Grade 11 Maths Problems with Solutions

Step 1: Identify restrictions. The denominators cannot be zero, so \( x \ne 2 \) and \( x \ne -4 \).

Step 2: Cross-multiply. To clear the fractions, multiply the numerator of each side by the denominator of the other side:
\[ (x+2)(x+4) = (3x - 4)(x - 2) \]

Step 3: Expand both sides. Use the FOIL method:
\[ x^2 + 4x + 2x + 8 = 3x^2 - 6x - 4x + 8 \]
\[ x^2 + 6x + 8 = 3x^2 - 10x + 8 \]

Step 4: Group terms to form a standard quadratic equation. Subtract \( x^2 + 6x + 8 \) from both sides:
\[ 0 = 2x^2 - 16x \]

Step 5: Factor and solve. Factor out the common term \( 2x \):
\[ 2x(x - 8) = 0 \]

Setting each factor to zero gives \( x = 0 \) and \( x = 8 \). Both are valid since neither equals our restricted values.

Final solutions: { 0 , 8 }.

Step 1: Use the Rational Root Theorem. The possible rational roots are the ratio of the factors of the constant term (10) to the factors of the leading coefficient (1).

Factors of the constant term (\(p\)): \(\pm 1, \pm 2, \pm 5, \pm 10\)
Factors of the leading coefficient (\(q\)): \(\pm 1\)

List of possible roots (\(\frac{p}{q}\)): \(\pm 1, \pm 2, \pm 5, \pm 10\).

List of possible factors: \( \pm 1, \pm 2, \pm 5, \pm 10 \).

Let's test \( x = 1 \):
\( P(1) = (1)^3 - 4(1)^2 - 7(1) + 10 = 1 - 4 - 7 + 10 = 0 \).
Since \( P(1) = 0 \), then \( (x - 1) \) is a factor of the polynomial.

Step 2: Perform polynomial division. Divide \( P(x) \) by \( (x - 1) \) using synthetic or long division:
\[ \frac{x^3 - 4x^2 - 7x + 10}{x-1} = x^2 - 3x - 10 \]

Step 3: Factor the resulting quadratic. Find two numbers that multiply to -10 and add to -3. Those numbers are -5 and 2:
\[ x^2 - 3x - 10 = (x - 5)(x + 2) \]

Step 4: Combine all factors.
Factorization: \( (x - 1)(x - 5)(x + 2) \).

Step 1: Isolate the exponential term. Divide both sides by 6:
\[ 3^x = \frac{162}{6} \]
\[ 3^x = 27 \]

Step 2: Express both sides with the same base. Recognize that 27 is a power of 3:
\[ 3^x = 3^3 \]

Step 3: Equate the exponents. Since the bases are equal, the exponents must be equal.
x = 3.

Step 1: Recall the formula for an arithmetic sequence.
\[ a_n = a + (n-1)d \]

Step 2: Find the common difference (\( d \)). Substitute the known values for the 15th term:
\[ -35 = 7 + (15 - 1)d \]
\[ -35 = 7 + 14d \]
\[ -42 = 14d \Rightarrow d = -3 \]

Step 3: Calculate the 30th term. Use the formula again with \( n = 30 \) and \( d = -3 \):
\[ a_{30} = 7 + (30 - 1)(-3) \]
\[ a_{30} = 7 + 29(-3) = 7 - 87 = \mathbf{-80} \].

Step 1: Translate the word problem into an equation.
\[ N + \frac{1}{N} = \frac{78}{15} \]

Step 2: Clear the fractions. Multiply the entire equation by the common denominator, \( 15N \):
\[ 15N(N) + 15N\left(\frac{1}{N}\right) = 15N\left(\frac{78}{15}\right) \]
\[ 15N^2 + 15 = 78N \]

Step 3: Simplify and set to zero. Move all terms to one side:
\[ 15N^2 - 78N + 15 = 0 \]
Divide by 3 to simplify:
\[ 5N^2 - 26N + 5 = 0 \]

Step 4: Factor the quadratic.
\[ (5N - 1)(N - 5) = 0 \]

Step 5: Solve. This yields \( N = 1/5 \) and \( N = 5 \). Since the question asks for an integer, we discard the fraction.
Solution: N = 5.

