EndsemCal_soln

8/7/2019 EndsemCal_soln

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Model solutions of End-semester Examination (MA 101)

Part II : Calculus

1. (a) If = limn→∞

xn, then 12

(1 − ) > 0 and so there exists n0 ∈ N such that |xn − | < 12

(1 − )

for all n ≥ n0. Hence 0 < xn < 12

(1 + ) for all n ≥ n0 ⇒ 0 < xnn < (1+

2)n for all n ≥ n0. Since

12(1 + ) < 1, limn→∞(1+2 )n = 0. Therefore by the sandwich theorem, limn→∞ xnn = 0.

Alternative: Since limn→∞

(xnn)

1n = lim

n→∞xn < 1, by the root test, the series

∞n=1

xnn converges

and hence limn→∞

xnn = 0.

(b) Since limn→∞

2n+1

(n+1)2· n2

2n= 2 > 1, the sequence (2

n

n2) is not convergent. Also, since 1

n→ 0,

the sequence (2n−nn2

) is not convergent (being the difference of a divergent and a convergentsequence). Hence the given series is not convergent.

Alternative: This can be also solved using the comparison test, limit comparison test, etc.

(c) For a given x ∈ R, let an = (−1)n(x+3)n

n5nfor all n ∈ N. Then lim

n→∞

an+1an

= 15|x + 3|.

Hence by the ratio test,∞n=1

an converges absolutely if 15|x + 3| < 1, i.e. if x ∈ (−8, 2) and

diverges if 15|x + 3| > 1, i.e. if x ∈ (−∞, −8) ∪ (2, ∞). If x = −8, then

∞n=1

an =∞n=1

1n

diverges.

If x = 2, then∞n=1

an =∞n=1

(−1)nn

converges by Leibniz’s test, but∞n=1

|an| =∞n=1

1n

diverges, i.e.

∞

n=1

an is conditionally convergent. Therefore the set of all x ∈ R for which∞

n=1

an converges

conditionally is {2}.

2. (a) Let f (x) =

(x − 2)(x − 4) if x ∈ Q,

0 if x ∈ R \Q.

Let (xn) be any sequence in R such that xn → 2. Since |f (xn)| ≤ |(xn − 2)(xn − 4)| → 0,f (xn) → 0 = f (2). This shows that f : R → R is continuous at 2. Similarly f is continuous at4.Let c ∈ (2, 4). Then there exist sequences (rn) in Q and (tn) in R \ Q such that rn → c andtn → c. Since f (rn) = (rn − 2)(rn − 4) → (c − 2)(c − 4) = 0 and since f (tn) → 0, it follows thatf cannot be continuous at c. Therefore the given statement is not true.

(b) We first show that there exist x1, y1 ∈ [a, b] such that |x1−y1| = 12

(b−a) and f (x1) = f (y1).

Let g(x) = f (x + b−a2

) − f (x) for all x ∈ [a, a+b2

]. Since f is continuous, g : [a, a+b2

] → R is

continuous. Also g(a) = f (a+b2

) − f (a) and g(a+b2

) = f (b) − f (a+b2

) = −g(a), since f (a) = f (b).

If g(a) = 0, then we can take x1 = a+b2

and x2 = a. Otherwise, g(a+b2

) and g(a) are of opposite

signs and hence by the intermediate value theorem, there exists c ∈ (a, a+b2

) such that g(c) = 0,

i.e. f (c + b−a2

) = f (c). We take x1 = c + b−a2

and x2 = c.Repeating the same procedure as above we get x2, y2 ∈ [a, b] such that |x2 − y2| = 1

2|x1 − y1| =

122

(b − a) and f (x2) = f (y2). Continuing in this way, for each n ∈ N, there exist xn, yn ∈ [a, b]such that |xn − yn| = 1

2n(b − a) and f (xn) = f (yn). If ε > 0, then there exists n0 ∈ N such that

1

2n0 (b − a) < ε. Hence the result follows by choosing x = xn0 and y = yn0 .

Alternative: By continuity of f on [a, b], there exist x0, y0 ∈ [a, b] such that f (y0) ≤ f (x) ≤ f (x0)for all x ∈ [a, b]. If both x0, y0 ∈ {a, b}, then f must be a constant function and so the resultis obvious. Hence we assume that x0 ∈ (a, b). (The case of y0 ∈ (a, b) is similar.) Let ε > 0.





f (x) ≤ 0 for all x ≥ 1, i.e. f : [1, ∞) → R is decreasing, limx→∞

f (x) = 0 and

x 1

g(t) dt

≤ 53

for

all x ≥ 1. Hence by Dirichlet’s test,∞ 1

f (x)g(x) dx, i.e.∞ 1

x cos3 x√1+x5

dx is convergent. Consequently

the given integral is convergent.

6. (a) Solving y2 = 4ax and x2 = 4ay, we obtain the x-coordinates of the points of intersections

of the two parabolas as 0 and 4a.

Hence the required volume is4a 0

π(4ax − x4

16a2) dx = 96

5πa3.

(b) The cardioids meet at three points corresponding to θ = 0, θ = π2

and θ = −π2

. By

symmetry, the required area is 4π/2 0

12

a2(1 − cos θ)2 dθ = 12

a2(3π − 8).

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