CUET PG Mathematics Mock Test - 4 - CUET PG MCQ

Concept:
y_P = 1/f(D) sinax
f(D) = ϕ (D²)

Substitute D² = -a²
Calculations:

Concept:

(i) Cauchy's 1st limit theorem: Let (an) be a sequence of positive terms then if = l then = l

(ii) Let ∑ a_n, ∑ bn be two series of positive terms such that an ≤ bn then

• If ∑ b_n is convergent then so is ∑ a_n.

• If ∑ a_n is divergent then so is ∑ b_n.

(iii) p- series test: is convergent if p > 1 and divergent if p ≤ 1

Explanation:

=

Then by using Cauchy's 1st limit theorem,

= = 0

So (an) is convergent

(4) is false

Now,

=

=

i.e., <

Now, is divergent by p-series test

So is divergent.

(2) is correct

Concept:
The volume of a solid Revolution: Disks and waster.
If a region in the plane is revolved about a line in the same plane, the resulting objects is shown as 'solid of revolution'.
The line about which we rotate the shape is called the "axis of revolution".
(1) Disk method
This is used when we rotate a single curve, y = f(x) around x (or) y-axis.
Assume y = f(x) is a continuous non-negative function in the interval [a, b]


The volume of solid formed by revolving the region bounded by curve and the x axis between x = a and x = b about x-axis is

The cross-section perpendicular to the axis of revolution has the form of a disk of radius R = f(x).
Similarly, we fund the volume of solid when region bounded by x = f(y) and y-axis between y = c and y = d, and rotated about y-axis.



Calculation:
Given function and a = 0, b =
Volume



= 76π/3

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6,0), (6, 8), and (0,5).
F = 4x + 6y

Maximum of F – Minimum of F = 72 - 12 = 60
The Correct option is (1).

Concept:

Method of finding the maxima & minima of z = f(x, y):

Step 1: Find for the given function

Step 2: Equate p & q to zero for obtaining stationary points (a, b).

Step 3: Calculate r, s, t at each stationary point (a, b).

Step 4:

i. If rt – s² > 0 & r > 0 then f (x, y) has a minimum at (a, b) & the minimum value is f(a,b).

ii. If rt – s² > 0 & r < 0 then f (x, y) has a maximum at (a, b) & the maximum value is f(a,b).

iii. If rt – s² < 0 then f (x, y) has neither a maximum nor a minimum value at (a, b) & (a, b) is a saddle point.

iv. If rt – s² = 0 & r < 0 then no conclusion by this method.

Calculation:

Given

f(x, y) =

Now, and

Putting and solving two equations we get

(x, y) = (a, a) or (-a, a)

Now at (a, a); ,

Hence, rt – s² = 3 > 0 and r > 0, hence it has a minimum value at (a, a)

Now, at (a, -a); ,

Hence, rt – s² = -5 < 0, hence it has neither minimum nor maximum at this point (a, -a)

Hence, (a, -a) is a saddle point

∴ Minimum value is, f(a, a) = 3a².

Concept:

Number of group homomorphisms from a cyclic group G of order n to a finite group G' of order m is,

and α_k is the number of elements of order k in G'

Let has the cycle decomposition {n1, n2...,n_k}, let a_i denote the number of cycles of length i in {n₁, n₂...,nk}.

Then, number of the elements similar to are .

Calculations:

The order of ℤ2 is two. So, k =1,2.

No. of elements of order 1 in S₅ is 1.

Elements of order 2 in S₅ are of the type (ab)(cd)(e) or (ab)(c)(d)(e).

Elements of the type (ab)(cd)(e) are

Elements of the type (ab)(c)(d)(e) are

No. of elements of order 2 in S5 is 15+10 = 25.

Thus, there are total 26 group homomorphisms from ℤ2 to S5.

Hence, option 3 is correct.