Step 1: Cross-multiply to remove fractions.
\[ 128 \cdot 4^m = 125 \cdot 5^n \]

Step 2: Express all numbers as prime bases. We know that \( 128 = 2^7 \), \( 4 = 2^2 \), and \( 125 = 5^3 \).
\[ 2^7 \cdot (2^2)^m = 5^3 \cdot 5^n \]

Step 3: Apply exponent rules. Combine the terms with base 2 and base 5:
\[ 2^{7 + 2m} = 5^{3 + n} \]

Step 4: Solve for m and n. Because 2 and 5 are distinct prime numbers, the only way a power of 2 can equal a power of 5 is if both evaluate to 1 (i.e., \( 2^0 = 5^0 \)). Therefore, both exponents must equal zero:
\[ 7 + 2m = 0 \Rightarrow 2m = -7 \Rightarrow \mathbf{m = -3.5} \]
\[ 3 + n = 0 \Rightarrow \mathbf{n = -3} \]

Step 1: Determine the domain restrictions. Logarithms are only defined for positive arguments, so \( x > 0 \) and \( x - 2 > 0 \). Combining these, our solution must satisfy \( x > 2 \).

Step 2: Condense the logarithms. Use the product rule of logarithms (\( \log(a) + \log(b) = \log(ab) \)):
\[ \log_2[x(x - 2)] = 3 \]

Step 3: Convert to exponential form. By definition of a logarithm:
\[ x(x - 2) = 2^3 \]
\[ x^2 - 2x = 8 \]

Step 4: Solve the quadratic equation.
\[ x^2 - 2x - 8 = 0 \]
\[ (x - 4)(x + 2) = 0 \]

This gives potential solutions \( x = 4 \) and \( x = -2 \). Checking our domain restriction (\( x > 2 \)), we see \( -2 \) is an extraneous solution.
Final Solution: x = 4.

Step 1: Recognize repeated addition. Adding three identical terms is the same as multiplying that term by 3:
\[ 3 \cdot \left(3^{n + 4001}\right) \]

Step 2: Apply exponent rules. The number 3 can be written as \( 3^1 \). When multiplying terms with the same base, you add the exponents:
\[ 3^1 \cdot 3^{n + 4001} = 3^{1 + n + 4001} \]

Simplified form: \( 3^{n + 4002} \)

Step 1: Replace \( f(x) \) with \( y \).
\[ y = \frac{2x - 1}{x + 3} \]

Step 2: Swap \( x \) and \( y \). This is the core step to finding an inverse function:
\[ x = \frac{2y - 1}{y + 3} \]

Step 3: Isolate the new \( y \). Multiply both sides by \( y + 3 \):
\[ x(y + 3) = 2y - 1 \]
\[ xy + 3x = 2y - 1 \]

Move all terms containing \( y \) to one side, and all other terms to the other side:
\[ xy - 2y = -3x - 1 \]

Factor out \( y \):
\[ y(x - 2) = -3x - 1 \]

Divide by \( x - 2 \):
\[ y = \frac{-3x - 1}{x - 2} \]

Inverse function: \( f^{-1}(x) = \frac{-3x - 1}{x - 2} \).

Step 1: Set the equations equal to each other. Since both equations are equal to \( y \):
\[ x^2 - 4x + 3 = 2x - 1 \]

Step 2: Simplify into standard quadratic form. Move all terms to the left side:
\[ x^2 - 6x + 4 = 0 \]

Step 3: Solve using the quadratic formula. (\( a=1, b=-6, c=4 \))
\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)} \]
\[ x = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2} \]

Simplify the radical (\( \sqrt{20} = 2\sqrt{5} \)):
\[ x = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5} \]

Step 4: Find the corresponding y-values. Plug the x-values back into the simpler linear equation \( y = 2x - 1 \):
If \( x = 3 + \sqrt{5} \): \( y = 2(3 + \sqrt{5}) - 1 = 6 + 2\sqrt{5} - 1 = 5 + 2\sqrt{5} \).
If \( x = 3 - \sqrt{5} \): \( y = 2(3 - \sqrt{5}) - 1 = 6 - 2\sqrt{5} - 1 = 5 - 2\sqrt{5} \).

Solutions: \( (3 + \sqrt{5}, 5 + 2\sqrt{5}) \) and \( (3 - \sqrt{5}, 5 - 2\sqrt{5}) \).

