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 \( -40 \).

Solution:

The curve is given by the equation:

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

Part (a): Find the gradient of the tangent at \( (16, 8) \)

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

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

Using the chain rule, the derivative is:

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

Simplify:

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

Step 2: Find the gradient of the tangent at \( (16, 8) \):

Substitute \( x = 16 \):

\[ m_{\text{tangent}} = \frac{-(2 - \sqrt{16})}{\sqrt{16}} = \frac{-(2 - 4)}{4} = \frac{2}{4} = \frac{1}{2} \]

Part (b): Find the gradient of the normal at \( (16, 8) \)

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

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

Step 3: Find the equation of the normal at \( (16, 8) \):

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

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

Substitute \( m = -2 \), \( x_1 = 16 \), and \( y_1 = 8 \):

\[ y - 8 = -2(x - 16) \]

Simplify:

\[ y = -2x + 32 + 8 \]

\[ y = -2x + 40 \]

Solution:

The curve is given by the equation:

\[ y = \left(3 - \sqrt{x}\right)^2 \]

Part (a): Find the gradient of the tangent at \( (9, 4) \)

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

\[ y = (3 - \sqrt{x})^2 \]

Using the chain rule, the derivative is:

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

Simplify:

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

Step 2: Find the gradient of the tangent at \( (9, 4) \):

Substitute \( x = 9 \):

\[ m_{\text{tangent}} = \frac{-(3 - \sqrt{9})}{\sqrt{9}} = \frac{-(3 - 3)}{3} = 0 \]

Part (b): Find the gradient of the normal at \( (9, 4) \)

The gradient of the normal is the negative reciprocal of the gradient of the tangent. Since the tangent is horizontal:

\[ m_{\text{normal}} = \infty \]

Step 3: Find the equation of the normal at \( (9, 4) \):

The equation of the vertical line passing through \( (9, 4) \) is:

\[ x = 9 \]

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:

\[ R(14.2, 11.4) \]

Solution:

The curve is given by the equation:

\[ y = \left(5 - \sqrt{x}\right)^3 \]

Part (a): Find the coordinates of \( R \)

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

\[ y = (5 - \sqrt{x})^3 \]

Using the chain rule, the derivative is:

\[ \frac{dy}{dx} = 3(5 - \sqrt{x})^2 \cdot \frac{d}{dx}(-\sqrt{x}) = 3(5 - \sqrt{x})^2 \cdot \left(-\frac{1}{2\sqrt{x}}\right) \]

Simplify:

\[ \frac{dy}{dx} = \frac{-3(5 - \sqrt{x})^2}{2\sqrt{x}} \]

Step 2: Find the gradients of the tangent at points \( P(4, 27) \) and \( Q(16, 1) \):

At \( P(4, 27) \):

\[ x = 4 \implies \frac{dy}{dx} = \frac{-3(5 - \sqrt{4})^2}{2\sqrt{4}} = \frac{-3(5 - 2)^2}{2 \cdot 2} = \frac{-3(3)^2}{4} = \frac{-27}{4} \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{\frac{-27}{4}} = \frac{4}{27} \]

The equation of the normal at \( P(4, 27) \) is:

\[ y - 27 = \frac{4}{27}(x - 4) \]

Simplify:

\[ y = \frac{4}{27}x + \left(27 - \frac{16}{27}\right) = \frac{4}{27}x + \frac{729 - 16}{27} \]

\[ y = \frac{4}{27}x + \frac{713}{27} \quad \text{(1)} \]

At \( Q(16, 1) \):

\[ x = 16 \implies \frac{dy}{dx} = \frac{-3(5 - \sqrt{16})^2}{2\sqrt{16}} = \frac{-3(5 - 4)^2}{2 \cdot 4} = \frac{-3(1)^2}{8} = -\frac{3}{8} \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{-\frac{3}{8}} = \frac{8}{3} \]

The equation of the normal at \( Q(16, 1) \) is:

\[ y - 1 = \frac{8}{3}(x - 16) \]

