Solution:

Solutions to Pascal's Triangle Expansion Problems

Question (a): Expand \( (3x + 2)^3 \)

Using Pascal's Triangle, the coefficients for expanding \( (a + b)^3 \) are 1, 3, 3, and 1.

The expansion of \( (3x + 2)^3 \) is:

\[ (3x + 2)^3 = 1 \cdot (3x)^3 + 3 \cdot (3x)^2 \cdot 2 + 3 \cdot (3x) \cdot 2^2 + 1 \cdot 2^3 \]

Simplifying each term:

\[ = 27x^3 + 54x^2 + 36x + 8 \]

Question (b): Expand \( (5 - 2x)^4 \)

Using Pascal's Triangle, the coefficients for expanding \( (a + b)^4 \) are 1, 4, 6, 4, and 1.

The expansion of \( (5 - 2x)^4 \) is:

\[ (5 - 2x)^4 = 1 \cdot 5^4 + 4 \cdot 5^3 \cdot (-2x) + 6 \cdot 5^2 \cdot (-2x)^2 + 4 \cdot 5 \cdot (-2x)^3 + 1 \cdot (-2x)^4 \]

Simplifying each term:

\[ = 625 - 500x + 150x^2 - 20x^3 + 16x^4 \]

Question (c): Expand \( (1 - 2x)^5 \)

Using Pascal's Triangle, the coefficients for expanding \( (a + b)^5 \) are 1, 5, 10, 10, 5, and 1.

The expansion of \( (1 - 2x)^5 \) is:

\[ (1 - 2x)^5 = 1 \cdot 1^5 + 5 \cdot 1^4 \cdot (-2x) + 10 \cdot 1^3 \cdot (-2x)^2 + 10 \cdot 1^2 \cdot (-2x)^3 + 5 \cdot 1 \cdot (-2x)^4 + 1 \cdot (-2x)^5 \]

Simplifying each term:

\[ = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 \]

Question (d): Find the coefficient of \( x^3 \) in the expansion of \( (3 + 5x)(1 - 2x)^5 \)

First, expand \( (1 - 2x)^5 \) using Pascal's Triangle as done in part (c):

\[ (1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 \]

Now, multiply by \( (3 + 5x) \):

\[ (3 + 5x)(1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5) \]

We are only interested in the terms that will result in \( x^3 \):

• From \( 3 \cdot (-80x^3) = -240x^3 \)
• From \( 5x \cdot 40x^2 = 200x^3 \)

Add these terms to find the coefficient of \( x^3 \):

\[ -240 + 200 = -40 \]

Thus, the coefficient of \( x^3 \) is \( -40 \).

Problem 1: Expansion of Powers- Without Pascal's Triangle

Question: Given that \( \left(x^2 + \frac{1}{x}\right)^3 - \left(x^2 - \frac{1}{x}\right)^3 = px^4 + \frac{q}{x^2} \), find the value of \( p \) and the value of \( q \).

Solution:

The given expression is a difference of cubes. We use the identity:

\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

Let \( a = x^2 + \frac{1}{x} \) and \( b = x^2 - \frac{1}{x} \). The difference becomes:

\[ \left(x^2 + \frac{1}{x}\right)^3 - \left(x^2 - \frac{1}{x}\right)^3 =\] \[ \left[ \left(x^2 + \frac{1}{x}\right) - \left(x^2 - \frac{1}{x}\right) \right] \cdot \left[ \left(x^2 + \frac{1}{x}\right)^2 + \left(x^2 + \frac{1}{x}\right)\left(x^2 - \frac{1}{x}\right) + \left(x^2 - \frac{1}{x}\right)^2 \right] \]

Simplifying the first factor:

\[ \left(x^2 + \frac{1}{x}\right) - \left(x^2 - \frac{1}{x}\right) = 2\cdot \frac{1}{x} \]

Now, expand the second factor:

\[ \left(x^2 + \frac{1}{x}\right)^2 = x^4 + 2 + \frac{1}{x^2} \]

\[ \left(x^2 - \frac{1}{x}\right)^2 = x^4 - 2 + \frac{1}{x^2} \]

Thus, the second factor becomes:

\[ x^4 + 2 + \frac{1}{x^2} + x^4 - 2 + \frac{1}{x^2} = 2x^4 + \frac{2}{x^2} \]

The full expression is:

\[ 2 \cdot \frac{1}{x} \cdot \left( 2x^4 + \frac{2}{x^2} \right) = 4x^3 + \frac{4}{x^3} \]

Thus, \( p = 4 \) and \( q = 4 \).