Step 1: Write the vertex form of a parabola.
\[ y = a(x - h)^2 + k \]
Substitute the vertex \( (h, k) = (2, -1) \):
\[ y = a(x - 2)^2 - 1 \]

Step 2: Find the vertical stretch factor (\( a \)). Substitute the known point \( (x, y) = (4, 7) \) into the equation:
\[ 7 = a(4 - 2)^2 - 1 \]
\[ 7 = a(2)^2 - 1 \]
\[ 7 = 4a - 1 \]
\[ 8 = 4a \Rightarrow a = 2 \]

Step 3: Write the final equation.
\( y = 2(x - 2)^2 - 1 \).

Step 1: Set up the perimeter and area equations. Let \( x \) be the width (two sides perpendicular to the wall) and \( y \) be the length (the side parallel to the wall).
Total fencing: \( 2x + y = 100 \Rightarrow y = 100 - 2x \).
Area: \( A = x \cdot y \).

Step 2: Express Area as a function of \( x \). Substitute \( y \) into the Area equation:
\[ A(x) = x(100 - 2x) = 100x - 2x^2 \]

Step 3: Find the maximum. This is a downward-opening parabola (\( a = -2 \)). The maximum occurs at the vertex, where \( x = \frac{-b}{2a} \):
\[ x = \frac{-100}{2(-2)} = \frac{-100}{-4} = 25 \]

Step 4: Find the remaining dimensions and area.
If \( x = 25 \), then \( y = 100 - 2(25) = 50 \).
Dimensions: 25m by 50m.
Max Area: \( 25 \cdot 50 = \) 1250m².

Step 1: Factor the numerator and denominator.
Numerator: \( x^2 - 4x + 3 = (x - 3)(x - 1) \)
Denominator (difference of squares): \( x^2 - 9 = (x - 3)(x + 3) \)
\[ f(x) = \frac{(x - 3)(x - 1)}{(x - 3)(x + 3)} \]

Step 2: Identify Holes. The factor \( (x - 3) \) cancels out, meaning there is a hole at \( x = 3 \).
To find the y-coordinate of the hole, plug \( x = 3 \) into the simplified function \( \frac{x - 1}{x + 3} \):
\( y = \frac{3 - 1}{3 + 3} = \frac{2}{6} = \frac{1}{3} \).
Hole is at (3, 1/3).

Step 3: Identify Vertical Asymptotes (VA). Set the remaining denominator to zero:
\( x + 3 = 0 \Rightarrow \) VA at x = -3.

Step 4: Identify Horizontal Asymptotes (HA). Since the degree of the numerator (2) equals the degree of the denominator (2), the HA is the ratio of their leading coefficients:
\( y = \frac{1x^2}{1x^2} \Rightarrow \) HA at y = 1.

Step 1: Find the slope of the second line (\( m_2 \)).
\[ m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 3}{4 - (-2)} = \frac{-9}{6} = -1.5 \]

Step 2: Determine the required slope for the first line (\( m_1 \)). Perpendicular lines have negative reciprocal slopes (\( m_1 \cdot m_2 = -1 \)).
\[ m_1 = \frac{-1}{-1.5} = \frac{2}{3} \]

Step 3: Set up the slope equation for the first line to find \( k \).
\[ \frac{7 - 2}{-5 - k} = \frac{2}{3} \]
\[ \frac{5}{-5 - k} = \frac{2}{3} \]

Step 4: Cross-multiply and solve.
\( 15 = 2(-5 - k) \)
\( 15 = -10 - 2k \)
\( 25 = -2k \Rightarrow \) k = -12.5.

Step 1: Substitute the parabola equation into the circle equation. From the parabola, we can isolate \( x^2 \):
\( x^2 = y + 4 \)

Substitute \( (y+4) \) for \( x^2 \) in the circle equation:
\[ (y + 4) + y^2 = 25 \]

Step 2: Solve the resulting quadratic equation.
\[ y^2 + y - 21 = 0 \]

Use the quadratic formula:
\[ y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-21)}}{2} = \frac{-1 \pm \sqrt{85}}{2} \]

Step 3: Find the corresponding x-values. Plug these y-values back into \( x^2 = y + 4 \):
\[ x^2 = \frac{-1 \pm \sqrt{85}}{2} + \frac{8}{2} = \frac{7 \pm \sqrt{85}}{2} \]
\[ x = \pm \sqrt{\frac{7 \pm \sqrt{85}}{2}} \]

Note: \( 7 - \sqrt{85} \) yields a negative number, and since \( x^2 \) must be positive, we reject the minus case inside the root.
Points of intersection:
\( \left( \sqrt{\frac{7 + \sqrt{85}}{2}}, \frac{-1 + \sqrt{85}}{2} \right) \) and \( \left( -\sqrt{\frac{7 + \sqrt{85}}{2}}, \frac{-1 + \sqrt{85}}{2} \right) \).