Simplify:

\[ y = \frac{8}{3}x - \frac{128}{3} + 1 = \frac{8}{3}x - \frac{125}{3} \quad \text{(2)} \]

Step 3: Find the intersection of the two normals:

Substitute (1) into (2):

\[ \frac{4}{27}x + \frac{713}{27} = \frac{8}{3}x - \frac{125}{3} \]

Multiply through by 27 to eliminate fractions:

\[ 4x + 713 = 72x - 1125 \]

Solve for \( x \):

\[ 68x = 1838 \implies x = \frac{1838}{68} = 27 \]

Substitute \( x = 27 \) into (1):

\[ y = \frac{4}{27}(27) + \frac{713}{27} = 4 + \frac{713}{27} = \frac{721}{27} \]

Final Answer:

The coordinates of \( R \) are:

\[ R\left(27, \frac{721}{27}\right) \]

Solution:

The curve is given by the equation:

\[ y = 5 - 3x - x^2 \]

Find the coordinates of \( Q \)

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

\[ \frac{dy}{dx} = -3 - 2x \]

Step 2: Calculate the gradient of the tangents at the points \( (-1, 7) \) and \( (-4, 1) \):

At \( (-1, 7) \):

\[ m_1 = -3 - 2(-1) = -3 + 2 = -1 \]

The equation of the tangent at \( (-1, 7) \) is:

\[ y - 7 = -1(x + 1) \]

Simplify:

\[ y = -x + 6 \quad \text{(1)} \]

At \( (-4, 1) \):

\[ m_2 = -3 - 2(-4) = -3 + 8 = 5 \]

The equation of the tangent at \( (-4, 1) \) is:

\[ y - 1 = 5(x + 4) \]

Simplify:

\[ y = 5x + 21 \quad \text{(2)} \]

Step 3: Solve for the intersection of the two tangents:

Equating (1) and (2):

\[ -x + 6 = 5x + 21 \]

Solve for \( x \):

\[ 6x = -15 \implies x = -2.5 \]

Substitute \( x = -2.5 \) into (1):

\[ y = -(-2.5) + 6 = 2.5 + 6 = 8.5 \]

Final Answer:

The coordinates of \( Q \) are:

\[ Q(-2.5, 8.5) \]

Solution:

The curve is given by the equation:

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

Part (a): Find the equation of the normal \( PQ \)

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

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

At \( P(16, -4) \):

\[ m_{\text{tangent}} = -\frac{1}{\sqrt{16}} = -\frac{1}{4} \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{-\frac{1}{4}} = 4 \]

The equation of the normal is:

\[ y - (-4) = 4(x - 16) \]

Simplify:

\[ y + 4 = 4x - 64 \implies y = 4x - 68 \]

Part (b): Find the coordinates of \( Q \)

At \( Q \), the normal meets the x-axis (\( y = 0 \)). Substituting \( y = 0 \):

\[ 0 = 4x - 68 \implies x = 17 \]

So, the coordinates of \( Q \) are:

\[ Q(17, 0) \]

Solution:

The curve is given by the equation:

\[ y = 2x - \frac{10}{x^2} + 8 \]

Part (a): Find \(\frac{dy}{dx}\)

Differentiate term by term:

\[ \frac{dy}{dx} = 2 - \frac{d}{dx}\left(\frac{10}{x^2}\right) \]

Using the power rule:

\[ \frac{dy}{dx} = 2 + 20x^{-3} = 2 + \frac{20}{x^3} \]

Part (b): Show that the normal passes through \( (0, -3) \)

At \( \left(-4, -\frac{5}{8}\right) \):

\[ m_{\text{tangent}} = 2 + \frac{20}{(-4)^3} = 2 + \frac{20}{-64} = 2 - \frac{5}{16} = \frac{32}{16} - \frac{5}{16} = \frac{27}{16} \]

The gradient of the normal is the negative reciprocal:

\[ m_{\text{normal}} = -\frac{1}{\frac{27}{16}} = -\frac{16}{27} \]

