Grade 11 Maths Problems with Solutions

We need to exclude values that make the denominators zero since they cannot be solutions: \[ x - 2 = 0 \Rightarrow x \ne 2 \] \[ x + 4 = 0 \Rightarrow x \ne -4 \] Cross-multiply: \[ (x+2)(x+4) = (3x - 4)(x - 2) \] Expand both sides \[ x^2 + 4x + 2x + 8 = 3x^2 - 6x - 4x + 8 \] Write the equation in standard frm: \[ 0 = 2x^2 - 16x \] Factor: \[ 2x(x - 8) = 0 \] Set each factor to zero: \[ 2x = 0 \Rightarrow x = 0 \] \[ x - 8 = 0 \Rightarrow x = 8 \] Check restrictions

None of the solutions found is equal to the excluded values \( 2 \) and \( -4 \), hence the solution set is given by: \[ \{ 0 , 8 \} \]

Let \[ y = \dfrac{2x - 1}{x + 3} \] Interchange \( x \) and \( y \) in the above equation: \[ x = \dfrac{2y - 1}{y + 3} \] Cross-multiply: \[ x(y + 3) = 2y - 1 \Rightarrow xy + 3x = 2y - 1 \] Bring terms involving \( y \) together: \[ xy - 2y = -3x - 1 \Rightarrow y(x - 2) = -3x - 1 \] Solve for \( y \): \[ y = \dfrac{-3x - 1}{x - 2} \] The inverse function is given by: \[ f^{-1}(x) = y = \dfrac{-3x - 1}{x - 2} \]

Set the two expressions for \( y \) equal: \[ x^2 - 4x + 3 = 2x - 1 \Rightarrow x^2 - 6x + 4 = 0 \] Use the quadratic formula: \[ x = \dfrac{6 \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)} = \dfrac{6 \pm \sqrt{36 - 16}}{2} = \dfrac{6 \pm \sqrt{20}}{2} = \dfrac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5} \] Now substitute \(x \) into \( y = 2x - 1 \): \[ \text{For} \quad x = 3 + \sqrt{5} \; , \; y = 2(3 + \sqrt{5}) - 1 = 5 + 2\sqrt{5} \] \[ \text{For} \quad x = 3 - \sqrt{5} \; , \; = 2(3 - \sqrt{5}) - 1 = 5 - 2\sqrt{5} \] The solutions of the system are: \[ (3 + \sqrt{5}, 5 + 2\sqrt{5}) \quad , \quad (3 - \sqrt{5}, 5 - 2\sqrt{5}) \]

Factors of the constant term 10 are : \[ \pm 1 , \pm 2 , \pm 5 , \pm 10 \] Factors of the leading coefficient 1 are : \[ \pm 1 \] According to rational root theorem, the possible rational solutions, if any, are the ratios of factors of the constant term and the factors of the leading coefficient term.

Try \( x = 1 \): \[ P(1) = 1^3 - 4(1)^2 - 7(1) + 10 = 1 - 4 - 7 + 10 = 0 \] Hence \( x = 1 \) is a root and \( x - \) is a factor of \( P(x) \). Use polynomial division: \[ \dfrac{x^3 - 4x^2 - 7x + 10 }{x - 1} = x^2 - 3x - 10 \] Now factor the quadratic expression: \[ x^2 - 3x - 10 = (x - 5)(x + 2) \] The factorisation of \( P(x) \) is given by: \[ P(x) = x^3 - 4x^2 - 7x + 10 = (x - 1)(x - 5)(x + 2) \]

Divide both sides by 2: \[ \dfrac{6 \cdot 3^x }{2} = \dfrac{162}{2} \] Simplify \[ 3 \cdot 3^x = 81 \] Use exponential rule to rewrite \( 3 \cdot 3^x \) as \( \cdot 3^{x+1} \) and substitute in the above equation: \[ \cdot 3^{x+1} = 81 \] Write 81 as a power of 3: \[ 3^{x + 1} = 3^4 \] Hence the algebraic equation: \[ x + 1 = 4 \] Solve for \( x \) \[ x = 3 \]

Use vertex form: \[ y = a(x - 2)^2 - 1 \] Substitute \( (4, 7) \) into the equation: \[ 7 = a(4 - 2)^2 - 1 \] Solve for \( a \): \[ 7 = a(4) - 1 \] \[ a = 2 \] Equation of the parabola is given by \[ y = 2(x - 2)^2 - 1 \]

We are given:

