1 The Eleventh Ray
TIFR-202
Part
Answer the following multiple choice questions.
1. For each positive integer n, let
1 1 1
sn = + + ... +
4n 2 − 1 4n 2 − 22 4n 2 − n 2
Then lim sn equal
n→∞
π π 1
(a) (b) (c) (d) ∞
2 6 2
Hint
n
1
∑
sn =
k=1 4n 2 − k 2
n
1 1
∑n
=
22 − ( n )
k 2
k=1
Consider the partition
1 1 n
n = {0, , , …, }, | n | → 0,
n n n
1
1 1 π
∫0 22 − x 2
lim sn = d x = sin−1 =
n→∞ 2 6
∞
g(n)
∑ n2
2. The number of bijective maps g : ℕ → ℕ such that < ∞ is
n=1
(a) 0 (a) 1 (a) 2 (a) ∞
Hint
For m < n if we have f (n) < f (m), by Rearrangement inequality we get,
f (m) f (n) f (m) f (n)
+ < + 2
n2 m2 m2 n
Since f (x) is bijective, interchanging the numerator accordingly we arrive at
𝒫 𝒫
:
A
1
s
,
,
,
, 2 The Eleventh Ray
∞ ∞ ∞
1 n g(n)
∑ n ∑ n2 ∑ n2
= <
n=1 n=1 n=1
https://en.wikipedia.org/wiki/Rearrangement_inequality
n
1
3. The value of
n→∞ ∏
lim (1 −
k 2)
is
k=2
1 1
(a) (a) 1 (a) (a)
2 4
Hint
n n
1 (k − 1) (k + 1)
∏( k2 ) ∏ ( )
sn = 1 − =
k=2 k=2
k2
1⋅3 2⋅4 (k − 2) ⋅ (k + 1) (k − 1) ⋅ (k + 1)
= ⋅ ⋅ … ⋅ ⋅
22 32 (k − 1)2 k2
1 ⋅ (k + 1) 1
= =
2⋅k 2
4. The set S = {x ∈ ℝ | x > 0 and (1 + x 2) tan(2x) = x} is
(a) empty (b) nonempty but nite (c) countably in nite (d) uncountabl
Hint
Consider,
(1 + x 2) tan(2x) = x
x
tan(2x) =
(1 + x 2)
x
Let, f (x) = tan(2x) and g(x) =
(1 + x 2)
For x > 0 we have g(x) > 0
1 − x2
g′(x) =
(1 + x 2)2
g′(x) > 0 for x < 1 and g′(x) < 0 for x > 1
So, g is increasing on 0 < x < 1 and g is decreasing on 1 < x. So f (x) = g(x) for
(2n − 1)π (2n + 1)π
atmost one x ∈ ( , ). Hence S is countable
4 4

 
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TIFR-202
Part
Answer the following multiple choice questions.
1. For each positive integer n, let
1 1 1
sn = + + ... +
4n 2 − 1 4n 2 − 22 4n 2 − n 2
Then lim sn equal
n→∞
π π 1
(a) (b) (c) (d) ∞
2 6 2
Hint
n
1
∑
sn =
k=1 4n 2 − k 2
n
1 1
∑n
=
22 − ( n )
k 2
k=1
Consider the partition
1 1 n
n = {0, , , …, }, | n | → 0,
n n n
1
1 1 π
∫0 22 − x 2
lim sn = d x = sin−1 =
n→∞ 2 6
∞
g(n)
∑ n2
2. The number of bijective maps g : ℕ → ℕ such that < ∞ is
n=1
(a) 0 (a) 1 (a) 2 (a) ∞
Hint
For m < n if we have f (n) < f (m), by Rearrangement inequality we get,
f (m) f (n) f (m) f (n)
+ < + 2
n2 m2 m2 n
Since f (x) is bijective, interchanging the numerator accordingly we arrive at
𝒫 𝒫
:
A
1
s
,
,
,
, 2 The Eleventh Ray
∞ ∞ ∞
1 n g(n)
∑ n ∑ n2 ∑ n2
= <
n=1 n=1 n=1
https://en.wikipedia.org/wiki/Rearrangement_inequality
n
1
3. The value of
n→∞ ∏
lim (1 −
k 2)
is
k=2
1 1
(a) (a) 1 (a) (a)
2 4
Hint
n n
1 (k − 1) (k + 1)
∏( k2 ) ∏ ( )
sn = 1 − =
k=2 k=2
k2
1⋅3 2⋅4 (k − 2) ⋅ (k + 1) (k − 1) ⋅ (k + 1)
= ⋅ ⋅ … ⋅ ⋅
22 32 (k − 1)2 k2
1 ⋅ (k + 1) 1
= =
2⋅k 2
4. The set S = {x ∈ ℝ | x > 0 and (1 + x 2) tan(2x) = x} is
(a) empty (b) nonempty but nite (c) countably in nite (d) uncountabl
Hint
Consider,
(1 + x 2) tan(2x) = x
x
tan(2x) =
(1 + x 2)
x
Let, f (x) = tan(2x) and g(x) =
(1 + x 2)
For x > 0 we have g(x) > 0
1 − x2
g′(x) =
(1 + x 2)2
g′(x) > 0 for x < 1 and g′(x) < 0 for x > 1
So, g is increasing on 0 < x < 1 and g is decreasing on 1 < x. So f (x) = g(x) for
(2n − 1)π (2n + 1)π
atmost one x ∈ ( , ). Hence S is countable
4 4

 
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0
fi .
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fi
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