IIT JAM Mathematics Mock Test- 4 - Mathematics MCQ

y = c₁x³

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

We solve

f : ℝ→ℝ is an odd function, so

f(-x) = -f(x) ∀ x ∈ ℝ

differeniating both side, we have

-f'(x) = -f'(x) i.e. f'(-x) = f'(x)

Consider 

Let 

Therefore , option(a) is incorrect.

For option (b) , let 

But 

Therefore option (b) is correct

For optin (c0 and (d)

Also in T.

Therefore option (c) and (d) are incorrect.

Given 

⇒ (D³ + 3D² + 3D + 1)y = 0

A.E. is m³ + 3m² + 3m + 1 = 0

⇒ (m + 1 )³ = 0

⇒ m = -1

therefore general solution 

y(x) = (c₁ + c₂x + c₃x²)e^-x

Number of elements of order d in Z_n where ���� is (d).

Therefore, number of elements of order p in Z_p²_q = (p) = p-1

Given, 

=  is independent of path

Therefore

Workdone= 0 (as curve is closed)

put x = a cos θ, y = a sin θ

sx = -a sin θdθ, dy = a cos θdθ

If σ is the n cycle then σⁱ is also n cycle iff(i, n) = 1

Now (12, 11) = 1

⇒ σ¹¹ is 12 cycle.

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)

If f(x) is continuous at x = 0. then

(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.

(I) 

Here f (x.y)  = 

hence (1. 1) is not in the Domain of function, hence along the curvey = x limit do not exist. So limit (I) do not exist.

(II)  

= 

Hence, series  converges and sum is 5e.

Given

replace x → -x

from(i) and (ii)

therefore,

from (iii) and (iv)

I = 0

Closure of  Which is uncountable

Becausw x√2 is an irrational number for 

Given (6y² + axy)dx + (6xy + bx²)dy = 0

I. F. = x³y²

multiplying by I.F. in (1). We get

(6y⁴x³ + ax⁴y³)dx + (6x⁴y³ + bx⁵y²)dy = 0 be exact. Then 

⇒ 24x³y³ + 3ax⁴y² = 24a³y³ + 5bx⁴y²

⇒ 3a = 5b

⇒ 3a - 5b = 0

Given rs = sr²

Now r = rs² = (rs)s = (sr²)s = (sr)(rs)

           = (sr)(sr²) = s(rs)r² = s(sr²)r² = s²y⁴ = r⁴

⇒ r= r⁴ i.e. r³ = 1

⇒ G is group of order atmost 6.

Given a_n = 3^-n a_n-1

Let 

 convergent if |z| < √3 and divergent if |z| > √3

So, radius of convergence is √3.

S₁; div curl  ⇒ S₁ is true

S₂ ; grad div 

So S₁ is true but S₂ is false.

Clearly G ≈ D₄

Also |z(D_n)| = 2 if n is even and |z(D_n)| = 1 if n is odd

Also G is non abelian

⇒ G/Z(G) can not be cyclic =G/Z(G) ≈ 

We can write general team for r ≥ 2

y" -2y' + y = e^-x

Where a₁ = -2, a₂ = 1 are constants and b (x) = e^-x is a continuous function an [0,∞)

Now, y(x) = e^x + xe^x +1/4e^-x is one of the solutions of given ODE

Since b(x) =e^-x is bounded on [0,∞)

but y(x) is unbounded on [0,∞) because 

Both statement are false

Given, 

⇒ (D² + 2cD + k)y = 0

A.E. is m² + 2cm + k =0

as c < 0, k > 0 and c² - k >0

⇒ m is real

General solution 

Thus lim sup (x_n) = 2

lim inf (x_n) = 0

lim sup (x_n) + lim inf (x_n)'

= 2 + 0 = 2

The matrix form.Ax =  B is

If α₁ = α₂ = α₃ = 1, then this system have a unique solution.

Hence rank (A) = rank (A  | B) = 3 (Number of variables)

If we change α ₁= 2, α₂ = 3, α₃ = 4, still rank (A) = rank (A|B)= 3

Hence plane intersect at a unique point.

If n > 3, then σ = (1, 2)(3, 4) ∈ An also 0(σ) = 2

⇒ A_n ∀n > 3 has a self inverse element.