The First term: \[ a = 7 \] The 15th term: \[ a_{15} = -35 \] The formula for the \(n\)th term of an arithmetic sequence is given by: \[ a_n = a + (n - 1)d \] Substitute into the formula for the 15th term: \[ a_{15} = a + (15 - 1)d = 7 + 14d \] We know \( a_{15} = -35 \), so: \[ 7 + 14d = -35 \] Solve for \( d \) \[ d = -3 \] Find the 30th term: \[ a_{30} = a + (30 - 1)d = 7 + 29(-3) = -80 \]

Use the formula find the slope of the line through the points \( (-2, 3) \) and \( (4, -6) \) \[ m_1 = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-6 - 3}{4 - (-2)} = \dfrac{-9}{6} = -\dfrac{3}{2} \] So the slope of this line through is the points \( (-2, 3) \) and \( (4, -6) \) is \( -\dfrac{3}{2} \)

The slope of the line through \( (k, 2) \) and \( (-5, 7) \) be \( m_2 \) is given by: \[ m_2 = \dfrac{7 - 2}{-5 - k} = \dfrac{5}{-5 - k} \] The product of the slopes of two perpendicular lines is equal to \( - 1 \): \[ m_1 \cdot m_2 = -1 \] Substitute \( m_1 \) and \( m_2 \): \[ \dfrac{5}{-5 - k} \cdot \left(-\dfrac{3}{2}\right) = -1 \] Simplify: \[ \dfrac{-15}{2(-5 - k)} = -1 \] Multiply both sides by \( 2(-5 - k) \): \[ -15 = -2(-5 - k) \] Solve for \( k \): \[ k = -\dfrac{25}{2} \]

Let:

\( x \) be width (the two sides perpendicular to the wall)

\( y \) be length (the side that needs fencing and is parallel to the wall) The length of the fencing is two withs and one length, hence: \[ 2x + y = 100 \] Solve for \( y \): \[ y = 100 - 2x \] Area \( A \) to be fenced: \[ A = x \cdot y = x(100 - 2x) = 100x - 2x^2 \] Now, maximize \( A \). This is a quadratic function: \[ A(x) = - 2x^2 + 100x \] The maximum occurs at: \[ x = \dfrac{-b}{2a} = \dfrac{-100}{2(-2)} = 25 \] Then: \[ y = 100 - 2(25) = 50 \] The fenced area \( A \) is maximum for:

Width \( x = 25 \) m

Length \( y = 50 \) m

and the maximum area is \[ A = 25 \times 50 = 1250 \, \text{m}^2 \]

a) Domain: The domain is all real numbers except those that make the denominator equal to zero: \[ x^2 - 9 = 0 \Rightarrow x = \pm 3 \] So, the domain is: \[ x \in \mathbb{R}, \quad x \ne -3, 3 \] b) Hole: First, factor numerator and denominator: \[ f(x) = \dfrac{(x - 1)(x - 3)}{(x - 3)(x + 3)} \] The factor \( x - 3 \) cancels and this means there's a hole at \( x = 3 \). To find the y-coordinate of the hole, cancel the term \( x - 3 \) : \[ f(x) = \dfrac{x - 1}{x + 3}, \quad x \ne 3 \] Substitute by \( x = 3 \): \[ f(3) = \dfrac{3 - 1}{3 + 3} = \dfrac{2}{6} = \dfrac{1}{3} \] The hole has the coordinates: \[ (3, \dfrac{1}{3}) \] c) Asymptotes: Vertical asymptote occurs where the denominator is zero and does not cancel: \[ x = - 3 \Rightarrow \text{Vertical asymptote} \] Horizontal asymptote:

Since degrees of numerator and denominator are equal (both degree 2), take the ratio of leading coefficients: \[ \text{Horizontal asymptote: } y = \dfrac{1}{1} = 1 \] d) Features of the Graph :

Hole at \[ (3, \dfrac{1}{3}) \] Vertical asymptote at \[ x = -3 \] Horizontal asymptote at \[ y = 1 \] Crosses the y-axis at: \[ f(0) = \dfrac{0^2 - 4(0) + 3}{0^2 - 9} = \dfrac{3}{-9} = -\dfrac{1}{3} \] X-intercepts (zeros) from the numerator (after cancelling): \[ f(x) = \dfrac{x - 1}{x + 3} \] \[ x - 1 = 0 \Rightarrow x = 1 \] The graph of : \[ f(x) = \dfrac{x^2 - 4x + 3}{x^2 - 9} \] is shown below

The minimum is at the vertex, therefore the vertex is at \( x = 3 \), we use vertex form: \[ f(x) = a(x - 3)^2 + k \] We need to find \( a \) and \( k \). Use the point \( (1, 2) \): \[ 2 = a(1 - 3)^2 + k \] Simplify \[ 4a + k = 2 \quad \text{(1)} \] Use the point \( (5, 10) \): \[ 10 = a(5 - 3)^2 + k \] Simplify \[ 4a + k = 10 \quad \text{(2)} \] Now subtract equations (1) and (2): \[ (4a + k) - (4a + k) = 10 - 2 \] Which gives \[ 0 = 8 \] No quadratic function that passes through points \( (1,2) \) and \( (5,10) \) and has a vertex at \( x = 3 \) can be found.