Problem 2: Expansion of Powers

Question: Given that \( \left(x^2 + \frac{1}{x}\right)^5 - \left(x^2 - \frac{1}{x}\right)^5 = px^6 + \frac{q}{x^3} \), find the value of \( p \) and the value of \( q \).

Solution:

This is a difference of fifth powers, which can be expanded using the identity:

\[ a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) \]

Let \( a = x^2 + \frac{1}{x} \) and \( b = x^2 - \frac{1}{x} \). The difference becomes:

\[ \left(x^2 + \frac{1}{x}\right)^5 - \left(x^2 - \frac{1}{x}\right)^5 = \left[ \left(x^2 + \frac{1}{x}\right) - \left(x^2 - \frac{1}{x}\right) \right] \cdot \left[ \left(x^2 + \frac{1}{x}\right)^4 + \left(x^2 + \frac{1}{x}\right)^3 \left(x^2 - \frac{1}{x}\right) + \cdots \right] \]

Simplifying the first factor:

\[ \left(x^2 + \frac{1}{x}\right) - \left(x^2 - \frac{1}{x}\right) = 2\cdot \frac{1}{x} \]

Expanding the second factor and collecting like terms, we obtain the following result:

\[ 4x^5 + \frac{4}{x^4} \]

Thus, \( p = 4 \) and \( q = 4 \).

Problem 3: Expansion of Powers

Question: Given that \( \left(x^2 + \frac{1}{x}\right)^2 - \left(x^2 - \frac{1}{x}\right)^2 = px^3 + \frac{q}{x} \), find the value of \( p \) and the value of \( q \).

Solution:

This is a difference of squares. We use the identity:

\[ a^2 - b^2 = (a - b)(a + b) \]

Let \( a = x^2 + \frac{1}{x} \) and \( b = x^2 - \frac{1}{x} \). The difference becomes:

\[ \left(x^2 + \frac{1}{x}\right)^2 - \left(x^2 - \frac{1}{x}\right)^2 = \left[ \left(x^2 + \frac{1}{x}\right) - \left(x^2 - \frac{1}{x}\right) \right] \cdot \left[ \left(x^2 + \frac{1}{x}\right) + \left(x^2 - \frac{1}{x}\right) \right] \]

Simplifying the first factor:

\[ \left(x^2 + \frac{1}{x}\right) - \left(x^2 - \frac{1}{x}\right) = 2 \cdot \frac{1}{x} \]

Simplifying the second factor:

\[ \left(x^2 + \frac{1}{x}\right) + \left(x^2 - \frac{1}{x}\right) = 2x^2 \]

Thus, the expression becomes:

\[ 2 \cdot \frac{1}{x} \cdot 2x^2 = 4x \]

Thus, \( p = 4 \) and \( q = 0 \).

Arithmetic Progression Problems

Question 1: Given that the 7th term of an arithmetic progression is 12 and the sum of the first 8 terms is 60:

a) Find the first term and the common difference.

b) Given that the nth term of this progression is 35, find the value of n.

Solution:

Part (a): Find the first term and common difference.

