GATE Mathematics Mock Test - 4 - Mathematics MCQ

So, SOB THE BOSS→ PRNADGSANR

There are 10_C4 4! = 5040 sequences of 4 distinct digits, out of which there is only one sequence in which the lock opens.

Required probability = 1/5040

(AUB)=A+B—(AnB) * (A u B) = 40 + 70 — 15 = 95%

That is 95% of the students have enrolled for at least one of the two subjects Math or Economics.

Therefore, the balance (100 — 95)% = 5% of the students have not enrolled for either of the two subjects.

1. data inadequate

2. definitely false

3. definitely true

Since T : R³ → R³

such that T(2, 1, 1) = (3, 2, 1) 

T(1, 0, 1) = (0, 1 , –1) 

But let (–1, –2 ,1) = a (2, 1, 1) + b(1, 0, 1) 

on solving – 1 = 2a + b 

– 2 = a + 0 

a+ b = 1

⇒ b = 3 

so on applying transformation T on (1) bothsides 

T(1, -2, 1) = T(–2, (2, 1,1) + 3(1, 0, 1) = –2 T(2, 1, 1) + 3T(1, 0, 1) 

= –2 (3, 2, 1)+ 3(0, 1, –1) = (–6, –1, –5) 

When origin O is inside S. In this case, divergence theorem cannot be applied to the region V enclosed by S, since  has a point to discontinuity at the origin. To remove this  difficulty, let us enclose the origin by a small sphare Σ of radius ε. 

The function F is continuously differentiable at the points of the region v´ enclosed between S and Σ. Therefore applying divergence theorem for this region V´, we have 

Now on the sphere Σ, the outward drawn normal n is directed towards the centre. Therefore on Σ, we have 

By a well known result we know that if A be an n × n matrix with coefficients in F, having rows {a₁, a₂, ....., a_n }, then the following statements are true. 

 (a) if A’ be a matrix obtained from A by an elementary row operation (interchanging two rows). Then 

 D(A’) = – D(A) 

 (b) if A’ be a matrix obtained from a by an elementary row operation (replacing the row a_i by λa_j , with λ ∈ F, i ≠ j). Then 

D(A’) = D(A) 

(c) if A’ be a matrix obtained from A by an elementary row operation (replacing ai by µai , for µ ≠ 0 in F). Then 

 D(A’) = µD(A)

i.e. all the three options are correct. 

Let L(w) = w 

and x,y ∈ w,   α, β ∈ FE 

then x,y∈ L(w) 

x,y are linear combination of members of w. 

⇒ αx + βy is a linear combination of members of w 

⇒ αx + βy ∈ L(w) 

⇒ αx + βy ∈w 

⇒ w is a subspace of v. 

Given differential equation 

now on observing we can see that. The differential equation is neither Linear and homogeneous nor separable now putting y² = v - x

The given equation reduces to 

which is a homogeneous equation 

Since given f(x) = x⁴ - 4x + 6 and f(-x)  =  x⁴ - 4x² + 6
so f(x) = f(-x)

If x² = e then either o(x) is lor 2.

We know that the number of elements of order 2 in S₃ = 3.

Thus the mumber ofelements in S₃ that satisfy the equation.x² = e is 4.

Clearly ℤ₈ ⊕ ℤ₂ = ℤ₄ ⊕ ℤ₄

If there exist an onto homomorphism from ℤ₈⊕ℤ₂, to ℤ₄⊕ℤ₄, then ℤ₈⊕ℤ₂≈ℤ₄⊕ℤ₄

But this is not possible since ℤ₈⊕ℤ₂ contains an element of order 8 but ℤ₄⊕ℤ₄ does not have.

⇒ There is no onto homomorphism between groups ℤ₈⊕ℤ₂ and ℤ₄⊕ℤ₄.

It is a very well known theorem that if f and g be continuous real – valued function on the metric space M and A be the set of all x ∈ M s.t. f(x) < g(x) then A is open set. 

A.E. is m² + 4 = 0 m = 2i, -2i 

C.F. is y = C₁cos 2x + c₂ sin 2x 

Given differential equation 

multiply both sides of the given equation by e^y we get 

Integrating factor = 

solution is 

where c is an arbitrary constant 

f(x) = x³ + x²f'(1) + xf''(2) + f'''(3)

f(0) = f'''(3)

f(2) = 8 + 4f'(1) + 2f''(2) + f'''(3)

f(1) = 1 + f'(1) + f''(2) + f'''(3)

Then f(2) - f(1) = 7 + 3f'(1) + f''(2)

Now, f'(x) = 3x² + 2x f;(1) + f''(2)

f''(x) = 6x + 2f'(1)

f'''(x) = 6   

f'''(3) = 6                  ...(1)

f''(2) = 12 + 2f'(1)        .....(2)

f'(1) = 3 + 2f'(1) + f''(2)

⇒ -f'(1) = 3 + 12 + 2f'(1)

⇒ -15 = 3f'(1)

⇒ f'(1) = -5  and f''(2) = 2

So, f(2) -f(1) = 7 + 3*(-5) + 2 

= 7 - 15 + 2

= -6 = -f(0)

Given, 

=  is independent of path

Therefore

Workdone= 0 (as curve is closed)

y = c₁x³

For orthogonal tarcjectory dy/dx replace by -dx/dy

We solve

(a) Let f(x) = x, x∈[0, ∞], then clearly f(x) is uniformly continuous but not bounded.

