Download MA1611 calculus 2022 past exam paper and more Exams Mathematics in PDF only on Docsity!
- a. Differentiate the following functions:
(i) f (x) = ln(2x 3 ). [2 marks]
Solution: f โฒ(x) = 6 x
2 2 x^3 =^
3 x.
(ii) g(x) = sin(x + ex). [2 marks]
Solution: g โฒ (x) = (1 + e x ) cos(x + e x ).
b. Evaluate the following limits:
(i) lim xโโ
x 2 โ 9
x + 3
โ 10. [2 marks]
Solution: lim xโโ
x 2 โ 9
x + 3
โ 10 = lim xโโ
x โ 3 โ 10 = โ.
(ii) lim xโโโ
x^2 โ 9
x^2 + 3
โ 10. [2 marks]
Solution: lim xโโโ
x^2 โ 9
x^2 + 3
โ 10 = lim xโโโ
1 โ 9 /x^2
1 + 3 /x^2
(iii) lim xโ 1
x^4 โ x^3 โ x + 1
x^3 + x^2 โ 5 x + 3
. [3 marks]
Solution:
lim xโ 1
x^4 โ x^3 โ x + 1
x^3 + x^2 โ 5 x + 3
= lim xโ 1
(x โ 1)(x^3 โ 1)
(x โ 1)(x^2 + 2x โ 3)
lim xโ 1
(x โ 1)^2 (x^2 + x + 1)
(x โ 1)^2 (x + 3)
Correct use of LโHopitalโs rule (after checking that the function is indeed โ0/0โ form) is also acceptable.
c. Find any constant x such that the following function f (y) is continuous at y = x:
f (y) = x 2
= x, if y โฅ x.
[3 marks]
Solution: Consider the two one-sided limits:
lim yโx+^
f (y) = x, and lim yโxโ^
f (y) = x 2 .
The function is thus continuous at x if x = x^2 , i.e. if x = 0 or x = 1. Either answer is fine.
d. Use the mean value theorem for f (x) = e x to show that
1 < ln
eb^ โ e
b โ 1
< b [6 marks]
for any b > 1.
Solution: Let f (x) = ex. f (x) is continuous over [1, b] and differentiable over (1, b). Then applying the Mean Value Theorem to f on the interval
[1, b], we see that there exists c โ (1, b) such that f โฒ(c) = ec^ =
eb^ โ e^1
b โ 1
(2 marks) But we know that 1 < c < b (1 mark) and hence 1 <
ln
ebโe bโ 1
< b, by substituting for c. (3 marks)
- a. Find the intervals of increase and decrease for h : R โ R, h(x) =
a + 2x^2 , where a > 0 is a constant. [6 marks]
Solution: We have hโฒ(x) =
2 x โ a + 2x^2
(2 marks). Consequently
hโฒ(x) > 0 for all x > 0, hโฒ(x) < 0 for all x < 0 and hโฒ(0) = 0 ( marks). Hence h(x) in increasing over [0, โ) and is decreasing over (โโ, 0] (2 marks).
b. Use the intermediate value theorem to show that f (x) = 2 cos xโx^2 โ ฯ 2 has only one root in (0, ฯ/ 2 ). [6 marks]
Both cos x and โx^2 โ ฯ/ 2 are continuous everywhere, so that the func- tion f (x) = 2 cos x โ x^2 โ ฯ/ 2 is continuous on [0, ฯ/2] (1 mark). Since f (0) = 2 โ ฯ 2 > 0 and f (ฯ/2) = โ(ฯ/ 2 )^2 โ ฯ/ 2 < 0, by intermediate value theorem, f (x) = 0 has a root in (0, ฯ/2) (2 marks). Lastly, since f โฒ(x) = โ2 sin x โ 2 x < 0 for x โ (0, ฯ/ 2 ), f (x) is decreasing over (0, ฯ/ 2 ) and hence has only one root (3 marks).
c. Find the absolute minimum and the absolute maximum for the func- tion g : [โ 1 , 2] 7 โ R, g(x) = x
4 โ x^2. [8 marks]
where A, B are constants to be determined (2 marks).
Multiplying out, 1 = A(u^2 +3)+B(u^2 +2). Comparing coefficients
of u^2 and u^0 leads to A + B = 0 and 3A + 2B = 1. This gives
A = 1 and B = โ1. (3 marks) Finally,
โซ 1
u^4 + 5u^2 + 6
du =
โซ (^
u^2 + 2
u^2 + 3
du
tan โ 1
u โ 2
tan โ 1
u โ 3
tan โ 1
sin ฮธ โ 2
tan โ 1
sin ฮธ โ 3
where c is constant.