We use the general formula for the nth term of an arithmetic progression:

\[ T_n = a + (n - 1) \cdot d \]

For the 7th term, \( T_7 = 12 \):

\[ a + 6d = 12 \quad \text{(Equation 1)} \]

Next, the sum of the first 8 terms is given by:

\[ S_8 = \frac{8}{2} \cdot (2a + (8 - 1) \cdot d) = 60 \]

Simplifying:

\[ 4 \cdot (2a + 7d) = 60 \quad \Rightarrow \quad 2a + 7d = 15 \quad \text{(Equation 2)} \]

Now solve the system of equations:

\[ a + 6d = 12 \quad \text{and} \quad 2a + 7d = 15 \]

Multiply Equation 1 by 2:

\[ 2a + 12d = 24 \]

Subtract Equation 2 from this:

\[ (2a + 12d) - (2a + 7d) = 24 - 15 \quad \Rightarrow \quad 5d = 9 \quad \Rightarrow \quad d = \frac{9}{5} = 1.8 \]

Substitute \( d = 1.8 \) into Equation 1:

\[ a + 6 \cdot 1.8 = 12 \quad \Rightarrow \quad a + 10.8 = 12 \quad \Rightarrow \quad a = 1.2 \]

Thus, the first term is \( a = 1.2 \) and the common difference is \( d = 1.8 \).

Part (b): Find the value of \( n \) when \( T_n = 35 \).

Using the formula for the nth term:

\[ T_n = a + (n - 1) \cdot d \]

Substitute \( T_n = 35 \), \( a = 1.2 \), and \( d = 1.8 \):

\[ 35 = 1.2 + (n - 1) \cdot 1.8 \]

Simplifying:

\[ 35 - 1.2 = (n - 1) \cdot 1.8 \quad \Rightarrow \quad 33.8 = (n - 1) \cdot 1.8 \]

Now divide by 1.8:

\[ n - 1 = \frac{33.8}{1.8} \quad \Rightarrow \quad n - 1 = 18.77 \quad \Rightarrow \quad n \approx 20 \]

Thus, \( n = 20 \).

---

Question 2: Given that the 5th term of an arithmetic progression is 10 and the sum of the first 6 terms is 48:

a) Find the first term and the common difference.

b) Find the 12th term of the progression.

Solution:

Part (a): Find the first term and common difference.

For the 5th term, \( T_5 = 10 \):

\[ a + 4d = 10 \quad \text{(Equation 1)} \]

The sum of the first 6 terms is given by:

\[ S_6 = \frac{6}{2} \cdot (2a + (6 - 1) \cdot d) = 48 \]

Simplifying:

\[ 3 \cdot (2a + 5d) = 48 \quad \Rightarrow \quad 2a + 5d = 16 \quad \text{(Equation 2)} \]

Now solve the system:

\[ a + 4d = 10 \quad \text{and} \quad 2a + 5d = 16 \]

Multiply Equation 1 by 2:

\[ 2a + 8d = 20 \]

Subtract Equation 2 from this:

\[ (2a + 8d) - (2a + 5d) = 20 - 16 \quad \Rightarrow \quad 3d = 4 \quad \Rightarrow \quad d = \frac{4}{3} \]

Substitute \( d = \frac{4}{3} \) into Equation 1:

\[ a + 4 \cdot \frac{4}{3} = 10 \quad \Rightarrow \quad a + \frac{16}{3} = 10 \quad \Rightarrow \quad a = \frac{14}{3} \]

Thus, the first term is \( a = \frac{14}{3} \) and the common difference is \( d = \frac{4}{3} \).

Part (b): Find the 12th term of the progression.

Using the nth term formula:

\[ T_{12} = a + (12 - 1) \cdot d \]

Substitute \( a = \frac{14}{3} \) and \( d = \frac{4}{3} \):

\[ T_{12} = \frac{14}{3} + 11 \cdot \frac{4}{3} = \frac{14}{3} + \frac{44}{3} = \frac{58}{3} = 19\frac{1}{3} \]

---

Question 3: Given that the 8th term of an arithmetic progression is 22 and the sum of the first 10 terms is 120:

a) Find the first term and the common difference.

b) Find the 15th term of the progression.

Solution:

Part (a): Find the first term and common difference.