Concept:

Green's theorem: Let C be a positively oriented, piecewise smooth, simple closed curve in the plane and let D be a region bounded by C. If P(x,y) and Q(x,y) have continuous first partial derivative on an open region that contains D, Then

Explanation:

∫c[(x2015 y2016 + 2014y) dx + (x2016 y2015 + 2017x) dy]

Use green's theorem

∫ ∫_c (2016 x²⁰¹⁵ y²⁰¹⁵+ 2017 - 2016 x2015 y2015 -2014) dx dy

= 3 ∫ ∫_c ds

= 3 x area of circle

= 3π

Concept:
By stokes theorem:

Calculation:
Given:
Let ABCD be the given rectangle

By stokes theorem:


∴ î(0) – ĵ(0) + k̂(-2y – 2y)
∴ -4yk̂
n̂ = k̂, ds = dx dy


Consider a vertical strip, then the limits are
X-varies from -a to a & Y-varies from 0 to b

∴ -4ab²

Concept:
Finite Oscillatory Sequence: A bounded sequence which is not convergent is said to oscillate finitely
Solution:
Given, (an) be a sequence of real numbers defined by

Let b_n = a_n/n for n ∈ ℕ. Then
here, a_n is an oscillatory sequence that oscillates between -1 to 1
= which is bounded lies between - 1 to 1
and = 0
Hence, it is a convergent sequence and converges to zero.
Therefore, b_n is a convergent sequence but a_n is not a convergent sequence (3) is correct

Concept:

To find the orthogonal trajectories of any curve

Step 1: Find the differential equation of the family of circles

Step 2: Replace dy/dx with -dx/dy to get the differential equation of the orthogonal trajectories

Step 3: Solve the new differential equation

Explanation:

We have , − 1 < h < 1 .

Differentiating both side with respect to x:
⇒

from given

⇒

⇒

⇒

Replacing dy/dx by -dx/dy:

⇒

⇒

Hence Option(1) is the correct Answer.

The given equation is

This is a Bernoulli’s equation

Let

The region for the given constraint x > y² and y > x² is drawn as:

Limit of y: y = 0 to y = 1
Limit of x:
x = y² to x² = y
Or x = y to x = √y
Now, the volume under f(x, y) will be:

Concept:
Residue of f(z) at z = 0 is the coefficient of 1/z in the Maclaurin series expansion of f(z)
Explanation:
f(z) = exp (z + (1/z))
=
=
Hence the coefficient of 1/z in the above expression
=
=
Hence option (3) is correct

Given :-
The function f(z) is continuous at z = z₀ 
Concept used :-
If the function f(z) is analytic at z = z₀ then so it is continuous.
Solution :-
If the function f(z) is analytic at z = z₀ then so it is continuous. But converse is not always true
Which means if f(z) is continuous at is analytic at z = z₀

We know that some results about the subgroup
(i) H is proper normal subgroup of
(ii) H is infinitely generated subgroup of
(iii) H is non cyclic subgroup of
Hence option (3) is true.

Concept:
Normal subgroup:
If N is a normal subgroup of a group G, g ∈ G and n ∈ N, then the gng^-1 ∈ N'.
Cyclic group:
There exist a element in G which generates the group and the order of the element is same as order of the group.
Abelian group:
A group G with any binary operation satisfy the commutativity.
Explanation:
(1) This is true by definition.
For 'H' a normal subgroup of 'G', for every 'g' in 'G', the left coset 'gH' is equal to the right coset 'Hg'.
Thus, every left coset is also a right coset.
(2) This is false.
Subgroup 'H' can be a proper, nontrivial subgroup of 'G' and still be a normal subgroup.
The simplest example would be the group 'Z₂' inside the group 'Z₄'.
(3) This is false.
Normal subgroups need not be cyclic.
An example would be the Klein four-group K₄ inside the symmetric group S₃.
K4 is not cyclic, but it is a normal subgroup of S3 because it is of index 2.
(4) take a counter-example - Quaternion group Q₈ of order 8

The expansion of power series ∑ a_n(x - a)ⁿ having R as the radius of convergence and x, a be any two points in R, converges if |x - a| < R
(4) is correct

Concept:

L’Hospital’s Rule: If = = 0 or ± ∞ and g'(x) ≠ 0 for all x in I with x ≠ c and exist then =

Explanation:

(0/0 form so using L'hospital rule)

=

=

Again using L'hospital rule

=

=

It will be 0/0 form if

x - 2a = 0

⇒ a = 0

Option (1) is correct

Given:



For the given region, straight-line joining (0, 0) to (1, 1)

x = y ⇒ dx = dy


∴ The value (up to two decimal places) of a line integral is 0.667

Concept:

Divergence of the vector function:

The net outward flux from a volume element around a point is a measure of the divergence of the vector field at that point.

Divergence of a function

Calculation:

Given that,

Therefore, the divergence of a given vector is

-----(1)

Here,

f₁ = 6x², f₂ = 3xy², f₃ = xyz³

Therefore, from equation (1)

⇒ = 12x + 6xy + 3xyz²

Therefore, divergence at point (2, 3, 4)

= 12(2) + 6(2)(3) + 3(2)(3)(4)²

⇒ = 24 + 36 + 288 = 348

Hence, option 3 is correct.

