step ( $n = k + 1$ ):
- We need to show:
$2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + ((k + 1)^2 + 1)(k + 1)! = (k + 1)(k + 2)!$
- Starting with the LHS:
$LHS = k(k + 1)! + ((k + 1)^2 + 1)(k + 1)!$
- Factor out $(k + 1)!$ :
$LHS = (k + 1)! \left[k + (k + 1)^2 + 1\right] = (k + 1)! \left[k + k^2 + 2k + 1 + 1\right]$
- Simplifying:
$= (k + 1)!(k^2 + 3k + 3) = (k + 1)(k + 2)!$
- Therefore, LHS = RHS:
- Hence the statement is true for $n = k + 1$ . Thus, by induction, it is true for all positive integers $n$ .

Step 2

Find the rate at which the height is increasing when the height of water is 9 cm.

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Answer

We are given the volume of the cone $V = \frac{1}{3} \pi r^2 h$ , where $h$ is the height of water and $r$ is the radius of the water’s surface.

From part (i), we established that $r = \frac{1}{3}h$ . Substituting:

$V = \frac{1}{3} \pi \left(\frac{1}{3}h\right)^2 h = \frac{1}{3} \pi \frac{1}{9} h^2 h = \frac{\pi}{27} h^3$

Differentiating with respect to time $t$ :

$\frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt}$

Set $rac{dV}{dt} = 3$ (the rate at which the cup is being filled).
Thus:

$3 = \frac{\pi}{9} h^2 \frac{dh}{dt}$

When $h = 9$ cm:

$3 = \frac{\pi}{9} (9)^2 \frac{dh}{dt}$

Solve for $rac{dh}{dt}$ :

$3 = \frac{\pi}{9} \cdot 81 \frac{dh}{dt} \Rightarrow 3 = 9\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{1}{3\pi}$

Therefore, the rate at which the height is increasing when the height of water is 9 cm is $\frac{1}{3\pi} \; \text{cm/s}$ .

Step 3

Show that $f'(x) = 0$ for $0 < x < 1$.

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Answer

To find $f'(x)$ , we take the derivative of the function:

Using Chain Rule:

$f(x) = 2 \sin^{-1} \sqrt{x} - \sin^{-1}(2x - 1)$

Derivative:

$f'(x) = 2 \cdot \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} - \frac{2}{\sqrt{1 - (2x - 1)^2}}$

= $\frac{1}{\sqrt{x(1 - x)}} - \frac{2}{\sqrt{4x(1 - x)}}$
- Setting this equal to zero and solving will show that the resulting equations yield critical points at $0 < x < 1$ .
- Thus we find $f'(x) = 0$ .

Step 4

Sketch the graph of $y = f(x)$.

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Answer

To sketch the graph of $y = f(x)$ , consider the following:

Domain: The function is defined for $0 \leq x \leq 1$ .
Behavior at endpoints: Evaluate:
- At $x = 0: f(0) = 0$
- At $x = 1: f(1) = 2\pi - \frac{\pi}{2}$ (from evaluating the function)
Critical points: The result from part (i) suggested that $f'(x) = 0$ leading to constant behavior or possible local max/min in between.

Using these aspects, sketch the curve starting from $(0, 0)$ and rising to its max within the given domain to $(1, 2\pi - \frac{\pi}{2})$ while ensuring continuous representation.

Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$ for all positive integers $n$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2002 - Paper 1

Worked Solution & Example Answer:Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$ for all positive integers $n$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2002 - Paper 1

Use the principle of mathematical induction to show that $2 \times 1! + 5 \times 2! + 10 \times 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$

Find the rate at which the height is increasing when the height of water is 9 cm.

Show that $f'(x) = 0$ for $0 < x < 1$.

Sketch the graph of $y = f(x)$.

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