Calculus 1 Answer key for the Midterm Page 1 of 5
PART 1: MULTIPLE CHOICE QUESTIONS
No calculator is allowed in this part. x2023 − 1 1. The limit lim is x→1 x − 1 (A) ∞ (B) 2023 (C) 2021 (D) 2022 Answer: (B) xn − 1 f (x) − f (1) Generally, the limit lim = lim
= f ′(1) = n by the definition of derivative. The x→1 x − 1 x→1 x − 1
value of n can be n = 2021, n = 2023 or n = 2024 in some exam versions.
2. A bacteria culture initially contains 100 cells and grows at a rate proportional to its size, that is,
dP = kP, where k is a positive constant. After an hour the population has increased to 400. The dt
number of bacteria after t hours is (A) P(t) = 100e(ln4)t (C) P(t) = 300tet−1 + 100 (B) P(t) = 100 + 300t (D) P(t) = 300t2 + 100 Answer: (A)
3. If the point (−1,2) lies on the graph of y = f (x) then which point must lie on the graph of y = f −1(x)? 1 1 (A) (−1 1, −2) (D) (2, −1 . , ) (B) ( , −1) (C) ( ) 2 2 Answer: (D)
4. Which of the following is the inverse of f (x) = ln(x − 2) for x > 2? (A) f −1(x) = ex−2 1 (B) y = (C) f −1(x) = ex + 2 (D) f −1(x) = ex − 2 ln(x − 2) Answer: (C)
5. If x6 − 2 f f (x) f x2 − 2 for −1 f x f 1, then lim f (x) is x→0 (A) undefined (B) 2 (C) 0 (D) −2 Answer: (D)
6. Given a curve with equation 2x2 + xy − 4y3 = 10. The slope of the tangent line to this curve at the point (2, −1) is 1 7 (A) (B) 1 (C) (D) 2 2 10 Answer: (C)
7. The displacement s of a particle (in meters) at time t seconds is given by s(t) = t4 − 3t3 + 4t2 + 2t − 1.
The acceleration of the particle (in m/s2) at time t = 1 (s) is Calculus 1 Answer key for the Midterm Page 2 of 5 (A) 2 (B) 3 (C) 4 (D) None of the above Answer: (A)
8. If the side x of a cube (a cubical box) is increased by 3%, then the volume is increased approximately by (A) 0.03x (B) 0.09x3 (C) 0.06x (D) 8% Answer: (B)
(Note: V (x) = x3. Thus, dV = V ′(x)dx = 3x2dx = 0.09x3 with dx = 0.03x.) √
9. The cost C in dollars of producing x units of a product is C(x) = 4000 + 20 x + 10x + 0.01x2. At a
production level of 100 units, the marginal cost is (A) $4000 /unit (B) $12 /unit (C) $4300 /unit (D) $13 /unit Answer: (D)
(the marginal cost is C′(x) at x = 100 units).
10. For the function f whose graph is given below. Find lim f (x) and lim f (x) x→1 x→5
(A) lim f (x) = 1 and lim f (x) = 2
(C) lim f (x) is undefined and lim f (x) = 2 x→1 x→5 x→1 x→5
(B) lim f (x) is undefined and lim f (x) = 1
(D) lim f (x) = 4 and lim f (x) = 2 x→1 x→5 x→1 x→5 Answer: (C)
11. The derivative of f (x) = x2 sin−1 x is x2 x2
(A) f ′(x) = 2x sin−1 x + √ (C) f ′(x) = 2x sin−1 x + 1 − x2 1 + x2 2x
(B) f ′(x) = x sin−1 x + √1−x2 (D) None of them Ans: (A) Calculus 1 Answer key for the Midterm Page 3 of 5
12. Consider the following piecewise function: x + 2 if x f 1   f (x) = x2 − 2x + 4 if 1 < x f 2  4 − x if x > 2.
(A) f is discontinuous at x = 1
(C) f is continuous at both x = 1 and x = 2
(B) f is discontinuous at x = 2 (D) None of these. Ans: (B) 13. Let g(x) = 6 + 4 f (x),
f (1) = 5 and f ′(1) = 1. Find g′(1). 1 1 2 (A) √ (B) √ (C) 0 (D) √ 2 26 2 26 26 Ans: (D) 4 (An answer in form of √
is also correct. Additionally, (A) and (B) are the same but the answer of 2 26 1 √ is incorrect). 2 26 |x + 2| 14. The value of lim is x→−2− x + 2 (A) 0 (B) 1 (C) −1 (D) undefined Ans: (C) |x + 2|
(Note: |x + 2| = −(x + 2) if x < −2. Therefore, = −1 if x < −2.) x + 2
15. A particle moves along a horizontal line with its displacement s(t) given by s = 1 + 10t − 5t2, where
s is measured in meters and t in seconds. At which time below is the particle moving forward? (A) t = 0.5 (B) t = 2 (C) t = 3 (D) t = 4 Ans: (A) √ √ 16. Let f (x) = 3 x and g(x) = x + 14. Find (g ◦ f )(8). √ (A) 3 (B) 2 (C) 4 (D) 14 Ans: C
17. Let f (x) = 1 + x + e3x. Find f −1(2). (A) −1 (B) 2 (C) 4 (D) 0 Ans: (D) (clearly, f (0) = 2.)
18. Suppose that f is continuous on [0,5] and the only solutions of the equation f (x) = 5 are x = 1 and x = 2. If f (3) = 4 then Calculus 1 Answer key for the Midterm Page 4 of 5 (A) f (4) < 5 (B) f (4) > 6 (C) f (4) > 5 (D) f (4) > 7 Answer: (A)
19. Let f (x) = x3 + 3x . Find  f −1′ (4). 1 1 1 (A) (B) (C) (D) None of them 2 3 6 Ans: (C) 1 1 1
(By derivative of the inverse,  f −1′ (4) = = = .) f ′( f −1(4)) f ′(1) 6
20. Consider the functions f and g shown in the figure. The relationship between these functions is: (A) g(x) = f (x + 1) − 2 (C) g(x) = f (x − 1) + 2 (B) g(x) = f (x + 2) + 1 (D) g(x) = f (x − 2) + 1 Ans: (D) —END OF PART 1— Calculus 1 Answer key for the Midterm Page 5 of 5 Answer key for Part 2 1 √ 1
1. Since f ′(x) = √ , the linearization of the function f (x) =
x at x = 16 is L(x) = 4 + (x − 16). 2 x 8 2. Given (x2 + 1) x4 − x2 + 15 f (x) = √ for x > −4. x + 4 2x 4x3 − 2x 1 1
(a) Use logarithmic differentiation: f ′(x) = f (x) × ( + 5 − ). x2 + 1 x4 − x2 + 1 2 x + 4 1 1 1
(b) The tangent line at (0, ) is y − = − (x − 0). 2 2 16 4x
3. (a) Differentiate 4x2 + y2 = 4 with respect to x, to obtain 8x + 2yy′ = 0 so y′ = − . y 4x y − xy′
(b) To obtain y′′, differentiating both sides of y′ = −
with respect to x, it yields: y′′ = −4 . y y2
Hence, at (0, −2) we have y′ = 0 and, therefore, y′′ = 2. -The end-