GATE Mathematics Mock Test - 2 - Mathematics MCQ

Adding (i) and (ii), we get

Given,

By integrating both sides we get,

Given differential equestion is 

multiply bothe of side of equation by e^y we get

Take e^y = t

Integrating factor = 

solutions is

where c is an arbitrary constant

where u = e^x

so 

now 

when x = log2

= not defind

Leibnitz’s linear equation:

Solution is,

The Cylindrical coordinates is of the form ( ρ, φ, z) where ρ =  and   and z = z where (x, y, z) is the Cartesian coordinates. The Spherical coordinates is of the form (r, θ, φ) where   and  .Now, substituting the values for x as 2, y as 6 and z as 9, we get the answer as (6.32, 71.565., 9) and (11, 35.097., 71.565.).

Hence F is irrotational field as Curl 

Let 

Put cos 2θ = t and -2 sin 2θ dθ = dt

Where θ = 0, t = 1

θ = 

= 0.75

Analytic Method : since 

Replacing x by 2x and y by 0, then 

= f’(0) = –1 x ∈ R   (given) 

Integrating, we get f(x) = –x + c 

Putting x = 0, then f(0) = 0 + c = 1     (given)

∴ c = 1 

then f(x) = 1 – x 

∴ f(2) = 1 – 2 = –1 

Let H be any subgroup of order 12 in S₄

Let H ≠ A₄

and let if possible that H contains and odd permutation thus H has 6 odd and 6 even permutation

 ⇒ H ∩ A₄ is a subgroup of A₄ of order 6 

⇒ A₄ has subgroup of order 6 contradiction by a well know theorem 

Hence A₄ is the only subgroup of S₄ of order 12. 

We have

⇒ for every closed path .

The formula for converting a vector from Cartesian coordinates to Cylindrical coordinates is

Substituting the column matrix in right hand side by the given the vector, and solving the matrix we get a vector in cylindrical coordinates. Now change the point P which is in Cartesian coordinates to Cylindrical coordinates. Now, we should substitute the point P in X thus finding the value of X at P. Hence we get the value 100 a_x – 398 a_y + 108 a_z.

In conics, the generator is revolving to form two anti-parallel cones joined at the apex. The plane is then made to cut these cones and we get different conic sections. If we cut at right angles with respect to the axis of the cone we get a circle.

For a discrete probability function, the variance is given by

Variance(V) = 

Where µ is the mean, substitute P(x)=ⁿC_x p^x q^(n-x) in the above equation and put µ = np to obtain

V = npq.

First convert each point which is in cylindrical coordinates to Cartesian coordinates. Now using the formula, distance =   and substituting the values of x, y, and z in it, we get the required answer as 6.19 units. This sum can also be solved using a direct formula to find distance using two points in Cylindrical coordinates.

Since, a^log_a^x = x . The given equation may be written as: 3x² x = 348 ⇒ x = (116)^1/3 = 4.8.

From Vietas formulas we can deduce that the x² coefficient of the monic polynomial is zero (Sum of roots = zero). Hence, we can rewrite our third degree polynomial as c

Now the question asks us to relate α and c

Where c is indeed the cyclic sum of two roots taken at a time by Vietas formulae

As usual, Rolles point in the rotated domain equals the Lagrange point in the existing domain. Hence, we must have

y ‘ = tan(α)

3x² + c = tan(α)

To find the minimum angle, we have to find the minimum value of α

such that the equation formed above has real roots when solved for x So, we can write

tan(α) – c > 0

tan(α) > c

α > tan^-1(c)

Thus, the minimum required angle is

α = tan^-1(c).

Hence, series  converges and sum is 5e.

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

The area is present above the x-axis. Area above the x-axis is positive. The area is bounded by x-axis, curve y = f(x), straight lines x=a and x=b. Hence, area is found by integrating the curve with the lines as limits.

Given that f(a) = f(b) for all integral values of a, b, one additional constraint is that a +  b / 2 = 10

This means that the graph is symmetric about the line x = 10

This means that the even degree polynomial has a Rolle point(the only Rolle point) at x = 10

We need to differentiate the functions given in the options and observe which of these has a single(OR repeated root), which is equal to 10.

The first Option when differentiated yields

f'(x) = x³ – 30x² + 300x – 1000

Equating to zero, we see that x = 10

satisfies the equation

(10)³ – 30(10)² + 300(10) – 1000 = 0.

The slope of the tangent line is -2x -3 means f0(x) = -2x -3. Thus, f(x) =

-x² -3x + C for some C. Plugging in the poing (-1; 4) gives C = 2 and thus f(x) =

-x²  -3x + 2 and f(0) = 2.

We must have . Iff we let  we have 

The lagrangian function is given by

L = xy+ λ (x + y - 1)

Then L_x = y + λ = 0 i.e. λ = -y             ........(i)

L_y = x + λ  = 0 i.e. λ = -x                   ........(ii)

and x+ y - 1= 0                                  .......(iii)

From (i) and (ii), we have x = y

The from (iii), we have 2x - 1= 0 i.e. x=1/2

The critical point is x = y = 1/2, hence