IB Mathematics SL 5.6 Derivatives of xn , sinx , cosx , tanx AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
For \( y = \frac{\ln x}{x^4} \), use the quotient rule. Let \( u = \ln x \), \( v = x^4 \), so \( u’ = \frac{1}{x} \), \( v’ = 4 x^3 \).
\[ \frac{dy}{dx} = \frac{u’ v – u v’}{v^2} = \frac{\frac{1}{x} \cdot x^4 – \ln x \cdot 4 x^3}{x^8} = \frac{x^3 – 4 x^3 \ln x}{x^8} = \frac{x^3 (1 – 4 \ln x)}{x^8} = \frac{1 – 4 \ln x}{x^5} \]
Answer: \( \frac{dy}{dx} = \frac{1 – 4 \ln x}{x^5} \).
Part (b)
A horizontal tangent occurs where \( \frac{dy}{dx} = 0 \):
\[ \frac{1 – 4 \ln x}{x^5} = 0 \Rightarrow 1 – 4 \ln x = 0 \Rightarrow \ln x = \frac{1}{4} \Rightarrow x = e^{1/4} \]
Compute the y-coordinate at \( x = e^{1/4} \):
\[ y = \frac{\ln (e^{1/4})}{(e^{1/4})^4} = \frac{\frac{1}{4}}{e^1} = \frac{1}{4e} \]
Answer: The coordinates of P are \( \left( e^{1/4}, \frac{1}{4e} \right) \).
Part (c)
Given \( f”(x) = \frac{20 \ln x – 9}{x^6} \), evaluate at \( x = e^{1/4} \):
\[ f”(e^{1/4}) = \frac{20 \ln (e^{1/4}) – 9}{(e^{1/4})^6} = \frac{20 \cdot \frac{1}{4} – 9}{e^{6/4}} = \frac{5 – 9}{e^{3/2}} = \frac{-4}{e^{3/2}} \]
Since \( f”(e^{1/4}) < 0 \), and \( \frac{dy}{dx} = 0 \) at \( x = e^{1/4} \), point P is a local maximum.
Answer: P is a local maximum point.
Part (d)
Solve \( f(x) = \frac{\ln x}{x^4} > 0 \):
Since \( x^4 > 0 \) for \( x > 0 \), we need \( \ln x > 0 \), which implies \( x > 1 \).
Answer: \( x > 1 \).
Part (e)
Find the x-intercept: \( f(x) = 0 \Rightarrow \frac{\ln x}{x^4} = 0 \Rightarrow \ln x = 0 \Rightarrow x = 1 \).
Point P is at \( \left( e^{1/4}, \frac{1}{4e} \right) \approx (1.284, 0.092) \). The graph is positive for \( x > 1 \), negative for \( 0 < x < 1 \), with a local maximum at P. As \( x \to 0^+ \), \( \ln x \to -\infty \), so \( f(x) \to -\infty \). As \( x \to \infty \), \( x^4 \) grows faster than \( \ln x \), so \( f(x) \to 0^+ \).
The graph below shows the x-intercept at \( (1, 0) \) and the approximate position of point P at \( \left( e^{1/4}, \frac{1}{4e} \right) \).
Answer: The graph has an x-intercept at \( (1, 0) \) and a local maximum at \( \left( e^{1/4}, \frac{1}{4e} \right) \).
▶️ Answer/Explanation
Part (a)(i)
For \( f(x) = \frac{\ln x}{x} \), use the quotient rule. Let \( u = \ln x \), \( v = x \), so \( u’ = \frac{1}{x} \), \( v’ = 1 \).
\[ f'(x) = \frac{u’ v – u v’}{v^2} = \frac{\frac{1}{x} \cdot x – \ln x \cdot 1}{x^2} = \frac{1 – \ln x}{x^2} \]
Solve \( f'(x) = 0 \):
\[ \frac{1 – \ln x}{x^2} = 0 \Rightarrow 1 – \ln x = 0 \Rightarrow \ln x = 1 \Rightarrow x = e \]
Answer: \( x = e \).
