Home / IB Mathematics SL 5.6 Derivatives of xn , sinx , cosx , tanx AA SL Paper 1- Exam Style Questions

IB Mathematics SL 5.6 Derivatives of xn , sinx , cosx , tanx AA SL Paper 1- Exam Style Questions

IB Mathematics SL 5.6 Derivatives of xn , sinx , cosx , tanx AA SL Paper 1- Exam Style Questions- New Syllabus

Question

Let \( y = \frac{\ln x}{x^4} \) for \( x > 0 \). Consider the function defined by \( f(x) = \frac{\ln x}{x^4} \) for \( x > 0 \) and its graph \( y = f(x) \).

Part (a):
Show that \( \frac{dy}{dx} = \frac{1 – 4 \ln x}{x^5} \). [2]

Part (b):
The graph of \( f \) has a horizontal tangent at point P. Find the coordinates of P. [5]

Part (c):
Given that \( f”(x) = \frac{20 \ln x – 9}{x^6} \), show that P is a local maximum point. [3]

Part (d):
Solve \( f(x) > 0 \) for \( x > 0 \). [2]

Part (e):
Sketch the graph of \( f \), showing clearly the value of the x-intercept and the approximate position of point P. [3]

▶️ Answer/Explanation
Solutions

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 \( \left( e^{1/4}, \frac{1}{4e} \right) \).

Graph of f(x) = ln x / x^4

Answer: The graph has an x-intercept at \( (1, 0) \) and a local maximum at \( \left( e^{1/4}, \frac{1}{4e} \right) \).

Question

Consider the function \( f(x) = \frac{\ln x}{x} \), \( 0 < x < e^2 \).

Part (a):
(i) Solve the equation \( f'(x) = 0 \). [2]
(ii) Hence show the graph of \( f \) has a local maximum. [2]
(iii) Write down the range of the function \( f \). [1]

Part (b):
Show that there is a point of inflection on the graph and determine its coordinates. [5]

Part (c):
Sketch the graph of \( y = f(x) \), indicating clearly the asymptote, x-intercept, and the local maximum. [3]

Part (d):
Now consider the functions \( g(x) = \frac{\ln |x|}{x} \) and \( h(x) = \frac{\ln |x|}{|x|} \), where \( 0 < x < e^2 \).
(i) Sketch the graph of \( y = g(x) \). [2]
(ii) Write down the range of \( g \). [1]
(iii) Find the values of \( x \) such that \( h(x) > g(x) \). [3]

▶️ Answer/Explanation
Solutions

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 \( x > e^{3/2} \).

Graph of f(x) = ln x / x

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).

Graph of g(x) = ln |x| / x

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 \(x < – 1\) ) 

Graph comparison of h(x) and g(x)

Answer: \( – {{\text{e}}^2} < x <  – 1\) (accept \(x < – 1\) ) 

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