(b) Note that  here f is continuous over [0, 1], so f(x) is uniformly continuous.

(c) Note that f(x) = 

Clearly f(x) is not continuous at x = 0, so f(x) is not uniformly continuous.

Consider 

given α₄ = 2 α₂ so that the 

vectors α₂ and α₄ are linearly dependent 

If S₁ = {α_1, α₂, α₃} then 

by subspace of R₃ spanned by S₁ is the same as that spanned by S. 

There is no real number c s.t. 

α₁ = cα₂ therefore the vectors α1 and α2 are linearly independent.

Now,examine whether the vector α₃ lie in the subspace of R₃ spanned by the vector α₁ and α₂ or not

Let α₃ = aα₁ + bα₂. 

where a, b ∈ R

then (1, –1, –4) = a(1, 2, 3) + b(2, 1, –1)

a + 2b = 1 ................, ,... (i) 

2a + b = –1 .............. (ii) 

3a – b = –4 ............. (iii) 

Solving the eq. (i) and (ii) we get a = –1, b = 1 these values of a and b also satisfy the eq. (iii). 

so α₃ = –α₁ + α₂. 

thus the vector α₃ has be expressed as a linear combination of α₁ and α₂ so that the subspace of R₃ spanned by the vectors α1, α₂ and α₃. 

Hence T = {α₁, α₂} is a linear independent subset of S which spans the same subspace of R₃ as a is spanned by S. 

The Set

of the cube roots of 1 forms a group w.r.t. multiplication set of complex number c. since 

(I) The product of any two elements of the set is an element of the set. 

(II) The associative law holds in c and hence in A.

(III) w₃ is the identity element. 

(iv) The inverse of w₁, w₂ and w₃ are w₂, w₁ and w₃ respectively therefore A is group w.r.t. multiplication.

In this case we differentiate from left to right. Here are the derivatives for this part.

Given eq. can be written as 

Where

D = d/dx

which is a homogeneous eq. 

Put x = e^z

then   then eq (2) become

[D₁(D₁ – 1) + D₁– 1] y = 0 

[D₁² – 1] y = 0 .........(3) 

its A E is m² – 1 = 0 

m = ± 1

then solu. of eq (3) is 

y = ae^z + be^-z

y = ae^z + b(e^z)^–1

y = Ax + Bx^–1 ................(4) 

be the complete solution of eq (1) then A and B are function of x which are so chosen that (1) will be satisfied. Differentiating (4) w.r.t x we have

y₁ = A₁x + A + B₁x^–1 – Bx^–2........................(5) 

Choose A and B s.t. 

A₁x + B₁x^–1 = 0 ......................(6) 

then by eq (4) we get 

y₁ = A – Bx^–2 .................(7) 

Differentiating (7) 

y₂ = A₁ – (B₁x^–2 – 2Bx^–3) ................(8) 

using (4), (7) and (8), (1) reduces to 

A₁ – B₁x^–2 = e^x  ...............(9) 

Solving (6) and (9) 

Integrating 

Substitute the value of A and B in eq. (4) we have the required solution is 

(c) & (d) follows from intermediate value theorem.

These are standard result and can be proved by contradiction.

All option are correct statement.

Given A is anon zero and non identity matrix such that.A² = A

⇒ Eigen values of A are 0 and 1.

Since A is a 3x3 matrix, then eigen value of A are either 0, 1, 1 or 0,0,1

⇒ A has repeated eigen value.

Also minimal polynomial of A contains only linear factors. Hence A is diagonaizable,

Since 1 is the eigen value ofA

⇒ ∃0 ≠ v ∈ ℝ3 such that Av = v.

Also 0 is the eigen value of A ⇒ A is a singular matrix

The Augmented matrix of given system is

Apply R₂ → R₂ - 2R₁ , R₃ → R₃ - R₁ , we have

Apply R₃→ R₃- R₂, we have

⇒ The corresponding system has no solution if rank (A) ≠ rank [A :b].

⇒ λ - 1 = 0 i.e. λ = 1

It is cauechy euler-equation let z = In x

then x2y'' = D(D-1)y, D = d/d₂

xy' = Dy , D = d/dz

then (D(D-1) - 2D - 4)y = 0

⇒ (D² - 3D - 4)y = 0

The characteristic equation is

r² -3r - 4 = 0

i.e. (r - 1)(r + 4) = 4 i.e. r = 1, -4

∴ y(z) = c₁e^z + c₂e^-4z

= c¹x + c²x^-4

Using y(1) = 1, we have c₁ + c₂ =1