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Calculus Exam Preparation
English 10 leveled MCQs Free
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10 marks Long answer
Find the maxima and minima of $f(x) = 2 + 2x^2 - x^4$.
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Use the first and second derivative tests to find critical points and determine their nature.
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10 marks Long answer
State and explain the Mean Value Theorem.
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The Mean Value Theorem states that if a function is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in (a, b) such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
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10 marks Long answer
State and prove Euler's theorem on homogeneous functions.
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Euler's theorem states that for a homogeneous function of degree n, $f(tx, ty) = t^n f(x, y)$, and can be proved using partial derivatives.
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10 marks Long answer
Derive the reduction formula for $\int \sin^n x \,dx$.
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The reduction formula can be derived using integration by parts: $\int \sin^n x \,dx = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} \int \sin^{n-2} x \,dx$.
The full exam-length answer is in the app.
10 marks Long answer
Find the maxima and minima of functions of a single variable.
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Apply the first and second derivative tests to identify critical points and their nature.
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10 marks Long answer
Explain the derivative as a function.
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The derivative represents the rate of change of a function at a point.
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10 marks Long answer
Explain the chain rule with an example.
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The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$; for example, if $y = (3x + 2)^2$, then $\frac{dy}{dx} = 2(3x + 2) \cdot 3$.
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10 marks Long answer
Find the total derivatives of a function.
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The total derivative of a function $f(x, y)$ with respect to $x$ is given by $\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}$.
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10 marks Long answer
Find the maxima and minima of functions of two variables.
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Use the second derivative test for functions of two variables to classify critical points.
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10 marks Long answer
Explain Lagrange's multiplier method with an example.
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Lagrange's multiplier method finds the extrema of a function subject to constraints by setting $\nabla f = \lambda \nabla g$.
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