MATH 2100 C958 Calculus I

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Free MATH 2100 C958 Calculus I Questions

1.

In class: d/dx[sec⁡x]?

  • sec⁡ x tan⁡x

  • sec⁡2x

  • −csc⁡ x cot ⁡x

  • tan⁡ x

Explanation

​​​​​​​


2.

Volume: rotate y = √x, x= 0 to 4, about y-axis (shells)?

  • 8π

  • 4π

  • 16π/3

  • 128π/5

Explanation

​​​​​​​


3.

A calculus teacher shows discontinuity: f(x) = (x2−1)/(x−1)​, removable at?

  • x = -1

  • x = 1

  • x = 0

  • None

Explanation

​​​​​​​


4.

Volume by shells: rotate y = x2 from 0 to 1 about y-axis?

  • 2π/5

  • π/5

  • 2π/3

  • π/2

Explanation

​​​​​​​


5.

Implicit: exy = y, find dy/dx​.

  • y/1−xy

  • −xy/1+y

  • yexy /1−xyexy

  • exy/y

Explanation

​​​​​​​


6.

A student asks: “Why do we need the chain rule?” The teacher responds it is used when:

  • Differentiating a composite function

  • Finding critical points

  • Evaluating definite integrals

  • Applying L’Hôpital’s rule

Explanation

Explanation:

The chain rule is a fundamental rule in calculus used to differentiate composite functions. If a function can be expressed as f(g(x)), where one function is nested inside another, the derivative is found by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function: d/dx [f`(g(x))] = f'(g(x)).g'(x). This allows us to handle more complex functions that cannot be differentiated directly using simple rules.

​​​​​​​Correct Answer:

Differentiating a composite function


7.

Differential: estimate √25.1​ using f(x) = √x at x = 25.

  • 5.1

  • 5.02

  • 5.005

  • 5.01

Explanation

​​​​​​​


8.

Apply the Mean Value Theorem to f(x) = x2 on [0,2]. What is the value of (c) in (0, 2) that satisfies the theorem?

  • √2

  • 2

  • 0

  • 1

Explanation

​​​​​​​


9.

Area under y = 4 − x2 from -2 to 2?

  • 32/3

  • 8

  • 16

  • 16/3

Explanation

​​​​​​​


10.

In teaching the Intermediate Value Theorem, a teacher considers f(x) = x3 - x - 2 on [1,2]. Does it guarantee a root?

  • Yes

  • No

  • Only if differentiable

  • Depends on limit

Explanation

Explanation

Polynomials are continuous on all real numbers. Evaluate the endpoints: f(1) = 1 - 1 - 2 = -2 and f(2) = 8 - 2 - 2 = 4. Since f(1) and f(2) have opposite signs and (f) is continuous on [1,2], the Intermediate Value Theorem ensures there exists c ∈ (1,2)) with f(c) = 0. Differentiability is not required—continuity on the closed interval suffices.

​​​​​​​Correct Answer

Yes


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