This TSA checks differentiation formulae, implicit differentiation, parametric differentiation, tangents and normals, locus problems, maxima and minima applications, rate of change, and derivative graph interpretation.
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Differentiation Formulae and Technique Selection
1. For \(a>0,\ a\ne 1\), which derivative is correct?
Tested fundamental: Differentiating \(a^{f(x)}\).
Explanation: For \(a^{f(x)}\), apply the chain rule and multiply by \(\ln a\). Hence the derivative is \(a^{f(x)}f'(x)\ln a\).
Common wrong choice: A misses the factor \(\ln a\).
2. For \(a>0,\ a\ne 1\), which derivative is correct?
Tested fundamental: Differentiating logarithms with general base.
Explanation: Since \(\log_a f(x)=\frac{\ln f(x)}{\ln a}\), differentiating gives \(\frac{f'(x)}{f(x)\ln a}\).
Common wrong choice: A misses the denominator \(f(x)\).
3. For differentiable \(f\), which derivative is correct?
Tested fundamental: Differentiating inverse tangent functions.
Explanation: The derivative of \(\tan^{-1}u\) is \(\frac{1}{1+u^2}\frac{du}{dx}\). Here \(u=f(x)\).
Common wrong choice: B forgets to differentiate the inside function \(f(x)\).
4. For \[ y=\sec(3x), \] which expression gives \(\frac{dy}{dx}\)?
Tested fundamental: Differentiating \(\sec f(x)\) using the chain rule.
Explanation: \(\frac{d}{dx}\sec u=\sec u\tan u\cdot \frac{du}{dx}\). Here \(u=3x\), so \(\frac{du}{dx}=3\).
Common wrong choice: D misses the chain-rule factor \(3\). B and C are traps from differentiating \(\mathrm{cosec}\,x\) and \(\cot x\).
5. Let \[ y=x^2\tan^{-1}(2x). \] Which expression gives \(\frac{dy}{dx}\)?
Tested fundamental: Product rule with inverse trigonometric differentiation.
Explanation: Use the product rule. The derivative of \(\tan^{-1}(2x)\) is \(\frac{2}{1+4x^2}\).
Common wrong choice: B misses the chain-rule factor \(2\) when differentiating \(\tan^{-1}(2x)\).
Implicit Differentiation
6. When differentiating implicitly with respect to \(x\), \[ \frac{d}{dx}(y^3)=3y^2. \]
Tested fundamental: Applying the chain rule in implicit differentiation.
Explanation: Since \(y\) is a function of \(x\), \(\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}\).
Common wrong choice: Choosing True ignores the factor \(\frac{dy}{dx}\).
7. For the curve \[ xy+y^2=5, \] which expression gives \(\frac{dy}{dx}\)?
Tested fundamental: Implicit differentiation involving the product rule.
Explanation: Differentiating gives \(y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\). Hence \((x+2y)\frac{dy}{dx}=-y\).
Common wrong choice: B comes from confusing which terms multiply \(\frac{dy}{dx}\).
8. For the curve \[ x^2+y^2=25, \] find \(\frac{dy}{dx}\) at the point \((3,4)\).
Answer format: Give your answer as an exact fraction. Use an improper fraction if applicable.
Answer: \(-\frac34\)
Tested fundamental: Finding the gradient of an implicit curve at a point.
Explanation: Differentiating gives \(2x+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x}{y}\). At \((3,4)\), this is \(-\frac34\).
Common error: Forgetting that \(y\) is a function of \(x\) when differentiating \(y^2\).
Parametric Differentiation
9. A curve is given by \[ x=f(t),\qquad y=g(t). \] Which formula gives \(\frac{dy}{dx}\), assuming \(\frac{dx}{dt}\ne0\)?
Tested fundamental: Parametric differentiation formula.
Explanation: For parametric curves, \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\).
Common wrong choice: A reverses the numerator and denominator.
10. A curve has parametric equations \[ x=t^2+1,\qquad y=t^3. \] Find \(\frac{dy}{dx}\) when \(t=2\).
Answer format: Give your answer as an integer or exact fraction. Use an improper fraction if applicable.
Answer: \(3\)
Tested fundamental: Evaluating \(\frac{dy}{dx}\) for a parametric curve.
Explanation: \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^2\). Hence \(\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}\). At \(t=2\), this gives \(3\).
Common error: Substituting \(t=2\) before forming \(\frac{dy/dt}{dx/dt}\) can lead to careless mistakes.
11. A curve has parametric equations \[ x=\sin 2t,\qquad y=2\cos t. \] Which expression gives \(\frac{dy}{dx}\)?
Tested fundamental: Differentiating trigonometric parametric equations.
