Grade 11 Maths Problems with Solutions

Detailed Questions and Step-by-Step Solutions

Explore a variety of Grade 11 math questions with clear, step-by-step solutions. Topics include quadratic functions, trigonometry, circles, parabolas, rational functions, exponential and logarithmic functions, systems of equations, polynomials, and series.

Algebra, Equations, and Sequences

Question 1: Solve for \( x \) the rational equation: \( \frac{x+2}{x-2} = \frac{3x - 4}{x + 4 } \)

Exclude values making denominators zero: \( x \ne 2, x \ne -4 \).

Cross-multiply: \[ (x+2)(x+4) = (3x - 4)(x - 2) \]

Expand: \[ x^2 + 6x + 8 = 3x^2 - 10x + 8 \]

Standard form: \[ 2x^2 - 16x = 0 \Rightarrow 2x(x - 8) = 0 \]

Final solutions: { 0 , 8 }.

Question 3: Factor the polynomial: \( P(x) = x^3 - 4x^2 - 7x + 10 \)

Try \( x = 1 \): \( P(1) = 1 - 4 - 7 + 10 = 0 \). Thus \( (x - 1) \) is a factor.

Division: \( \frac{P(x)}{x-1} = x^2 - 3x - 10 \).

Factor quadratic: \( (x - 5)(x + 2) \).

Factorization: (x - 1)(x - 5)(x + 2).

Question 4: Solve for \( x \): \( 6 \cdot 3^x = 162 \)

Divide by 6: \( 3^x = 27 \). Since \( 27 = 3^3 \), then x = 3.

Question 5: Arithmetic Sequence: \( a = 7 \), \( a_{15} = -35 \). Find \( a_{30} \).

Using \( a_n = a + (n-1)d \): \( 7 + 14d = -35 \Rightarrow d = -3 \).

Find term: \( a_{30} = 7 + 29(-3) = \mathbf{-80} \).

Question 7: Sum of integer \( N \) and reciprocal is \( 78/15 \).

Equation: \( 15N^2 - 78N + 15 = 0 \Rightarrow 5N^2 - 26N + 5 = 0 \).

Factor: \( (5N - 1)(N - 5) = 0 \). Integer solution: N = 5.

Question 8: Solve \( \frac{4^m}{125} = \frac{5^n}{128} \) for rational \( m, n \).

Cross-multiply: \( 2^7 \cdot 2^{2m} = 5^3 \cdot 5^n \Rightarrow 2^{2m+7} = 5^{n+3} \).

Result: \( 2m+7=0 \Rightarrow m = -3.5 \), \( n+3=0 \Rightarrow n = -3 \).

Question 9: Solve: \( \log_2(x) + \log_2(x - 2) = 3 \)

Domain: \( x > 2 \). Property: \( \log_2(x^2 - 2x) = 3 \Rightarrow x^2 - 2x = 8 \).

Factor: \( (x - 4)(x + 2) = 0 \). Solution: x = 4.

Question 10: Simplify: \( 3^{n + 4001} + 3^{n + 4001} + 3^{n + 4001} \)
\[ 3 \cdot 3^{n + 4001} = 3^{1 + n + 4001} = \mathbf{3^{n + 4002}} \]

Functions and Graphs

Question 11: Find the inverse of \( f(x) = \frac{2x - 1}{x + 3} \)

Swap \( x, y \): \( x = \frac{2y-1}{y+3} \Rightarrow xy + 3x = 2y - 1 \).

Isolate \( y \): \( y(x-2) = -3x-1 \Rightarrow f^{-1}(x) = \mathbf{\frac{-3x-1}{x-2}} \).

Question 12: Solve the system: \( y = x^2 - 4x + 3 \) and \( y = 2x - 1 \)

Set equal: \( x^2 - 6x + 4 = 0 \). Formula: \( x = 3 \pm \sqrt{5} \).

Solutions: (3 + √5, 5 + 2√5) and (3 - √5, 5 - 2√5).

Question 13: Parabola vertex \( (2, -1) \) passing through \( (4, 7) \).

Vertex form: \( 7 = a(4-2)^2 - 1 \Rightarrow 8 = 4a \Rightarrow a = 2 \).

Equation: y = 2(x - 2)² - 1.

Question 14: Maximize area: 100m fencing, 3 sides (one wall).

Equation: \( 2x + y = 100 \). Area: \( A = x(100-2x) \).

Vertex at \( x = 25 \). Dimensions: 25m by 50m. Area: 1250m².

Question 15: Analysis of \( f(x) = \frac{x^2 - 4x + 3}{x^2 - 9} \)

Hole at \( (3, 1/3) \). VA: \( x = -3 \). HA: \( y = 1 \).

Rational Graph

Coordinate Geometry and Circles

Question 16: Find \( k \) so line through \( (k, 2), (-5, 7) \) is \(\perp\) to line through \( (-2, 3), (4, -6) \).

Slopes: \( m_1 = -1.5 \), \( m_2 = 2/3 \). Solve: \( \frac{5}{-5-k} = \frac{2}{3} \).

Result: k = -12.5.

Question 17: Intersections: \( x^2 + y^2 = 25 \) and \( y = x^2 - 4 \).

Substitute: \( y + 4 + y^2 = 25 \Rightarrow y^2 + y - 21 = 0 \).

Points: (±√[(7+√85)/2], (-1+√85)/2).

Question 18: Concentric circle area: \( x^2+y^2-2x+4y+1=0 \) and \( -11=0 \).

Radii: \( r_1 = 2 \), \( r_2 = 4 \). Area: \( 16\pi - 4\pi = \mathbf{12\pi} \).

Question 19: Find \( r \) so \( x+y=r \) is tangent to \( x^2+y^2=4 \).

Discriminant \( \Delta = 32 - 4r^2 = 0 \). Result: r = ±2√2.

Trigonometry

Question 20: Boat Distance Problem

Let \( h = 80 \). Distance \( d_1 = 80/\tan(28^\circ) \). Distance \( d_2 = 80/\tan(18^\circ) \).

\[ \Delta d \approx 246.30 - 150.44 = \mathbf{95.86 \text{ m}} \]

Trig Diagram

Question 21: Solve \( 2\cos^2 x + 3\sin x = 3 \) for \( 0^\circ \leq x < 360^\circ \).

Equation: \( 2\sin^2 x - 3\sin x + 1 = 0 \Rightarrow (2\sin x - 1)(\sin x - 1) = 0 \).

Solutions: 30°, 90°, 150°.

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