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Math101 January 2011 Solutions
Jon Woolf
January 31, 2012
[Marginal notes: HW means similar to homework, L means appeared in lectures, NP means new, unseen, problem.]
- The natural domain of f (x) = 1 โ 1 /(x + 2) is (โโ, โ2) โช (โ 2 , โ). The natural [HW] codomain is (โโ, 1) โช (1, โ). [2 marks] Here is a sketch of the graph, obtained by shifting the graph of โ 1 /x left by 2 and up by 1, also showing the two asymptotes x = โ2 and y = 1: - - -
1
2
3
4
y
โ6 โ4 โ2 2 x
[2 marks] [4]
- To find the inverse f โ^1 we set y = f (x) and solve for x: [HW]
y = 3 x + 2 2 x โ 1
โโ (2x โ 1)y = 3x + 2 provided x 6 =
โโ (2y โ 3)x = 2 + y โโ x = 2 + y 2 y โ 3 provided y 6 =
Hence the inverse is f โ^1 (y) = (2 + y)/(2y โ 3). [3 marks] To check we compute
f (f โ^1 (x)) = f ((2 + x)/(2x โ 3))
3(2 + x)/(2x โ 3) + 2 2(2 + x)/(2x โ 3) โ 1 = 3(2 + x) + 2(2x โ 3) 2(2 + x) โ (2x โ 3) = 7 x/ 7 = x
as required. [2 marks] [5]
- Using the angle sum formulae for sine and cosine and the identity cos^2 (x) = 1 โ sin^2 (x) we have [HW]
sin(3x) = sin(x + 2x) = sin(x) cos(2x) + cos(x) sin(2x) = sin(x)
( cos^2 (x) โ sin^2 (x)
)
- cos(x) ร 2 sin(x) cos(x) = 3 sin(x) cos^2 (x) โ sin^3 (x) = 3 sin(x) โ 4 sin^3 (x). [4 marks]
[4]
- The limits can be obtained by using the โalgebra of limitsโ and standard examples. [HW]
(i) We evaluate directly since the expression is continuous at x = 2:
xlimโ 2
3 x + 2 cos(ฯx)
cos(2ฯ) = 8. [1 mark]
(ii) Since ex^ โ1 and sin(x) vanish at x = 0 and are differentiable there we use LโHหopitalโs Theorem and then evaluate directly:
xlimโ 0
ex^ โ 1 sin(x) = lim xโ 0 ex cos(x)
e^0 cos(0) = 1. [3 marks]
- Given that g(x) = x^3 + 12x^2 + 21x + 5 we have [HW]
gโฒ(x) = 3x^2 + 24x + 21 = 3(x + 7)(x + 1) and gโฒโฒ(x) = 6x + 24. [2 marks]
Hence the extrema occur when x = โ7 and x = โ1. Since gโฒโฒ(โ7) < 0 and gโฒโฒ(โ1) > 0 we see that they are a local maximum and a local minimum respectively. [2 marks] [4]
- (i) Using the substitution t = x^2 + 4x, so that dt/dx = 2x + 4 = 2(x + 2), we have [HW]
โซ (x+2) sin(x^2 +4x) dx =
โซ sin(t) dt = โ
cos(t)+k = โ
cos(x^2 +4x)+k. [3 marks]
(ii) Using integration by parts we have [L] โซ (^3) 1
ln(x) dx = [x ln x]^31 โ
โซ (^3) 1
x
x dx = 3 ln(3) โ [x]^31 = 3 ln(3) โ 2. [3 marks] [6]
- Let an = exp(n^2 )/n! = en^2 /n! for n โ N. Then [HW]
an+ an
e(n+1)^2 n! en^2 (n + 1)!
e(n+1)^2 โn^2 n + 1
e^2 n+ n + 1 = en^ en+ n + 1
as n โ โ because et/t โ โ as t โ โ. Hence by the ratio test the series โ^ โ n=
exp(n^2 ) n!