Explanation: \(\frac{dx}{dt}=2\cos2t\) and \(\frac{dy}{dt}=-2\sin t\). Thus \(\frac{dy}{dx}=-\frac{\sin t}{\cos2t}\).
Common wrong choice: C forgets that both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) contain a factor of \(2\), which cancels.
12. A curve has parametric equations \[ x=t^2,\qquad y=t^3-3t. \] Find the positive value of \(t\) at which the tangent is horizontal.
Answer format: Give your answer as an integer or exact fraction. Use an improper fraction if applicable.
Answer: \(1\)
Tested fundamental: Horizontal tangent condition for parametric curves.
Explanation: A horizontal tangent occurs when \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\ne0\). Since \(\frac{dy}{dt}=3t^2-3\), we get \(t=\pm1\). The positive value is \(1\).
Common error: Setting \(\frac{dx}{dt}=0\) instead gives the condition for a vertical tangent, not a horizontal tangent.
13. A curve has parametric equations \[ x=t^3-kt,\qquad y=t^2+1. \] At \(t=2\), the tangent is parallel to the \(y\)-axis. Find \(k\).
Answer format: Give your answer as an integer.
Answer: \(12\)
Tested fundamental: Vertical tangent condition for parametric curves.
Explanation: For a tangent parallel to the \(y\)-axis, \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}\ne0\). Here \(\frac{dx}{dt}=3t^2-k\). At \(t=2\), \(12-k=0\), so \(k=12\).
Common error: Setting \(\frac{dy}{dt}=0\) instead gives a horizontal tangent condition.
Tangents, Normals and Locus Problems
14. A curve has parametric equations \[ x=t^3,\qquad y=t^2+1. \] Find the gradient of the normal when \(t=1\).
Answer format: Give your answer as an exact fraction. Use an improper fraction if applicable.
Answer: \(-\frac32\)
Tested fundamental: Finding a normal gradient from parametric equations.
Explanation: \(\frac{dx}{dt}=3t^2\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=\frac{2}{3t}\). At \(t=1\), the tangent gradient is \(\frac23\), so the normal gradient is \(-\frac32\).
Common error: Giving the tangent gradient instead of the normal gradient.
15. In a parametric locus-of-midpoint problem, which sequence is usually the most appropriate?
Tested fundamental: Method structure for parametric locus problems.
Explanation: A midpoint locus usually requires finding two moving points first, forming their midpoint in terms of the parameter, then eliminating the parameter.
Common wrong choice: B treats a locus problem like a rate or integration problem, which is not the usual method here.
16. In a locus problem, once a moving midpoint is expressed as \((X(t),Y(t))\), the final Cartesian locus is usually obtained by eliminating the parameter.
Tested fundamental: Eliminating parameters in locus problems.
Explanation: The Cartesian equation of a locus should usually be written without the parameter.
Common wrong choice: Leaving the answer in terms of the parameter does not give the final Cartesian locus.
Maxima and Minima Applications
17. If \(f'(a)=0\) and \(f”(a)>0\), what can be concluded about \(f\) at \(x=a\)?
Tested fundamental: Second derivative test.
Explanation: If \(f”(a)>0\), the curve is concave up near \(x=a\), so a stationary point there is a local minimum.
Common wrong choice: A reverses the second derivative test.
18. In an optimisation word problem involving several variables, what should usually be done before differentiating?
Tested fundamental: Setting up optimisation problems.
Explanation: Optimisation usually requires reducing the objective to a one-variable function before differentiating.
Common wrong choice: A differentiates before properly using the constraint.
19. The curve \(C\) is defined by \[ x^2+xy+y^2=3. \] Find the maximum value of \(y\) on \(C\).
Answer format: Give your answer as an integer or exact fraction. Use an improper fraction if applicable.
Answer: \(2\)
Tested fundamental: Finding a stationary value using implicit differentiation.
Explanation: Differentiating implicitly gives \((x+2y)\frac{dy}{dx}=-(2x+y)\). At a stationary point of \(y\), \(\frac{dy}{dx}=0\), so \(2x+y=0\), giving \(y=-2x\). Substituting into \(x^2+xy+y^2=3\) gives \(3x^2=3\), so \(x=\pm1\). The corresponding \(y\)-values are \(-2\) and \(2\), so the maximum value is \(2\).
Common error: Solving for \(x\) only and forgetting that the question asks for the maximum value of \(y\).
Rate of Change
20. The variables \(x\) and \(y\) are connected by \[ x^2+y^2=25. \] At the instant when \((x,y)=(3,4)\), \(x\) is decreasing at \(2\) units per second. Find \(\frac{dy}{dt}\).
Answer format: Give your answer as an exact fraction. Use an improper fraction if applicable.
Answer: \(\frac32\)
Tested fundamental: Connected rate of change using implicit differentiation.
