IIT Jam Mathematics Mock Test - 5 - Mathematics MCQ

Re-wrirte the equation as

Compare with general form y" + py' +Qy = 0, we have p(t) = 

Now using W (1) = 1, we have 1 =ce i.e. c = 1/e

⇒ W(t) = t²e^t-1

Since 

Let r be a number such that L > r > 1, and ∈ = L - r > 0.Then by definition, there exist a number k ∈ N such

Therefore, if n ≥ k, we obtain

Let  then we have

since , and hence 

Now, 

By stock theorem = 

Consider an example, as f : (0,1)→ ℝ as f(x) = 1/x clearly f (x) is continuous over (0,1).

Also  or, hence both (a) and (b) are incorrect

Also x_n = 1n is a cauchy sequence in (0, 1). But f(x_n) = n is not cauchy.

If o(G) = pq where p and q are distinct prime such that p>q then the number of elements of order p in G is p-1

If 

⇒ g⁸ = g

⇒ g⁷ = e

⇒ either o(g) = 7 or g = e

So, number of non identity elements in S = number of elements of order 7=6.

Clearly. A is a symmetric matrix. Also we know that eigen values ofasymmetric matrix are always real

⇒ Eigen values of A are real

Now Trace (A) =0 = sum of all eigen values.

⇒ A has both positive and negative eigen values

Given o(G) = 65

We know thatevery group of order 65 is cyclic.

⇒ every subgroup of G is normal

Also number of subgroup of finite cyclic group G = number of positive divisior of o (G)

⇒ Number of subgroup of G of order 65

⇒ (1+1)(1+1)= 4 =number of normal subgroup of G

Therefore, number of proper normal subgroup of G is 2.

Consider the series 

Comparing with 

Then we have

Hence, radius of convergence R = 1/P = 1/5

S₁: Take G=K₄

Clearly G is an abelian group of order 4 and for every divisior of 4, G has a subgroup but G is not cyclic.

⇒ S₁ is false

S₂ : take G = Q₈

Everyproper subgroup of G is cyclic but G is not abelian.

⇒ S₂ is false

Given 

Hence, the line integral is given by

To calculate 

Let 

Therefore 

For (A), Let (S_n) be the sequence of partial sum of .

Then

Since  converges, hence its sequence of partial sum is convergent.

Let 

For (B), Let (S_n) and (t_n) are the sequence of partial sums of , and respectively. Then

 be the sequence of partial sum of .

Since , converges, hence (s_n) is convergent. Also diverges, so (t_n) is divergent.

then (s_n + t_n) is divergent, henve diverges.

But by the statement of p-series test  diverges.

For (d), Note that an infinite geometric series  converges only when |r|<1.

hence statement (c) is incorrect

Let 

Using variable seprable, we have

log(z - 1) = t + log c

⇒ z - 1 = ce^t ⇒ y = e^t = exp(ce^t +1)

 (by definition this is resolution of  into rectangular components.)

Given.A is a 3 x 3 matrix whose columns are linearly dependent.

⇒ Rank (A) ≤ 2

Let

Clearly columns of A are linearly dependent

Let

Also Ax = b does not have any solution

⇒ Statement (II) is false

Now, let

Since columns of A are linearly dependent.

⇒ (c₁, c₂, c₃) = a(a₁, a₂, a₃) + b(b₁, b₂, b₃)

⇒ (c₁, c₂, c₃) =(aa₁+bb₁, aa₂+bb₂, aa₃+bb₃)

Let v be any vector which is a linear combination of column of A ie.

v = α(a₁, a₂, a₃) + β(b₁, b₂, b₃) + r(c₁, c₂, c₃)

  = (α + ra)(a₁, a₂, a₃) + (β + rb)(b₁, b₂, b₃)

⇒ v lies in the same plane

⇒ Statement (I) is true.

The shaded region represents the given region. The area in the polar form is given by,

So, we have symmetry in the function as well as the given. Hence,

Put r² = m ⇒ when r = 0, m=0

⇒ 2r dr = dm, when r = 2, m = 4

Weknow that {1,x} is the standard basis for P₁(R)

T(1) = (1-2, 1)=(-1, 1)

T(x) = (0-2,1) = (-2,1)

Clearly T(1) and T(x) are linear independent vectors.

⇒ Rank (T) = 2

By Rank-Nullity theorem, we have

Rank (T)+ mullity (T) = 2

⇒ nullity(T) = 0

⇒T is one-one.

Also rank (T) = dim (ℝ²)

⇒ Tis onto

Since   do not exist

Let 

Also (x_n is an increasing sequence, diverges to positive infinity. Hence the set of all limit points is empty.

The equation of the curve above the x-axis is given by 

Volume of revolution = 

We know that

The number of cyclic subgroups of order p in a group G = (number of elements of order p/p - 1)

Now, the number of elements of order p in  is p³ -1

Thus the number of cyclic subgroup of order p in  = p² + p + 1.

The length of th curve y = x^3/2, 0 ≤ x ≤4 is given by = 

We know that {1, x, x²} is the standard basis for P₂ (ℝ)

Now, T (1) = 0 - g(x) + 2·1 = 2

T(x) = 1(x + 1) + 2x =3x + 1

T(x²) = 2x(x + 1) + 2x² = 2x² + 2x + 2x² = 4x² + 2x

⇒ The matrix of T with respect to standard basis is A = 

Clearly.A is a triangular matrix

So eigen values of A are diagonal entries of A

⇒ Eigen values of A are 2,3,4 ⇒ Tr (T) = Tr(A) = 2 + 3 + 4 = 9

Given

x⁵ + x³ -2 = 0

Let f(x) = x⁵ + x³ -2 

⇒ f(x) = 5x⁴ +3x² ≥0 ∀x ∈ ℝ ⇒ f is increasing and degree of f is odd

Therefore, f has only one real root.

Hence no. of real root is 1.

Since f : ℝ→ℝ is a continuous map.

take f (x) = x, clearly 'f' is unbounded

∴ option (a) is incorrect

take f(x) = sin x

clearly f (R) = [-1,1] which is closed

∴ option (b) is incorrect

also take A = [-1.1] which is compact.

But f^-1 (A) = R, ⇒ f^-1(A) is not compact.

Therefore option (d) is incorrect.

For (a),

Then |f'(x)| =|2 sin x cos x| = |sin 2x| ≤ 1∀x ∈ ℝ

hence f(x) is lipschitz continuous over ℝ.Therefore,f (x) is uniformly continuous.

For (b),

Then 

since both limit exist and finite, hence f(x) is uniformly continuous.

For (c), f(x) = x², x∈ℝ

If x_n = n+ 1/n and y_n = n, then |x_n - y_n| → 0

But 

since both limit exist and finite, hencef(x) is uniformly continuous.

Hence, f (x) = x² is not uniformly contineous

For (d),

Since both limit exist and finite, hence f(x) is uniformly continuous.

For continuity at (0,0)

Let x = r cos θ, y = r sinθ

If (x,y)→ (0,0) then r→0

= cosθ.sinθ, which depends on the choice of θ

hence limit do not exist at (0,0). Therefore f(x, y) is discontinuous at (0,0).

Partial derivatives at (0,0)

∴ both partial derivatives exist at (0,0).

Also at any other point

Hence partial derivatives exist at every point of ℝ².

The characteristic equation is given by r² + 2r + 6 = 0

using y(0) = 2, we have c₁ = 2

Now y'(0) = α, we have  

Therefore

Smallest positive value of x for which y= 0 is x(α) = 

Let us consider A_n = 

then A_n is open for each n ∈ ℕ. But

 is not open.