diverges. [5 marks] [5] [Section A comprises 55 marks]
SECTION B
- The derivative of f (x) = exp(x/3) is f โฒ(x) = exp(x/3)/ 3 > 0, so f is strictly increasing. [1 mark] [HW] The sequence xn is defined by x 0 = 0 and xn+1 = f (xn) for n โฅ 0. Clearly x 0 = 0 < x 1 = f (0) = 1. Since f is increasing it preserves strict inequalities. In particular if xnโ 1 < xn then xn = f (xnโ 1 ) < f (xn) = xn+1. Hence we see that x 0 < x 1 < x 2 < x 3 < ยท ยท ยท. [4 marks] Note that f (3) = exp(1) = e < 3. If xn < 3 then xn+1 = f (xn) < f (3) < 3 too, again because f is strictly increasing. Since x 0 < 3 we see that x 1 = f (x 0 ) < f (3) < 3, therefore x 2 < 3, therefore x 3 < 3 and so on. Hence xn < 3 for all n โฅ 0. [4 marks] By the above, the sequence xn is increasing and bounded above (by 3) and hence it converges to a limit, say xn โ x. [1 mark] Since f is continuous we have xn+1 = f (xn) โ f (x). Since the limit of a sequence is well-defined (when it exists) we must therefore have x = f (x) = exp(x/3). [3 marks] To approximate x we compute iterates xn until they remain fixed to 2 significant figures: x 0 = 0, x 1 = 1, x 2 = 1. 3956 x 3 = 1. 5923 x 4 = 1. 7002 x 5 = 1. 7625 x 6 = 1. 7995 x 7 = 1. 8218 x 8 = 1. 8354 x 9 = 1. 8437 x 10 = 1. 8489 x 11 = 1. 8521 x 12 = 1. 8540... So, to 2 significant figures, x = 1.9. [2 marks] [15]
- Let [HW] f (x) = 1 +
x + 4
x โ 3
The first derivative is
f โฒ(x) = โ
(x + 4)^2
(x โ 3)^2
2 x + 1 (x + 4)^2 (x โ 3)^2 which is < 0 for x < โ 1 /2, vanishes at x = โ 1 /2 and is > 0 for x > โ 1 /2 (where throughout we assume x 6 = โ 4 , 3). Therefore f is decreasing for x โ (โโ, โ4) and for x โ (โ 4 , โ 1 /2), increasing for x โ (โ 1 / 2 , 3) and for x โ (3, โ) and has a local extremum at x = โ 1 /2. [5 marks] (Note that it is incorrect, and from the graph can be seen to be untrue, to say either that f is decreasing for x > โ 1 /2 or that it is decreasing for x โ (โโ, โ4) โช (โ 4 , โ 1 /2) because the derivative does not exist at x = โ4. A similar remark applies for the range on which f is increasing.) The second derivative is
f โฒโฒ(x) =
(x + 4)^3
(x โ 3)^3
3 x^2 + 3x + 13 (x + 4)^3 (x โ 3)^3
Hence A(f, x) = L(f, x) as claimed. [5 marks] Now suppose that A(f, x) = L(f, x) for x โ (0, โ). Differentiating this equation, using [NP] the fundamental theorem of calculus, gives
f (x) = d dx
A(f, x) = d dx
L(f, x) =
โ 1 + f โฒ(x)^2.
Squaring both sides we obtain f (x)^2 = 1 + f โฒ(x)^2. Differentiating again, using implicit differentiation, we obtain 2 f (x)f โฒ(x) = 2f โฒ(x)f โฒโฒ(x). By assumption f โฒ(x) > 0 on (0, โ) so that f โฒโฒ(x) = f (x) as claimed. [5 marks] It follows (as explained in the hint) that f (x) = Aex^ + Beโx^ for some constants A and B. Substituting into the first differential equation f (x)^2 = 1 + f โฒ(x)^2 we have ( Aex^ + Beโx
) 2 = 1 +
( Aex^ โ Beโx
) 2 โโ 2 AB = 1 โ 2 AB โโ 4 AB = 1.
In order that f (0) = 1 we must have A + B = 1. Hence
4 A(1 โ A) = 1 โโ 4 A^2 โ 4 A + 1 = 0 โโ (2A โ 1)^2 = 0 โโ A = 1/ 2
and so B = 1/2 too. Therefore f (x) = cosh(x) as claimed. [5 marks] [15]
- The Mean Value Theorem states that if f : [a, b] โ R is continuous and differentiable [L] on (a, b) then there is some s โ (a, b) such that
f (b) โ f (a) b โ a = f โฒ(s). [4 marks]
Let f (t) = ln(1 + t). This is continuous on [0, x] (where x > 0) and differentiable on (0, x) with derivative f โฒ(t) = 1/(1 + t). Hence by the Mean Value Theorem there is some s โ (0, x) such that
ln(1 + x) x
ln(1 + x) โ ln(1 + 0) x โ 0 = f โฒ(s) =
1 + s
. [2 marks]
Since 0 < s < x we have [NP] 1 1 + x
1 + s
and therefore 1 1 + x
ln(1 + x) x < 1. [2 marks]
Applying the strictly increasing function exp to the above inequality gives
exp
1 + x
) < (1 + x)^1 /x^ < exp(1)
because exp (ln(1 + x)/x) = exp (ln(1 + x))^1 /x^ = (1 + x)^1 /x. Taking limits as x โ 0 and using the Sandwich Theorem gives
xlimโ 0 (1 +^ x)^1 /x^ = exp(1) =^ e.^ [5 marks] Finally, since 1/n โ 0 as n โ โ, we conclude that
nlimโโ
( 1 +
n
)n = e. [2 marks] [15]
- The integral test states that for a decreasing, continuous function f : [1, โ) โ [0, โ) [L] the series โโ n=
f (n) converges โโ
โซ (^) โ 1
f (x) dx converges. [4 marks]
Let f (x) = 1/xk. This is continuous and differentiable on (0, โ) and is decreasing for k > 0. For k 6 = 1 we have โซ (^) R 1
f (x) dx =
[ x^1 โk 1 โ k
]R
1
R^1 โk^ โ 1 1 โ k โ
{ โ 1 /(1 โ k) = 1/(k โ 1) k > 1 โ k < 1 as R โ โ. Hence by the integral test โโ n=1 1 /nk^ converges if^ k >^ 1 and diverges if^ k <^ 1. [3 marks] When k = 1 we have (^) โซ (^) R
1
f (x) dx = [ln(x)]R 1 = ln(R) โ โ as R โ โ. So โโ n=1 1 /n k (^) diverges when k = 1. [2 marks] For a decreasing, continuous function f : [1, โ) โ [0, โ) the diagrams
y = f (x)
f (1) f (2) f (3)
y = f (x)
f (1) f (2) f (3) 1 2 3 4