Part (a)(ii)
Compute the second derivative:
\[ f'(x) = \frac{1 – \ln x}{x^2} \]
\[ f”(x) = \frac{\frac{-1}{x} \cdot x^2 – (1 – \ln x) \cdot 2x}{x^4} = \frac{-x – 2x + 2x \ln x}{x^4} = \frac{2 \ln x – 3}{x^3} \]
At \( x = e \):
\[ f”(e) = \frac{2 \ln e – 3}{e^3} = \frac{2 \cdot 1 – 3}{e^3} = \frac{-1}{e^3} < 0 \]
Since \( f'(e) = 0 \) and \( f”(e) < 0 \), there is a local maximum at \( x = e \).
Answer: The graph of \( f \) has a local maximum at \( x = e \).
Part (a)(iii)
At \( x = e \), \( f(e) = \frac{\ln e}{e} = \frac{1}{e} \), the local maximum. As \( x \to 0^+ \), \( \ln x \to -\infty \), so \( f(x) \to -\infty \). As \( x \to e^2 \), \( f(x) = \frac{\ln e^2}{e^2} = \frac{2}{e^2} < \frac{1}{e} \). Since \( f'(x) > 0 \) for \( x < e \) and \( f'(x) < 0 \) for \( x > e \), the range is \( y \leq \frac{1}{e} \).
Answer: \( y \leq \frac{1}{e} \).
Part (b)
From part (a)(ii), \( f”(x) = \frac{2 \ln x – 3}{x^3} \). Solve \( f”(x) = 0 \):
\[ 2 \ln x – 3 = 0 \Rightarrow \ln x = \frac{3}{2} \Rightarrow x = e^{3/2} \]
Test concavity: \( f”(x) < 0 \) for \( x < e^{3/2} \), \( f”(x) > 0 \) for \( x > e^{3/2} \), indicating a point of inflection.
Coordinates at \( x = e^{3/2} \):
\[ f(e^{3/2}) = \frac{\ln e^{3/2}}{e^{3/2}} = \frac{\frac{3}{2}}{e^{3/2}} = \frac{3}{2 e^{3/2}} \]
Answer: Point of inflection at \( \left( e^{3/2}, \frac{3}{2 e^{3/2}} \right) \).
Part (c)
X-intercept: \( f(x) = 0 \Rightarrow \frac{\ln x}{x} = 0 \Rightarrow \ln x = 0 \Rightarrow x = 1 \).
Local maximum: At \( x = e \), \( f(e) = \frac{1}{e} \).
Asymptote: As \( x \to 0^+ \), \( f(x) \to -\infty \), so there is a vertical asymptote at \( x = 0 \).
Shape: \( f'(x) > 0 \) for \( x < e \), \( f'(x) < 0 \) for \( x > e \); \( f”(x) < 0 \) for \( x < e^{3/2} \), \( f”(x) > 0 \) for \( x > e^{3/2} \).
Answer: The graph has a vertical asymptote at \( x = 0 \), x-intercept at \( (1, 0) \), and local maximum at \( \left( e, \frac{1}{e} \right) \).
Part (d)(i)
For \( g(x) = \frac{\ln |x|}{x} \), since \( x > 0 \), \( |x| = x \), so \( g(x) = \frac{\ln x}{x} = f(x) \). The graph is identical to part (c).
Answer: The graph of \( g(x) \) has a vertical asymptote at \( x = 0 \), x-intercept at \( (1, 0) \), and local maximum at \( \left( e, \frac{1}{e} \right) \).
Part (d)(ii)
all real values
Part (d)(iii)
\( – {{\text{e}}^2} < x < – 1\) (accept \(x < – 1\) )
Answer: \( – {{\text{e}}^2} < x < – 1\) (accept \(x < – 1\) )