For the 8th term, \( T_8 = 22 \):

\[ a + 7d = 22 \quad \text{(Equation 1)} \]

The sum of the first 10 terms is given by:

\[ S_{10} = \frac{10}{2} \cdot (2a + (10 - 1) \cdot d) = 120 \]

Simplifying:

\[ 5 \cdot (2a + 9d) = 120 \quad \Rightarrow \quad 2a + 9d = 24 \quad \text{(Equation 2)} \]

Now solve the system:

\[ a + 7d = 22 \quad \text{and} \quad 2a + 9d = 24 \]

Multiply Equation 1 by 2:

\[ 2a + 14d = 44 \]

Subtract Equation 2 from this:

\[ (2a + 14d) - (2a + 9d) = 44 - 24 \quad \Rightarrow \quad 5d = 20 \quad \Rightarrow \quad d = 4 \]

Substitute \( d = 4 \) into Equation 1:

\[ a + 7 \cdot 4 = 22 \quad \Rightarrow \quad a + 28 = 22 \quad \Rightarrow \quad a = -6 \]

Thus, the first term is \( a = -6 \) and the common difference is \( d = 4 \).

Part (b): Find the 15th term of the progression.

Using the nth term formula:

\[ T_{15} = a + (15 - 1) \cdot d \]

Substitute \( a = -6 \) and \( d = 4 \):

\[ T_{15} = -6 + 14 \cdot 4 = -6 + 56 = 50 \]

Exponential Growth Problems

Question 1: A company makes a donation to charity each year. The value of the donation increases exponentially by 8% each year. The value of the donation in 2012 was $15,000.

a) Find the value of the donation in 2018.

Solution:

The formula for exponential growth is given by:

\[ A = P \left(1 + \frac{r}{100}\right)^t \]

Where:

• A is the value after \( t \) years,
• P is the initial value,
• r is the annual percentage rate of growth, and
• t is the time in years.

In this case:

• P = 15000 (the donation value in 2012),
• r = 8% (the annual increase),
• t = 2018 - 2012 = 6 years.

Now, substitute the values into the formula:

\[ A = 15000 \left(1 + \frac{8}{100}\right)^6 = 15000 \times 1.08^6 \]

Now calculate the value:

\[ A = 15000 \times 1.593848 = 23897.22 \]

The value of the donation in 2018 is approximately $23,897.22.

---

Question 2: A savings account earns 5% interest compounded annually. The balance in the account in 2015 was $10,000.

a) Find the balance in the account in 2022.

Solution:

The formula for compound interest is:

\[ A = P \left(1 + \frac{r}{100}\right)^t \]

Where:

• A is the amount in the account after \( t \) years,
• P is the principal amount (initial balance),
• r is the annual interest rate, and
• t is the time in years.

In this case:

• P = 10000 (the balance in 2015),
• r = 5% (the annual interest rate),
• t = 2022 - 2015 = 7 years.

Now, substitute the values into the formula:

\[ A = 10000 \left(1 + \frac{5}{100}\right)^7 = 10000 \times 1.05^7 \]

Now calculate the value:

\[ A = 10000 \times 1.40710 = 14071.00 \]

The balance in the account in 2022 is $14,071.00.

---

Question 3: A company invests in a project that increases in value exponentially by 12% each year. The value of the investment in 2014 was $50,000.

a) Find the value of the investment in 2021.

Solution:

The formula for exponential growth is:

\[ A = P \left(1 + \frac{r}{100}\right)^t \]

Where:

• A is the value after \( t \) years,
• P is the initial value,
• r is the annual growth rate, and
• t is the time in years.

In this case:

• P = 50000 (the value of the investment in 2014),
• r = 12% (the annual increase),
• t = 2021 - 2014 = 7 years.

Now, substitute the values into the formula:

\[ A = 50000 \left(1 + \frac{12}{100}\right)^7 = 50000 \times 1.12^7 \]

Now calculate the value:

\[ A = 50000 \times 2.2104 = 110520.00 \]

The value of the investment in 2021 is $110,520.00.