Explanation: Differentiating with respect to \(t\), \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). Hence \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\). Since \(\frac{dx}{dt}=-2\), \(x=3\), and \(y=4\), we get \(\frac{dy}{dt}=\frac32\).
Common error: Treating “decreasing at \(2\)” as \(\frac{dx}{dt}=2\) instead of \(\frac{dx}{dt}=-2\).
21. A moving point has coordinates \[ x=t^2,\qquad y=t^3-3t. \] Find the rate of change of \(x+y\) with respect to \(t\) when \(t=2\).
Answer format: Give your answer as an integer or exact fraction. Use an improper fraction if applicable.
Answer: \(13\)
Tested fundamental: Rate of change with respect to the parameter \(t\).
Explanation: \(\frac{d}{dt}(x+y)=\frac{dx}{dt}+\frac{dy}{dt}\). Here \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^2-3\). At \(t=2\), this gives \(4+9=13\).
Common error: Finding \(\frac{dy}{dx}\) instead of the rate of change with respect to \(t\).
Derivative Graphs, Concavity and Interpretation
22. If \(f\) has a stationary point at \(x=a\), what feature should usually appear on the graph of \(y=f'(x)\)?
Tested fundamental: Relationship between stationary points of \(f\) and zeros of \(f’\).
Explanation: At a stationary point, \(f'(a)=0\), so the graph of \(y=f'(x)\) crosses or touches the \(x\)-axis at \(x=a\).
Common wrong choice: D confuses a stationary point of \(f\) with a turning point of \(f’\).
23. If \(y=k\), where \(k\ne0\), is a horizontal asymptote of \(y=f(x)\), then \(y=k\) is also a horizontal asymptote of \(y=f'(x)\).
Tested fundamental: Horizontal asymptote behaviour when sketching \(y=f'(x)\).
Explanation: If \(f(x)\) tends to a constant \(k\), then the gradient tends to \(0\). So \(y=f'(x)\) tends to the horizontal asymptote \(y=0\), not \(y=k\).
Common wrong choice: Choosing True assumes asymptotes are copied unchanged from \(f\) to \(f’\).
24. If the graph of \(y=f(x)\) has oblique asymptote \[ y=3x-2, \] then \(y=f'(x)\) approaches the horizontal asymptote \(y=c\). Find \(c\).
Answer format: Give your answer as an integer or exact fraction. Use an improper fraction if applicable.
Answer: \(3\)
Tested fundamental: Connecting an oblique asymptote of \(f\) to a horizontal asymptote of \(f’\).
Explanation: The gradient of the oblique asymptote \(y=3x-2\) is \(3\). Hence \(f'(x)\) tends to \(3\).
Common error: Using the \(y\)-intercept \(-2\) instead of the gradient \(3\).
25. At \(x=a\), the graph of \(y=f(x)\) has a non-stationary point of inflexion. The curve changes from concave down to concave up at \(x=a\). Which statement about \(y=f'(x)\) is correct?
Tested fundamental: Interpreting non-stationary points of inflexion on derivative graphs.
Explanation: When \(f\) changes from concave down to concave up, \(f’\) changes from decreasing to increasing. Hence \(f’\) has a local minimum.
Common wrong choice: A is not necessarily true because the point of inflexion is non-stationary, so \(f'(a)\ne0\).
26. If \(f(x)\) is increasing over an interval, where should the graph of \(y=f'(x)\) lie over that interval?
Tested fundamental: Linking increasing behaviour of \(f\) to the sign of \(f’\).
Explanation: If \(f(x)\) is increasing, its gradient is positive, so \(f'(x)>0\). Thus \(y=f'(x)\) lies above the \(x\)-axis.
Common wrong choice: A would mean \(f'(x)=0\), which corresponds to a stationary point, not an interval of increase.
27. If \(x=h\) is a vertical asymptote of \(y=f(x)\), what is the corresponding asymptote behaviour usually expected for \(y=f'(x)\)?
Tested fundamental: Vertical asymptote behaviour when sketching derivative graphs.
Explanation: A vertical asymptote \(x=h\) remains at the same \(x\)-value when sketching \(y=f'(x)\).
Common wrong choice: A confuses a vertical asymptote \(x=h\) with a horizontal asymptote \(y=h\).
28. The graph of \(y=f(x)\) is decreasing, and the slope becomes more steeply downward as \(x\) increases. What happens to \(f'(x)\)?
Tested fundamental: Interpreting gradient magnitude from the graph of \(f\).
Explanation: A steeper downward slope means the gradient is negative with increasing magnitude. Hence \(f'(x)\) becomes more negative.
Common wrong choice: B describes a decreasing graph that is becoming less steep, not more steep.
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