TN TRB PG Assistant Mock Test- 1 (Mathematics) - TN TET MCQ

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

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)

Concept:

The divergence of any vector field is defined as:

The nabla operator is defined as:

Calculation:

Given:

vector

Divergence of u will be

Concept:
Cauchy's theorem:
If f(z) is an analytic function and f'(z) is continuous at each point within and on a closed curve C, or if the simple closed curve does not contain any singular point of f(z) then,

Analysis:

f(z) exists for all values of z inside the closed curve C.
∴ f(z) is analytic and also f'(z) is continuous at all points.

Importatnt Point
Cauchy's integral formula:
If f(z) be an analytic function within an on the closed curve C and let 'a' is a point inside C then,

We can solve the given problem by this method as follows:
z₀ is enclosed within the closed curve C

∴ f(z) = (z - z0)2
The value of the integral is
2πi f(z0) = 2πi (z0 - z0)2
= 0

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)

The function f(x) is defined piecewise as:
1. f(x) = ,
2. f(x) = 0 , if x = 0 ,
3. f(x) = 1 , if x = 1
At x = 1/2 :
x = 1/2 lies in the interval (1/3) ≤ x < 1/2 for r = 2
Thus, f(x) = (-1)² = 1
The left-hand limit ( LHL ) as x → 1/2^-  and the right-hand limit ( RHL ) as x → 1/2⁺ do not match because the function is piecewise and discontinuous at x = 1/2
Therefore, f(x) is discontinuous at x = 1/2
On the Interval (1/2, 1) :
For x ∈ (1/2, 1), f(x) is defined piecewise over disjoint intervals
Within each interval, f(x) is constant and hence continuous
Therefore, f(x) is continuous on (1/2, 1)
On the Entire Interval [0, 1] :
At x = 0, f(x) = 0 , which matches the limit as x → 0⁺ . Thus, f(x) is continuous at x = 0
At x = 1, f(x) = 1 , which matches the limit as x → 1^-. Thus, f(x) is continuous at x = 1
However, f(x) is discontinuous at points like x = 1/2 , x = 1/3 , etc. Hence, f(x) is not continuous on [0, 1]
Verifying Statements:
Statement A: f(x) is continuous at x = 1/2 . Incorrect, as f(x) is discontinuous at x = 1/2
Statement B: f(x) is continuous on [0, 1] . Incorrect, as f(x) is discontinuous at points like x = 1/2 , x = 1/3 , etc
Statement C: f(x) is discontinuous at x = 1/2 . Correct, as LHL RHL at x = 1/2
Statement D: f(x) is continuous on (1/2, 1) . Correct, as f(x) is piecewise constant and continuous within each interval in (1/2, 1)
The correct statements are C and D Only
Hence Option (4) is the correct answer.

Concept:
Let us assume a determinate A , whose elements are given below:
A =  ,
then co-factor of the element a₃₁ of determinate A is given by:
 C₃₁ = (−1)³ + 1 = 

Solution:
Given, determinate A =

Cofactor of 7 for the determinate A is given by;
C₃₁ = (−1)³ + 1 | −2 4
1 0
⇒ (−1)⁴ × (−2 × 0 − 1 × 4) = −4

Concept:
A function is written in the form of the ratio of two polynomial functions is called a rational function.
Rational functions are continuous at all the points except for the points where the denominator becomes zero.
f(x) = P(x) / Q(x)
where P(x) and Q(x) are polynomials and Q(x) ≠ 0.
f(x) will be discontinuous at points where Q(x) = 0.
Calculation:
Given:
f(x) = (4 - x²) / (4x - x³)
This is a rational function, so it will be discontinuous at points where the denominator becomes zero.
4x - x³ = 0
x(4 - x²) = 0
x(2² - x2) = 0
x(2 + x)(2 - x) = 0
x = 0, x = - 2 and x = 2
Hence the function f(x) = (4 - x²) / (4x - x³) will be discontinuous at exactly three points 0, - 2 and 2.
Mistake Points
There may be a doubt that some factors are eliminating each other so first, we have to simplify this.
Note that,
If (4 - x2) = 0 then the f(x) will come indeterminant or 0/0 form.
So, the function will not have any value for x = ± 2. So, these will also be the points of discontinuity.
Also, if (4 - x2) ≠ 0
⇒ f(x) = (4 - x²) / (4x - x³) = 1/x
Here, x = 0 is also the point of discontinuity.
There will be exactly three points 0, - 2 and 2.

By a well known result we know that if A be an n × n matrix with coefficients in F, having rows {a₁, a₂, ....., a_n }, then the following statements are true. 

 (a) if A’ be a matrix obtained from A by an elementary row operation (interchanging two rows). Then 

 D(A’) = – D(A) 

 (b) if A’ be a matrix obtained from a by an elementary row operation (replacing the row a_i by λa_j , with λ ∈ F, i ≠ j). Then 

D(A’) = D(A) 

(c) if A’ be a matrix obtained from A by an elementary row operation (replacing ai by µai , for µ ≠ 0 in F). Then 

 D(A’) = µD(A)

i.e. all the three options are correct. 

(I) FALSE: But trace of LHS = RHS.
(II). TRUE
(III). FALSE: Inverse of a matrix exists only iff determinant of the matrix in non- zero.
(IV). TRUE

Calculation:
Given:
Y = (x + 2)(x – 1) (x + 3)
Y = (x² - x + 2x - 2)(x + 3)
Y = x³ + 3x² + x² + 3x - 2x - 6
Y = x³ + 4x² + x - 6
dy/dx = 3x² + 8x + 1

When origin O is inside S. In this case, divergence theorem cannot be applied to the region V enclosed by S, since  has a point to discontinuity at the origin. To remove this  difficulty, let us enclose the origin by a small sphare Σ of radius ε. 

The function F is continuously differentiable at the points of the region v´ enclosed between S and Σ. Therefore applying divergence theorem for this region V´, we have 

Now on the sphere Σ, the outward drawn normal n is directed towards the centre. Therefore on Σ, we have 

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 = 

Condition 1: 

(It is Homogeneous)

Condition 2: 

Equation (1) can be written as 

It is not a linear form.

It is in linear form.

Condition 3: 

So, it is an exact equation.

Consider 

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.

2014 starts with a Wednesday

1234567

WTFSSM T

So if Wednesday is day n = 1 then Sundays will come on days n = 5,12,19,…96, total 15 Sundays.

Similarly if we start with-

Thursday - 14 sundays 

Friday - 14 sundays 

Saturday - 15 sundays 

Sunday - 15 sundays 

Monday - 14 sundays 

Tuesday - 14 sundays

So, probability 

⇒ P = 2/7

Let y = mx


it depends on m.
⇒ limit does not exists
Hence Option (4) is the correct answer.

Concept:
To find the value of f(a,b) we need to find the critical points and then by second derivatives test we can check Maxima or Minima
Given:
f(x, y) = 12xy e−(2x + 3y − 2
Explanation:




Similarly,




Now To Find Critical Points Set Partial Derivatives to Zero
1.
2
Since, we get the conditions:
1. y = 0 or 1 - 2x = 0 .
2. x = 0 or 1 - 3y = 0 .
Solving these equations:
1. From 1 - 2x = 0, we get x = 1/2.
2. From 1 - 3y = 0, we get y = 1/3.
Thus, one of the critical points is.
Evaluate f(x, y) at


Conclusion:
The value of f(a, b) at the local maximum point (a, b) = is 2.
Hence Option (1) is correct.