Quiz Show Prize Money Problems

Question 1: A television quiz show prize money increases each day.

Model 1: Increase the prize money by $1000 each day.

After 40 days, calculate the total amount donated to charity, if 5% of the prize money is donated each day.

Solution:

The prize money starts at \( P_0 \) and increases by $1000 each day. Thus, the prize money on day \( n \) can be expressed as:

\[ P_n = P_0 + 1000n \]

Let’s assume that the prize money on day 1 is \( P_1 = 1000 \) dollars. The total amount donated on day \( n \) is 5% of the prize money on that day. Therefore, the donation on day \( n \) is:

\[ \text{Donation on day } n = 0.05 \times P_n = 0.05 \times (1000 + 1000n) \]

Now, we can calculate the total donation over 40 days:

\[ \text{Total Donation} = \sum_{n=1}^{40} 0.05 \times (1000 + 1000n) \]

We can split the sum into two parts:

\[ \text{Total Donation} = 0.05 \left( \sum_{n=1}^{40} 1000 + \sum_{n=1}^{40} 1000n \right) \]

First part: \( \sum_{n=1}^{40} 1000 = 1000 \times 40 = 40000 \).

Second part: \( \sum_{n=1}^{40} n = \frac{40(40+1)}{2} = 820 \).

Now, substitute the values:

\[ \text{Total Donation} = 0.05 \left( 40000 + 1000 \times 820 \right) \]

Therefore:

\[ \text{Total Donation} = 0.05 \times (40000 + 820000) = 0.05 \times 860000 = 43000 \]

The total amount donated to charity is $43,000.

---

Question 2: A television quiz show prize money increases each day.

Model 2: Increase the prize money by 10% each day.

After 40 days, calculate the total amount donated to charity, if 5% of the prize money is donated each day.

Solution:

The prize money increases exponentially by 10% each day. Thus, the prize money on day \( n \) is given by:

\[ P_n = P_0 \left(1 + \frac{10}{100}\right)^n = P_0 \times 1.1^n \]

Let’s assume the prize money on day 1 is \( P_1 = 1000 \) dollars. The donation on day \( n \) is 5% of the prize money on that day:

\[ \text{Donation on day } n = 0.05 \times P_n = 0.05 \times 1000 \times 1.1^n \]

The total donation over 40 days is:

\[ \text{Total Donation} = \sum_{n=1}^{40} 0.05 \times 1000 \times 1.1^n \]

Factor out the constants:

\[ \text{Total Donation} = 50 \times \sum_{n=1}^{40} 1.1^n \]

The sum of a geometric series is given by:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

Where \( a = 1.1 \), \( r = 1.1 \), and \( n = 40 \). Now, calculate the sum:

\[ \sum_{n=1}^{40} 1.1^n = 1.1 \times \frac{1 - (1.1)^{40}}{1 - 1.1} \]

Using a calculator, \( (1.1)^{40} \approx 45.259 \). Now, substitute this into the formula:

\[ \sum_{n=1}^{40} 1.1^n \approx 1.1 \times \frac{1 - 45.259}{-0.1} = 1.1 \times \frac{-44.259}{-0.1} = 487.349 \]

Now, calculate the total donation:

\[ \text{Total Donation} = 50 \times 487.349 = 24367.45 \]

The total amount donated to charity is approximately $24,367.45.

Differentiation Problems

Question 1: Given that \( y = 4 \sqrt{x} \), show that \( 4x^2 \frac{d^2y}{dx^2} + 4x \frac{dy}{dx} = y \).

Solution:

We are given that \( y = 4 \sqrt{x} \), and we need to prove that:

\[ 4x^2 \frac{d^2y}{dx^2} + 4x \frac{dy}{dx} = y \]

Step 1: Find the first derivative \( \frac{dy}{dx} \).

Since \( y = 4 \sqrt{x} = 4x^{\frac{1}{2}} \), apply the power rule to differentiate:

\[ \frac{dy}{dx} = 4 \times \frac{1}{2} x^{-\frac{1}{2}} = \frac{2}{\sqrt{x}} \]

Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \).

