The LNM Institute of Information Technology Jaipur, Rajasthan
Mid Semester Exam
MATH-I, September 8, 2015 Time: 60 minutes, Maximum Marks: 25
Part-II
Instructions: You should attempt all questions. Your writing should be legible and neat. Marks awarded are shown next to the question. Start a new question on a new page and answer all its parts in the same place. Please make an index showing the question number and page number on the front page of your answer sheet in the following format, otherwise you may be penalized by the deduction of 2 marks. Question No. Page No.
1. Consider the sequence (an ) defined by a1 := 1 and an+1 := 1 +
1 for n ∈ N. an
Show that (an ) converges and find its limit.
[05 marks]
Solution: We show that (an ) is a Cauchy sequence (a) an−1 − an 1 1 1 1 = [1mark] = − |an+1 − an | = 1 + − 1+ an an−1 an an−1 an−1 an (b) It is clear that an ≥ 1 for all n ∈ N and hence 1 an−1 = an−1 + 1 ≥ 2 ∀ n ∈ N with n ≥ 2.[1mark] an an−1 = 1 + an−1 (c) |an+1 − an | =
1 |an − an−1 | ≤ |an − an−1 | ∀n ∈ N with n ≥ 2. an−1 an 2
Hence (an ) is a Cauchy sequence
[1 mark]
(d) By Cauchy criterion, it is convergent. Let an → a. Then an+1 → a, and since an+1 = √ 1 + a1n , we have a = 1 + a1 =⇒ a = 1±2 5 . [1 mark] (e) Also, an ≥ 1 for all n ∈ N implies that a ≥ 1. Hence a =
√ 1+ 5 2 .
[1 mark]
2. Show that function f : Z → R defined by f (m) = (−1)m + 2m . is continuous everywhere on Z.
[5 marks]
Solution: (a) Let k ∈ Z, and (xn ) be sequence in Z such that xn → k. (b) Then for =
1 2
(any < 1 would work), there exist n0 such that 1 |xn − k| < , ∀ n ≥ n0 2
[1 mark]
[1 mark]
(c) Since xn ∈ Z for all n hence xn = k for all n ≥ n0 . [1 mark] (d) This implies that f (xn ) = f (k) for all n ≥ n0 , which means that f (xn ) → f (k). [1 mark] (e) Since k was arbitrary so f is continuous on Z.
[1 mark]
Alternate Solution: (a) Let k ∈ Z and > 0 be given.
[1 mark]
(b) Choose δ < 1.
[1 mark]
(c) m ∈ Z such that |m − k| < δ implies that m = k.
[1 mark]
(d) Hence m ∈ Z such that |m − k| < δ =⇒ 0 = |f (k) − f (k)| = |f (m) − f (k)| < . mark] (e) Since k was arbitrary so f is continuous on Z.
[1
[1 mark]
1 . Use the Picard Convergence Theorem to 1 + x2 show that f has a unique fixed point in [0, 1] and any Picard sequence with its initial point x0 ∈ [0, 1] will converge to this fixed point. Compute the first three values of the Picard sequence for f when x0 = 0. [5 marks] (Hint: Compute f 00 to estimate |f 0 (x)|)
3. Consider f : [0, 1] → R defined by f (x) =
Solution: 2x ≤0 (1 + x2 )2 for all x ∈ [0, 1], hence f is decreasing function on [0, 1]. Also f is continuous on [0, 1], hence f has IVP on [0, 1]. This shows that range of f is [ 21 , 1] ⊂ [0, 1]. [1 mark]
(a) We first show that range of f is contained in [0, 1]. Note that f 0 (x) = −
(b) Note that 4x2 2(1 − 3x2 ) 2(3x2 − 1) 1 − = − = f (x) = −2 (1 + x2 )2 (1 + x2 )3 (1 + x2 )3 (1 + x2 )3 00
Hence f 0 is decreasing on [0, √13 ] and increasing on [ √13 , 1]. (c) Also we note that f 0 (0) = 0, f 0 ( √13 ) =
√ − 3 8 3 , f 0 (1)
= − 12 .
[1 mark] [1 mark]
(d) Hence √ 3 3 <1 |f (x)| ≤ 8 0
∀x ∈ [0, 1].
So by Picard Convergence Theorem to f has a unique fixed point in [0, 1] and any Picard sequence with its initial point x0 ∈ [0, 1] will converge to this fixed point. [1 mark] (e) With x0 = 0 first three values of the Picard sequence for f are: x1 = f (x0 ) = f (0) = 1 1 x2 = f (x1 ) = f (1) = 2 1 4 x3 = f (x2 ) = f ( ) = 2 5 [1 mark] 4. Let n ≥ 2, r > 0, a ∈ R. Let f (n) be continuous on [a − r, a + r]. Assume that f (k) (a) = 0 for 1 ≤ k ≤ n − 1, but f (n) (a) < 0. If n is even, then show that a is point of local maximum for f. [5 marks] Solution:
(a) Continuity of f (n) at a and f (n) (a) > 0 implies that there exist δ > 0 with δ < r such that f (n) (x) < 0 for all x ∈ (a − δ, a + δ). [1 mark] (b) Now for x ∈ (a − δ, a + δ), with x 6= a, by Taylors theorem we have f (x) = f (a)+f 0 (a)(x−a)+· · ·+
f (n−1) (a) f (n) (c) f (n) (c) (x−a)n−1 + (x−a)n = f (a)+ (x−a)n , (n − 1)! n! n!
for some c between a and x. (c) Since c between a and x, hence f (n) (c) < 0. (d) When n is even (x −
a)n
≥ 0 for all x ∈ R.
[1 mark] [1 mark] [1 mark]
(e) f (x) ≤ f (a), for all x ∈ (a − δ, a + δ). Hence a is point of local maximum.
[1 mark]