TIFR GS 2019 Mathematics solved problems with hints

1 TIFR 2019 MATHEMATICS The Eleventh Ray




1. The following sum of numbers( expressed in decimal notation) 1 + 11 + 111 + … + 11…1 is
n
equal to
(10 n+1 − 10 − 9n) (10 n+1 − 10 + 9n) (10 n+1 − 10 − n) (10 n+1 − 10 + n)
(a) (b) (c) (d)
81 81 81 81
Hint:
9 + 99 + 999 + … + 99…9
n
1 + 11 + 111 + … + 11…1 =
9
n
(10 − 1) + (100 − 1) + (1000 − 1) + … + (1 0…0 − 1)
⏟
n
= s
9
{(10) + (10) + (10) + … + (10)n} − n
2 3
=
9
n+1
(10 − 10 − 9n)
=
81
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1 1
2. For n ≥ 1, the sequence {xn}∞
n=1 where: xn = 1 + +…+ − 2 n is
2 n
(a) decreasing (b) increasing (c) constant (d) oscillations

Hint:
1 + 2 n(n + 1)
| xn+1 − xn | = − 2 n + 1 (multiplying and dividing by (n + 1))
(n + 1)
n + 1 + 2 n + 1( n − n + 1)
= < 0 (Since x is a increasing function)
n+1
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π
x + x 2cos( ), x ≠ 0
{
3. De ne a function: f (x) = x
0, x = 0
Consider the following statements:
(i) f′(0) exists and is equal to 1
(ii) f is not increasing in any neighbourhood of 0
(iii) f′(0) does not exist
(iv) f is increasing on ℝ
How many of the above statements is/are true?
(a) 0 (b)1 (c) 2 (d) 3

Hint:
π
f′(0) = lim 1 + h cos( ) = 1 so (i) is true.
h→0 h
1 1 −1 −1
f( ) = → 0 and f (
2n ) 2n
= → 0 as n → ∞. So f is not increasing in any
2n 2n
neighbourhood of zero.
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4. Consider di erentiable function f : ℝ → ℝ with the property that for all a, b ∈ ℝ we have:

( 2 )
a+b
f (b) − f (a) = (b − a) f′ . Then which one of the following sentences is true?

 
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, 2 TIFR 2019 MATHEMATICS The Eleventh Ray




(a) Every such f is a polynomial of degree less or equal to 2 (b) There exists such a function f
which is a polynomial of degree bigger than 2 (c) There exists such a function f which is not a

( 2 )
a+b f (b) + f (a)
polynomial (d) Every such f satis es the condition f ≤ for all
2
a, b ∈ ℝ.
Hint:
Let m, n ∈ ℝ and m < n.
m+n
m< <n
2
Then we have,
m+n m+n m+n m+n
−( − m) = m and +( − m) = n
2 2 2 2
m+n m+n
Let, x = ,y = ( − m). Then f (x − y) + f (x + y) = − (2y . f′(x)) ∀x, y.
2 2
Di erentiating with respect to y, f′′(x − y) + f′′(x + y) = 0 ⟹ f′′(t) = 0, ∀t. So f is a
polynomial of degree utmost 2.
2(b + a)
f (x) = x 2 satis es the above property: (b 2 − a 2) = (b − a) . For a = b = 2 the
2
inequality in (d) is not satis ed.
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5. Let V be an n-dimensional vector space and let T : V → V be a linear transformation such that
Rank T ≤ Rank T 3. Then which one of the following statements is necessarily true?
(a) Null space(T ) = Range(T ) (b) Null space(T ) ∩ Range(T ) = {0} (c) There exists a nonzero
subspace W of V such that Null space(T ) ∩ Range(T ) = W (d).Null space(T ) ⊆ Range(T )

Hint:
r (T ) ≤ r (T 3) ⟹ r (T ) = r (T 3), since r (T ) ≥ r (T 3). By which we also have by rank-
nullity theorem ⟹ n(T ) = n(T 3). If x ∈ N(T ) then T (x) = 0 and if x ∈ R(T ) then there
exist y such that,
T (y) = x
⟹ T 2(y) = 0
⟹ y ∈ K(T 2)
⟹ y ∈ K(T )
⟹ T (y) = 0, x = 0

[0 0]
1 0
Consider T = satis es they hypothesis N(T ) = ⟨(0,1)⟩ ⊈ ⟨(1,0)⟩ = R(T ).

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1
1
∫0 (1 + x 2)n
2
6. The limit lim n d x is equal to
n→∞
1
(a) 1 (b) 0 (c) +∞ (d)
2
Hint:
Will be updated soon.
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7. Let A be an n × n matrix with rank k. Consider the following statements:
(i) if A has real entries, then A A t necessarily has rank k.
(ii) If A has complex entries, then A A t necessarily has rank k.
Then
(a) (i) and (ii) are true (b) (i) and (ii) are false (c) (i) is true and (ii) are false (d) (i) is false and (ii)
are true

   
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