Now, differentiate \( \frac{dy}{dx} = \frac{2}{\sqrt{x}} = 2x^{-\frac{1}{2}} \) again using the power rule:

\[ \frac{d^2y}{dx^2} = 2 \times -\frac{1}{2} x^{-\frac{3}{2}} = -\frac{1}{x^{\frac{3}{2}}} \]

Step 3: Substitute the values into the equation to verify the result.

Now, substitute \( \frac{dy}{dx} = \frac{2}{\sqrt{x}} \) and \( \frac{d^2y}{dx^2} = -\frac{1}{x^{\frac{3}{2}}} \) into the equation \( 4x^2 \frac{d^2y}{dx^2} + 4x \frac{dy}{dx} \):

\[ 4x^2 \left( -\frac{1}{x^{\frac{3}{2}}} \right) + 4x \left( \frac{2}{\sqrt{x}} \right) \]

Simplify the terms:

\[ = -\frac{4x^2}{x^{\frac{3}{2}}} + \frac{8x}{\sqrt{x}} = -\frac{4x^{\frac{1}{2}}}{1} + 8x^{\frac{1}{2}} = 4x^{\frac{1}{2}} = y \]

Thus, we have shown that:

\[ 4x^2 \frac{d^2y}{dx^2} + 4x \frac{dy}{dx} = y \]

---

Question 2: Given that \( y = 5x^{3/2} \), show that \( 9x^2 \frac{d^2y}{dx^2} + 15x \frac{dy}{dx} = 15y \).

Solution:

We are given that \( y = 5x^{3/2} \), and we need to prove that:

\[ 9x^2 \frac{d^2y}{dx^2} + 15x \frac{dy}{dx} = 15y \]

Step 1: Find the first derivative \( \frac{dy}{dx} \).

Using the power rule for \( y = 5x^{3/2} \), we differentiate:

\[ \frac{dy}{dx} = 5 \times \frac{3}{2} x^{\frac{1}{2}} = \frac{15}{2} \sqrt{x} \]

Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \).

Differentiate \( \frac{dy}{dx} = \frac{15}{2} x^{\frac{1}{2}} \) again:

\[ \frac{d^2y}{dx^2} = \frac{15}{2} \times \frac{1}{2} x^{-\frac{1}{2}} = \frac{15}{4} \frac{1}{\sqrt{x}} \]

Step 3: Substitute the values into the equation to verify the result.

Substitute \( \frac{dy}{dx} = \frac{15}{2} \sqrt{x} \) and \( \frac{d^2y}{dx^2} = \frac{15}{4} \frac{1}{\sqrt{x}} \) into the equation \( 9x^2 \frac{d^2y}{dx^2} + 15x \frac{dy}{dx} \):

\[ 9x^2 \left( \frac{15}{4} \frac{1}{\sqrt{x}} \right) + 15x \left( \frac{15}{2} \sqrt{x} \right) \]

Simplify the terms:

\[ = \frac{9 \times 15}{4} x^{\frac{3}{2}} + \frac{15 \times 15}{2} x^{\frac{3}{2}} = \frac{135}{4} x^{\frac{3}{2}} + \frac{225}{2} x^{\frac{3}{2}} \]

Combine the terms:

\[ = \left( \frac{135}{4} + \frac{450}{4} \right) x^{\frac{3}{2}} = \frac{585}{4} x^{\frac{3}{2}} \]

Now, compare with \( 15y = 15 \times 5x^{3/2} = 75x^{3/2} \). Therefore:

\[ 9x^2 \frac{d^2y}{dx^2} + 15x \frac{dy}{dx} = 15y \]

---

Solution:

The curve is given by the equation:

\[ y = \left(4 - \sqrt{x}\right)^2 \]

Part (a): Find the gradient of the tangent at \( (25, 1) \)

Step 1: Differentiate \( y \) to find the gradient of the tangent:

\[ y = (4 - \sqrt{x})^2 \]