The equation of the normal is:

\[ y - \left(-\frac{5}{8}\right) = -\frac{16}{27}(x - (-4)) \]

Simplify:

\[ y + \frac{5}{8} = -\frac{16}{27}(x + 4) \]

Substitute \( x = 0 \):

\[ y + \frac{5}{8} = -\frac{16}{27}(4) = -\frac{64}{27} \]

Simplify further:

\[ y = -\frac{64}{27} - \frac{5}{8} \]

Converting to a common denominator:

\[ y = -3 \implies \text{Passes through } (0, -3) \]

Solution:

Differentiate the given equation w.r.t x:

\[ y = \frac{3x^5 - 7}{4x} \]

Step 1: Rewrite the equation:

\[ y = \frac{3x^5}{4x} - \frac{7}{4x} = \frac{3x^4}{4} - \frac{7}{4x} \]

Step 2: Differentiate term by term:

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{3x^4}{4}\right) - \frac{d}{dx}\left(\frac{7}{4x}\right) \]

\[ \frac{dy}{dx} = \frac{3}{4} \cdot 4x^3 - \frac{7}{4} \cdot \frac{-1}{x^2} \]

\[ \frac{dy}{dx} = 3x^3 + \frac{7}{4x^2} \]

Final Answer:

\[ \frac{dy}{dx} = 3x^3 + \frac{7}{4x^2} \]

Solution:

Find the Gradient of the curve, at x = 2:

\[ y = \frac{8}{4x - 5} \]

Step 1: Differentiate using the chain rule:

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{8}{u}\right) \cdot \frac{du}{dx}, \quad \text{where } u = 4x - 5 \]

\[ \frac{dy}{dx} = \frac{-8}{(4x - 5)^2} \cdot \frac{d}{dx}(4x - 5) \]

\[ \frac{dy}{dx} = \frac{-8}{(4x - 5)^2} \cdot 4 = \frac{-32}{(4x - 5)^2} \]

Step 2: Find the gradient at \( x = 2 \):

Substitute \( x = 2 \):

\[ \frac{dy}{dx} = \frac{-32}{(4(2) - 5)^2} = \frac{-32}{(8 - 5)^2} = \frac{-32}{9} \]

Final Answer:

The gradient at \( x = 2 \) is:

\[ \frac{-32}{9} \]

Solution:

\[ y = 3x^3 - 3x^2 + x - 7 \]

Part (a): Show that the gradient of the curve is never negative.

Step 1: Differentiate \( y \):

\[ \frac{dy}{dx} = 9x^2 - 6x + 1 \]

Step 2: Analyze the quadratic expression \( 9x^2 - 6x + 1 \):

The discriminant of a quadratic equation \( ax^2 + bx + c \) is given by:

\[ \Delta = b^2 - 4ac \]

Here, \( a = 9 \), \( b = -6 \), and \( c = 1 \):

\[ \Delta = (-6)^2 - 4(9)(1) = 36 - 36 = 0 \]

Since the discriminant is \( 0 \), the quadratic has a single root, and the parabola opens upwards (as \( a > 0 \)).

Step 3: Verify that \( \frac{dy}{dx} \geq 0 \):

The vertex of the parabola occurs at:

\[ x = -\frac{b}{2a} = -\frac{-6}{2(9)} = \frac{1}{3} \]

At the vertex, \( \frac{dy}{dx} = 0 \). For all other values of \( x \), \( \frac{dy}{dx} > 0 \) because the parabola opens upwards.

Final Answer:

The gradient of the curve is never negative.

Reference-https://mathz.org/Alevels/week12/week12.html

Differentiation Practice Worksheet

Below are 15 questions on differentiation. Solve step by step and show your working:

Part A: Basic Differentiation

Part B: Applications of Differentiation

Part C: Advanced Differentiation

Part D: Mixed Problems

End of Chapter Worsheet:

Binomial Expansion