Other Related Points

Gauss divergence theorem:

It states that the surface integral of the normal component of a vector function taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function taken over a volume enclosed by the closed surface ‘S’.

Concept:

For; sin h (z + 2π)

z + 2π = -(z + 2π)

z + 2π = -z - 2π

2z = -4π

The general solution is given by:

where k = 0i, 1i............ni

CONCEPT:
There are two cases by which we can determine the area between the curves by using definite integrals.
Case 1: The region is bounded between the curves y₁ = f(x)  and y₂ = g(x) on the interval [a, b] assume that f(x) > g(x).

The area of the shaded region is given by:
Case 2: The region is bounded between the curves x₁ = f(y) and x₂ = g(y) on the interval [a, b] assume that f(y) > g(y).

The area of the shaded region is given by:

• Note that in both cases it is important to find the point the intersection so that the value of a and b can be determined.

CALCULATIONS:
Curves given are y = x² and y² = x.
∴ y = x² ⇒ y = √x

On putting the value of y in the equation we get -

Again integrating the function and putting the x values


on solving this we get 6/35 as answer, i.e. option (a) is correct answer.

CONCEPT:
If we consider a function f(z)= u+iv then it will satisfies the Cauchy – Riemann equation if and if it possesses all the partial derivatives then the function f(z) is analytic.
Solution:
Let f(z) = u + iv
f(z) = 1/z
where z = x + iy
now we can write
So here
,
So from here we can say that the Cauchy – Riemann equation is satisfied and all the partial derivatives are exist except origin(0,0)
If we take w = 1/z  then dw/dz = - (1/z²) also exists everywhere at z=0
So, 1/z, is analytic at z = 0.

Concept:

Since the above function f(x) satisfy the assumption of Rolle’s theorem in interval [-1, 1]

∴ f(x) is continuous in [-1, 1]

F(x) is differentiable in (-1, 1)

F(-1) = f(1)

∴ for continuity

LHL = RHL = Functional value at 0 i.e f(0)

= 1

= q

∴ LHL = RHL

⇒ q = 1

Also for differentiability

LHD = RHD

at x = 0

at x = 0

p – q = 1

⇒ p = 1 + q

⇒ p = 1 + 1

⇒ p = 2

The above equations is in the form of

Mdx + N dy = 0

N = - x²y

Hence it is not an exact equation. So we have to reduce this equation to exact equation. For that we need to calculate integrating factor.

This is the function of ‘x’ only.

Multiplying the given equation by I.F. we can reduce in to exact equation.

Now the solution is,

Given:

z = e^ax+by f (ax - by)

On partial differentiation of z w.r.t x

On partial differentiation of z w.r.t y

On substituting we get

= 2abz

Concept:

The method of finding Maxima and Minima of y = f(x)

• Find f’(x) and f”(x) for given function y = f(x)
• Equate f’(x) to zero to obtain stationary points x = a
• Calculate f”(x) at each stationary points x = a (i.e f”(a))

• The following three conditions are obtained:

• If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)
• If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)
• If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a

Calculation :

f(x) = x3 − 3x2 + 2x in [1, 2]

f(x)= x(x² - 3x +2)

f(x)= x(x-2)(x-1)

Critical Points : 0,1,2

f'(x) = 3x² - 6x +2

f''(x) =6x - 6

f''(0) = -6 < 0 ⇒ maximum

f''(1)= 0

f''(2) = 6 ⇒ minimum

Maxima at x = 0

maximum value of f(x)

⇒ f(0) = 0

Hence correct option is "4"

Concept:

The divergence of any vector field is defined as:

The nabla operator is defined as:

Calculation:

Given:

vector

Divergence of u will be

f = y^x
ln f = x lny
⇒
⇒
⇒
= y^x−1 + xy^x−1lny
= 1⁽²⁻¹⁾ + [2×1⁽²⁻¹⁾ ln(1)] = 1

Concept:
For simple poles at z = a, b, c…
Residue of
For multiple poles at z = a, a, a … n times
{Residue of f(z)}z=a
According to Cauchy’s residue theorem,

Calculation:
Given,
The poles of f(z) = -1
it lies inside of the given contour.
For multiple pole (n = 4) at z = -1

Residue will be