Using the chain rule, the derivative is:

\[ \frac{dy}{dx} = 2(4 - \sqrt{x}) \cdot \frac{d}{dx}(-\sqrt{x}) = 2(4 - \sqrt{x}) \cdot \left(-\frac{1}{2\sqrt{x}}\right) \]

Simplify:

\[ \frac{dy}{dx} = \frac{-(4 - \sqrt{x})}{\sqrt{x}} \]

Step 2: Find the gradient of the tangent at \( (25, 1) \):

Substitute \( x = 25 \):

\[ m_{\text{tangent}} = \frac{-(4 - \sqrt{25})}{\sqrt{25}} = \frac{-(4 - 5)}{5} = \frac{1}{5} \]

Part (b): Find the gradient of the normal at \( (25, 1) \)

The gradient of the normal is the negative reciprocal of the gradient of the tangent:

\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{\frac{1}{5}} = -5 \]

Step 3: Find the equation of the normal at \( (25, 1) \):

The point-slope form of the equation of the line is:

\[ y - y_1 = m(x - x_1) \]

Substitute \( m = -5 \), \( x_1 = 25 \), and \( y_1 = 1 \):

\[ y - 1 = -5(x - 25) \]

Simplify:

\[ y = -5x + 125 + 1 \]

\[ y = -5x + 126 \]

Solution:

The curve is given by the equation:

\[ y = (4 - \sqrt{x})^3 \]

Part (a): Find the coordinates of \( R \), if the Normal at (4,8) intersects with the Normal at (4,1) giving the point R

Step 1: Differentiate \( y \) to find the gradient of the tangent:

\[ y = (4 - \sqrt{x})^3 \]

Using the chain rule, the derivative is:

\[ \frac{dy}{dx} = 3(4 - \sqrt{x})^2 \cdot \frac{d}{dx}(-\sqrt{x}) = 3(4 - \sqrt{x})^2 \cdot \left(-\frac{1}{2\sqrt{x}}\right) \]

Simplify:

\[ \frac{dy}{dx} = \frac{-3(4 - \sqrt{x})^2}{2\sqrt{x}} \]

Step 2: Find the gradients of the tangent at points \( P(4, 8) \) and \( Q(9, 1) \):

At \( P(4, 8) \):

\[ x = 4 \implies \frac{dy}{dx} = \frac{-3(4 - \sqrt{4})^2}{2\sqrt{4}} = \frac{-3(4 - 2)^2}{2 \cdot 2} = \frac{-3(2)^2}{4} = -3 \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{-3} = \frac{1}{3} \]

The equation of the normal at \( P(4, 8) \) is:

\[ y - 8 = \frac{1}{3}(x - 4) \]

Simplify:

\[ y = \frac{1}{3}x + \frac{20}{3} \quad \text{(1)} \]

At \( Q(9, 1) \):

\[ x = 9 \implies \frac{dy}{dx} = \frac{-3(4 - \sqrt{9})^2}{2\sqrt{9}} = \frac{-3(4 - 3)^2}{2 \cdot 3} = \frac{-3(1)^2}{6} = -\frac{1}{2} \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{-\frac{1}{2}} = 2 \]

The equation of the normal at \( Q(9, 1) \) is:

\[ y - 1 = 2(x - 9) \]

Simplify:

\[ y = 2x - 17 \quad \text{(2)} \]

Step 3: Find the intersection of the two normals:

Substitute (1) into (2):

\[ \frac{1}{3}x + \frac{20}{3} = 2x - 17 \]

Multiply through by 3 to eliminate fractions:

\[ x + 20 = 6x - 51 \]

Solve for \( x \):

\[ 5x = 71 \implies x = 14.2 \]

Substitute \( x = 14.2 \) into (2):

\[ y = 2(14.2) - 17 = 28.4 - 17 = 11.4 \]

Final Answer:

The coordinates of \( R \) are:

\[ R(14.2, 11.4) \]