Substitute \( y = x^2 - 4 \) into the circle equation: \[ x^2 + (x^2 - 4)^2 = 25 \] Expand: \[ x^2 + (x^4 - 8x^2 + 16) = 25 \] Group like terms and write the equation in standard form: \[ x^4 - 7x^2 - 9 = 0 \] Let \( u = x^2 \), then: \[ u^2 - 7u - 9 = 0 \] Solve for \( u \): \[ u = \dfrac{7 \pm \sqrt{49 + 36}}{2} = \dfrac{7 \pm \sqrt{85}}{2} \] So: \[ x^2 = u = \dfrac{7 \pm \sqrt{85}}{2} \] No real solutions for \( x \) exist for \( u = x^2 = \dfrac{7 - \sqrt{85}}{2} \approx -1.10977\dots \) because is is negative. Hence the only value of \( u \) that gives real solutions for \( x \) is: \[ x^2 = u = \dfrac{7 +\sqrt{85}}{2} \] Solve for \( x \) by extracting the square root: \[ x = \pm \sqrt{ \dfrac{7 +\sqrt{85}}{2} } \] \[ y = x^2 - 4 = \dfrac{7 + \sqrt{85}}{2} - 4 = \dfrac{-1 + \sqrt{85}}{2} \] There are 2 points of intersection: \[ \left( \sqrt{ \dfrac{7 + \sqrt{85}}{2} }, \dfrac{-1 + \sqrt{85}}{2} \right), \quad \left( - \sqrt{ \dfrac{7 + \sqrt{85}}{2} }, \dfrac{-1 + \sqrt{85}}{2} \right) \]

Let \( x \) be the speed of airplane in still air and \( y \) be speed of wind, and \( D \) the distance between A and B. We need to find the ratio \( \frac{x}{y} \).

Against the wind: \[ D = 8(x - y) \] With the wind: \[ D = 7(x + y) \] Combine the two equations to write the equation: \[ 8x - 8y = 7x + 7y \] \[ x = 15y \] Which gives the ratio of the two speeds: \[ \frac{x}{y} = 15 \]

Rewrite equations of circles in standard form. Hence, the equation \[ x^2 + y^2 - 2x + 4y + 1 = 0 \] may be written as \[ (x - 1)^2 + (y + 2)^2 = 4 = 2^2 \] And the equation \[ x^2 + y^2 - 2x + 4y - 11 = 0 \] as \[ (x - 1)^2 + (y + 2)^2 = 16 = 4^2 \] Knowing the radii, the area of the ring is \[ \pi(4)^2 - \pi(2)^2 = 12\pi \]

Write equation in \( N \) as follows \[ N + \dfrac{1}{N} = \dfrac{78}{15} \] Multiply all terms by \( N \), obtain a quadratic equation: \[ N^2 + 1 = N \dfrac{78}{15} \] Multiply all terms by \( 15 \), obtain a quadratic equation: \[ 15 N^2 + 15 = 78 N \] Solve for \( N \) to obatin two solutions: \( N = 5 \) and \( N = 0.2 \), and since we are looking for an integer, the solution to the problem is: \[ N = 5 \]

We are given: \[ \frac{4^m}{125} = \frac{5^n}{128} \] Cross Multiply \[ 128 \cdot 4^m = 125 \cdot 5^n \] Express numbers as powers of primes

\( 128 = 2^7 \)

\( 4 = 2^2 \), so \( 4^m = (2^2)^m = 2^{2m} \)

\( 125 = 5^3 \) Substitute into the equation: \[ 2^7 \cdot 2^{2m} = 5^3 \cdot 5^n \] Which gives: \[ 2^{2m + 7} = 5^{n + 3} \] This equation: \[ 2^{2m + 7} = 5^{n + 3} \] can only be true if both sides equal the same number and that number is \( 1 \) (since 2 and 5 are distinct primes and no power of 2 equals a power of 5 except when both exponents are 0), hence \[ 2m + 7 = 0 \quad \text{and} \quad n + 3 = 0 \] Solve \[ 2m + 7 = 0 \Rightarrow m = -\frac{7}{2} \] \[ n + 3 = 0 \Rightarrow n = -3 \] The two rational numbers are: \[ m = -\frac{7}{2} \] and \[ n = -3 \]

