IB Mathematics SL 5.3 The derivative of functions AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
\( f'(x) = 6x^2 – 10x + 3 \)
Differentiate: \( \frac{d}{dx} (2x^3) = 6x^2 \), \( \frac{d}{dx} (-5x^2) = -10x \), \( \frac{d}{dx} (3x) = 3 \), \( \frac{d}{dx} (1) = 0 \)
Combine: \( f'(x) = 6x^2 – 10x + 3 \)
Result: \( f'(x) = 6x^2 – 10x + 3 \) [3]
(b)
\( f'(2) = 7 \)
Substitute: \( f'(2) = 6 \times 2^2 – 10 \times 2 + 3 = 6 \times 4 – 20 + 3 = 7 \)
Result: \( f'(2) = 7 \) [1]
(c)
\( y = 7x – 11 \)
Point: \( (2, 3) \), slope: \( f'(2) = 7 \)
Point-slope form: \( y – 3 = 7 (x – 2) \)
Simplify: \( y – 3 = 7x – 14 \implies y = 7x – 11 \)
Verify point: \( f(2) = 2 \times 2^3 – 5 \times 2^2 + 3 \times 2 + 1 = 16 – 20 + 6 + 1 = 3 \)
Result: \( y = 7x – 11 \) [2]
▶️ Answer/Explanation
(a)
\( f'(x) = -\frac{2}{x^2} + 6x \)
Rewrite: \( f(x) = 2x^{-1} + 3x^2 – 3 \)
Differentiate: \( \frac{d}{dx} (2x^{-1}) = -2x^{-2} = -\frac{2}{x^2} \)
\( \frac{d}{dx} (3x^2) = 6x \), \( \frac{d}{dx} (-3) = 0 \)
Combine: \( f'(x) = -\frac{2}{x^2} + 6x \)
Result: \( f'(x) = -\frac{2}{x^2} + 6x \)
(b)
\( x + 4y – 9 = 0 \)
Verify point: \( f(1) = \frac{2}{1} + 3 \times 1^2 – 3 = 2 \)
Tangent slope: \( f'(1) = -\frac{2}{1^2} + 6 \times 1 = -2 + 6 = 4 \)
Normal slope: \( m_{\text{normal}} = -\frac{1}{4} \)
Point-slope form at \( (1, 2) \): \( y – 2 = -\frac{1}{4} (x – 1) \)
Simplify: \( y = -\frac{1}{4} x + \frac{9}{4} \)
Convert: \( x + 4y – 9 = 0 \)
Result: \( x + 4y – 9 = 0 \)