IIT JAM Mathematics Mock Test- 2 - Mathematics MCQ

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.

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.

A discrete logarithm modulo of an integer to the base is an integer such that ax ≡ b (mod g). The problem of computing the discrete logarithm is a well-known challenge in the field of cryptography and is the basis of the cryptographic system i.e., the Diffie-Hellman key exchange.

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.

If the plane cuts at an angle with respect to the axis and does not cut all the generators then the conics formed is a parabola. If the plane cuts all the generators then the conic section formed is called as ellipse.

The variance (V) for a Binomial Distribution is given by V = npq

Standard Deviation = 

The Spherical coordinates is of the form (r, θ, φ) and Cartesian coordinates is of the form (x, y, z) where x = r sin⁡θ cos⁡ϕ and y = rsin⁡θ sin⁡ϕ and z=rcos⁡θ. Now, substituting the values for r as 10, θ as 90, and φ as 60, substituting the values we get

x = 10 sin90 cos60 = 5

y = 10 sin90 sin60 = 8.66

z = 10 cos90 = 0.

It is the formula for Binomial Distribution that is asked here which is given by P(X = x) = ⁿC_x p^x q^{(n - x)}.

To solve the logarithmic function ln 

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.

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.

For the transformed function to have a Rolles point is equivalent to the existing function having a Lagrange point somewhere in the real number domain, we are finding the point in the domain of the original function where we have f'(x) = tan(α)

Let the angle to be rotated be α

We have

f'(x) = 9x² + 5 = tan(α)

9x² = tan(α) – 5

For the given function to have a Lagrange point we must have the right hand side be greater than zero, so

tan(α) – 5 > 0

tan(α) > 5

α > tan^-1(5)

In degrees we must have,

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

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.).

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.

Circle is a conic section. When the plane cuts the right circular cone at right angles with the axis of the cone, the shape obtained is called as a circle. If the angle is oblique we get the other parts of the conic sections.

For a discrete probability function, the mean value or the expected value is given by

Mean(μ) =

For Binomial Distribution P(x)= ⁿC_x p^x q^(n-x), substitute in above equation and solve to get

µ = np.

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

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

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.

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).

The formula to convert any vector from Spherical coordinates to Cartesian coordinates is given by

after substituting the values of the vector. Now, solving the matrix we get the answer

By using the property if log_ax = log_ay then x = y, gives 2x ²- 3x = 10 - 6x. Now, to solve the equation x² - 3x - 5x + 15 = 0

⇒ x² - 8x + 15

⇒ x = 3, x = 5

For x = 3: log₂(3² - 3 x 3) = log₂(5 x 3 - 15) ⇒ true

For x = 5: log₂(5² - 3 x 5) = log₂(5 x 5 - 15) ⇒ true

The solutions to the equation are : x = 3 and x = 5.

Hence F is irrotational field as Curl 

 Apply the logarithmic law, that is logax = xlog(a). Now the function is ln(1 + 5x)^-5 = -5log(1 + 5x). By taking the series = 

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.