Use log addition rule \[ \log_2(x) + \log_2(x - 2) = \log_2(x(x - 2)) = \log_2(x^2 - 2x) \] So the equation becomes: \[ \log_2(x^2 - 2x) = 3 = \log_2(2^3) = \log_2(8) \] Hence the algebraic equation: \[ x^2 - 2x = 2^3 = 8 \] Solve the quadratic equation \[ x^2 - 2x - 8 = 0 \] Factor: \[ (x - 4)(x + 2) = 0 \] So: \[ x = 4 \quad \text{or} \quad x = -2 \] In the original log expressions:

\( \log_2(x) \) is defined only if \( x > 0 \)

\( \log_2(x - 2) \) is defined only if \( x - 2 > 0 \Rightarrow x > 2 \) So the domain is: \[ x > 2 \] Out of the two solutions, only \( x = 4 \) is a valid solution.

Since all three terms are equal, this simplifies to: \[ 3^{n + 4001} + 3^{n + 4001} + 3^{n + 4001} = 3 \cdot 3^{n + 4001} \] Use exponent rules \[ = 3^{1 + n + 4001} = 3^{n + 4002} \]

Solving the system of the equation of the line and the circle, we obtain the points of intersection of the line and the circle

Solve the equation \( x + y = r \) for \( y \): \[ y = r - x \] Substitute into the equation of the circle: \[ x^2 + (r - x)^2 = 4 \] Expand: \[ 2x^2 - 2rx + r^2 - 4 = 0 \] The line is tangent to the circle is they have one single point of intersection which is the point of tangency. This occur when the discriminant \( \Delta \) of the quadratic equation is zero. Hence, \[ \Delta = b^2 - 4 a c = (-2r)^2 - 4(2)(r^2 - 4) = 4(8 - r^2) = 0 \] Solve for \( r \) to obtain: \[ r = 8 \] Which give the solutions \[ r = 2\sqrt{2} \quad \text{and} \quad r = -2\sqrt{2} \]

Let:

\( h = 80 \) m (height of the cliff)

\( d_1 \) = horizontal distance when angle of depression is \( 28^\circ \)

\( d_2 \) = horizontal distance when angle of depression is \( 18^\circ \)

We're solving for \( \Delta d = d_2 - d_1 \) From each triangle: \[ \tan(28^\circ) = \frac{80}{d_1} \Rightarrow d_1 = \frac{80}{\tan(28^\circ)} \approx \frac{80}{0.5317} \approx 150.44 \text{ m} \] \[ \tan(18^\circ) = \frac{80}{d_2} \Rightarrow d_2 = \frac{80}{\tan(18^\circ)} \approx \frac{80}{0.3249} \approx 246.30 \text{ m} \] \[ \Delta d = d_2 - d_1 \approx 246.30 - 150.44 = 95.86 \text{ m} \] The boat travelled \( 95.86 \text{ m} \) between the two position.

We know that \( \cos^2 x = 1 - \sin^2 x \). So, substitute this into the equation: \[ 2(1 - \sin^2 x) + 3 \sin x = 3 \] Expand and simplify: \[ 2 - 2 \sin^2 x + 3 \sin x = 3 \] Now subtract 3 from both sides: \[ 2 - 2 \sin^2 x + 3 \sin x - 3 = 0 \] \[ -2 \sin^2 x + 3 \sin x - 1 = 0 \] This is a quadratic equation in \( \sin x \). Let \( y = \sin x \), so the equation becomes: \[ -2y^2 + 3y - 1 = 0 \] Multiply through by -1 to make the leading coefficient positive: \[ 2y^2 - 3y + 1 = 0 \] Now, we can solve for \( y \) using the quadratic formula: \[ y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)} \] \[ y = \frac{3 \pm \sqrt{9 - 8}}{4} \] \[ y = \frac{3 \pm \sqrt{1}}{4} \] \[ y = \frac{3 \pm 1}{4} \] So, the two solutions for \( y \) are: \[ y = \frac{3 + 1}{4} = 1 \quad \text{or} \quad y = \frac{3 - 1}{4} = \frac{1}{2} \] Solve for \( x \)

Case 1: \( \sin x = 1 \) When \( \sin x = 1 \), the solution is \( x = 90^\circ \).

Case 2: \( \sin x = \frac{1}{2} \) When \( \sin x = \frac{1}{2} \), the solutions are: \[ x = 30^\circ \quad \text{or} \quad x = 150^\circ \] Thus, the solutions to the equation \( 2 \cos^2 x + 3 \sin x = 3 \) for \( 0^\circ < x < 360^\circ \) are: \[ x = 30^\circ, 90^\circ, 150^\circ \]