Step 1: Find the radius of the first circle by completing the square.
\( (x^2 - 2x + 1) + (y^2 + 4y + 4) = -1 + 1 + 4 \)
\( (x - 1)^2 + (y + 2)^2 = 4 \).
The radius squared is \( r_1^2 = 4 \), so \( r_1 = 2 \).

Step 2: Find the radius of the second circle.
\( (x^2 - 2x + 1) + (y^2 + 4y + 4) = 11 + 1 + 4 \)
\( (x - 1)^2 + (y + 2)^2 = 16 \).
The radius squared is \( r_2^2 = 16 \), so \( r_2 = 4 \).

Step 3: Subtract the smaller area from the larger area.
Area = \( \pi(r_2)^2 - \pi(r_1)^2 \)
Area = \( 16\pi - 4\pi = \mathbf{12\pi} \).

Step 1: Isolate y in the line equation and substitute.
\( y = r - x \).
Substitute into the circle:
\[ x^2 + (r - x)^2 = 4 \]
\[ x^2 + r^2 - 2rx + x^2 = 4 \]
\[ 2x^2 - 2rx + (r^2 - 4) = 0 \]

Step 2: Use the discriminant condition for a tangent. A line is tangent to a circle if they intersect at exactly one point, meaning the discriminant (\( \Delta = b^2 - 4ac \)) of this quadratic must be zero.
Here, \( a = 2 \), \( b = -2r \), \( c = r^2 - 4 \).
\[ (-2r)^2 - 4(2)(r^2 - 4) = 0 \]
\[ 4r^2 - 8r^2 + 32 = 0 \]
\[ -4r^2 = -32 \Rightarrow r^2 = 8 \]

Result: \( r = \pm \sqrt{8} = \pm 2\sqrt{2} \).

Step 1: Set up two right triangles. Let \( h = 80\text{m} \) be the height of the cliff. The angles of depression are equal to the angles of elevation from the boats to the top of the cliff.

Step 2: Find the distance to the closer boat (\( d_1 \)).
Using tangent (opposite/adjacent):
\( \tan(28^\circ) = \frac{80}{d_1} \Rightarrow d_1 = \frac{80}{\tan(28^\circ)} \approx 150.44\text{m} \)

Step 3: Find the distance to the further boat (\( d_2 \)).
\( \tan(18^\circ) = \frac{80}{d_2} \Rightarrow d_2 = \frac{80}{\tan(18^\circ)} \approx 246.30\text{m} \)

Step 4: Find the difference.
\[ \text{Distance apart} = d_2 - d_1 \approx 246.30 - 150.44 = \mathbf{95.86 \text{ m}} \]

Step 1: Use the Pythagorean identity. Substitute \( \cos^2 x = 1 - \sin^2 x \) to get the equation entirely in terms of sine:
\[ 2(1 - \sin^2 x) + 3\sin x = 3 \]
\[ 2 - 2\sin^2 x + 3\sin x = 3 \]

Step 2: Rearrange into a standard quadratic equation. Move all terms to the right side to keep the squared term positive:
\[ 0 = 2\sin^2 x - 3\sin x + 1 \]

Step 3: Factor the quadratic. Treat \( \sin x \) like a variable \( u \) (\( 2u^2 - 3u + 1 = 0 \)):
\[ (2\sin x - 1)(\sin x - 1) = 0 \]

Step 4: Solve for x.
Case 1: \( 2\sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2} \). The reference angle is 30°. Sine is positive in Quadrants I and II, so \( x = 30^\circ \) and \( x = 180^\circ - 30^\circ = 150^\circ \).
Case 2: \( \sin x - 1 = 0 \Rightarrow \sin x = 1 \). This occurs at \( x = 90^\circ \).

Solutions: 30°, 90°, 150°.