GATE Mathematics Mock Test - 3 - GATE Mathematics MCQ

In figure (X), one of the dots lies in the region common to the square and the triangle only, another dot lies in the region common to the circle and the triangle only and the third dot lies in the region common to the triangle and the rectangle only. In figure (2), there is no region common to the square and the triangle only. In figure (3), there is no region common to the circle and the triangle only. In figure (4), there is no region common to the triangle and the rectangle only. Only figure (1) consists of all the three types of regions.

If the plane cuts all the generators and is at an angle to the axis of the cone, then the resulting conic section is called as an ellipse. If the cutting angle was right angle and the plane cuts all the generators then the conic formed would be circle.

The coordinate axes are rotated by 45 degree then the problem transforms into that of Lagrange mean value theorem where the point in some interval has the slope of tan(45).

Hence differentiating the function and equating to tan(45).

We have

f ‘(x) = tan(45) = 2x – 2

2x – 2 = 1

x = 3⁄2.

To find points such that f'(c) = 1

We need to check points on graph where slope remains the same ( 45 degrees)

In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points

Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.

Hence, in the interval [0, 16π] we have

= 1 + 3 + 5…….(16terms)

=(16)² = 256.

The sections cut by a plane on a right circular cone are called as conic sections or conics. The plane cuts the cone on different angles with respect to the axis of the cone to produce different conic sections.

The question is simply asking us to find if there is some open interval in the original function f(x)

where we have f'(x) = tan(60)

We have

f'(x) = 3x² + 1 = tan(60)

3x² = √3 – 1

As the value of ‘n’ increases, it becomes difficult and tedious to calculate the value of ⁿC_x.

We have.

y = 3e^2x + e^-2x - αx

⇒ y' = 6e^2x - 2e^-2x - α           ...(1)

y' (0) = 1      (Given)

.:. 1 = 6 - 2 - α

⇒ α = 3

Again from (1)

y" = 4(3e^2x + e^-2x)

⇒ y" -  4y = 4αx

⇒ β = - 4

.:. α + β= -1

Let the smallest order for a group to have a non abelian proper subgroup isn.

Clearly n≠1,2,3,4,5 because if o(G)<5 then G is abelian.

If o(G)= 6 then every proper subgroup ofG is cyclic. So, o(G) ≠ 6.

If o(G) = 7, 11 then G is cyclic. So. o(G) ≠ 7.11.

If o(G)= 8, 9, 10, then every proper subgroup of G is cyclic.

If o(G) = 12 then let G=S₃*ℤ₂

⇒ G has a proper subgroup which is non abelian.

Now 

⇒ -ax - y - a + 1 + x + y = 0

⇒ (1 - a)x + (1-a) = 0

⇒ a = 1

Let a be an element in a group G and i, k be positive integers then  is cyclic subgroup of <a> and 

∴ k = 12

Let Y = (y_n) be the sequence of real numbers given by 

Clearly, Y is not a monotone sequence. However, if m > n, then 

Since ,  2^r-1≤ r! it follows that if m > n, then 

Therefore, it follows that (y_n) is a Cauchy sequence. Hence it converges to a limit y. At the present moment we cannot evaluate y directly; however, passing to the limit (with respect to m) in the above inequality, we obtain 

Hence we can calculate y to any desired accuracy by calculating the terms yn for sufficiently large n. The reader should to this and show that y is approximately equal to 0.632 120559. 

Since, we know that every group of order p² is cyclic, where p is a prime integer.

∴  Group of order p² is abelian also

∴  Group of order 7² i.e, 49 is abelian and cyclic.

Given that 

and 

By a well known th. we know that 

Given ∈ > 0, it follows from (2), (3), and (1) that there exists N ∈ I such that 

and such that

Hence

For any n ≥ N, choose x such that 1 – 1/n < x < 1 – 1/(n + 1). Then 1 – 1/N ≤ 1 – 1/n < x < 1, and so, by (6),

Now I – x^k = (1 – x) (1 + x + x² + .... + x^{k – 1}) ≤ k(1 – x), for any k ∈ I. 

Hence, since 1 – x < 1/n, we have 

 1 – x^k ≤ k(1 – x) < k/n . .....(8) 

By (8) and (5) we then have (since n ≥ N)

To estimate I3 we have, using (4),

But x < 1 – 1/(n + 1) and so 1 – x > 1/(n + 1). Thus (n + 1) (1 – x) > 1 and so

From (7) we then have

which proves that   converges to L.

= 

= 

In this case the region D will be the region between these two circles and that will only change the limits in the double integral. 

Here is the work for this integral. 

T(x₀ , x₁ , x₂ ) = (x₀ , x₀ + x₁, x₀ + x₁ + x₂ )

Let basis are (1, 0, 0), (0, x, 0), and (0, 0, x² ) 

 Then 

T(1, 0, 0) = (1, 1, 1) 

 T (0, x, 0) = (0, x, x) 

 T(0, 0, x²) = (0, 0, x²) 

Cofactors of T

T₁₁ = 1       T₁₂ = 0         T₁₃ = 0

T₂₁ = – 1     T₂₂ = 1        T₂₃ = 0 

T₃₁ = 0        T₃₂ = – 1     T₃₃ = 1 

∴  adj. T = Transpose of co-factors matrix = 

Hence T^-1 = 

Let iof T : R² → R³ be a Linear transformation such that matrix a with respect to standard basis of T is 

here clearly the columns of A are linearly independent b ⇒ T is one one mapping since matrix is 3 × 2 the column of A span R³ if A has 3 pivot positions but it is contradiction as A has 2 columns only

⇒ Associated Linear transformation is not onto 

Rank of matrix = Rank of Linear transformation = 2 

Let f : (0, 2) → R be defined by

⇒ f(x) is differentiable only when x = 1  

i.e., f(x) is differentiable, exactly at one point

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

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

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)

The Chain Rule will have to be applied three times in this. 

p’(x) = 2(sin (h(2x))) · cos (h(2x)) · h’(2x) · 2 

Because 2x = 2(π/2) = p, you can write: 

Given, 

⇒ (AB)' = B'A'

Continuity at (0,0):

We have f(0,0) = |0| + |0| = 0 and g(0,0) = 0

Therefore, Both f and g are continuous at (0,0).

Differentiability at (0,0) :

Since, partial derivative does not exist at (0,0). This means f is not differentiable at (0,0).

⇒ g is differentiable at (0,0).

To generate an element of order 15 we need atleast 8 distinct symbols. But S₁ has only seven distinct symbols.

⇒ S₁ has no element of order 15.

Given curve: y =x+2 from (0,2) to (3,5)

⇒ dy = dx

= [-2(3) - (3)²]

= 6 - 9

= -15.

Let  be an arbitrary constant vector Then,

 for any arbitrary constant vector .

=