UP LT Grade Teacher Mathematics Mock Test - 4 Free Online Test 2026

The police are responsible for enforcing the laws and maintaining law and order in a country.

Roots are mainly of two types: tap root and fibrous roots.
Plants having leaves with reticulate venation have tap roots while plants having leaves with parallel venation have fibrous roots.

• Mahatma Gandhi worked as an editor in Indian Opinion.
• The newspaper existed between 1903 and 1915 and was a tool to fight racial discrimination and win civil rights for the Indian immigrant community in South Africa.
• The newspaper was published in English & Gujarati and for a short period in Hindi & Tamil.
• The other newspapers associated with Mahatma Gandhi are Navjeevan, Young India, and Harijan.
• The Leader was founded by Madan Mohan Malviya and C.Y. Chintamani was the editor.

Hence, the correct option is (D).

Given : 

A monoid is a group if 

Explanation: 

A group is a set and a binary operation such that. For all x, y ∈ G, x * y ∈ G (closure).

There exists an identity element e ∈ G with x * e = e * x = x for all x ∈ G (identity).

For all x , y , z ∈ G we have (x * y) * z = x * (y * z) (associativity).

There exists an Inverse element c such that x * c = e = c * x for all x ∈ G (Inverse)

Now,

Monoid ⇒ Associative binary operation and identity

∴ A monoid(B,*) is called Group if to each element there exists an element c such that (a*c)=(c*a)=e. Here e is called an identity element and c is defined as the inverse of the corresponding element.

∴ option 1 is correct 

Calculation:

Given

Applying substitution method of integration,

let x = sin(u)-------(1)

Differentiating equation (1) both sides with respect to u we get,

dx = cos(u) du

Now substituting the value of x and dx in the integral we get,

= 

We know, the trigonometric identity as,

sin²(u) + cos²(u) = 1

cos²(u) = 1 - sin²(u)

Substituting above identity in above integral we get,

= 

Formula used:

1. cot θ = Base / Perpendicular
2. sin θ = Perpendicular / Hypotenuse
3. cos θ = Base / Hypotenuse

In a right-angle triangle,

Hypotenuse² = Base² + Perpendicular²

Alternate Method

Concept:

The domain is the range of inverse trigonometric functions.

Calculation:

Given: f(x) = sin-1 (x + 1) 

⇒ y = sin-1 (x + 1) 

The given function can be written as sin y = x + 1

We know that sin y is a function that oscillates between -1 and 1 i.e it

can take all real numbers as a value in the interval [-1 , 1]

So, -1 ≤ sin y ≤ 1

⇒ -1 ≤ x + 1 ≤ 1

⇒ -1 - 1 ≤ x + 1 - 1 ≤ 1 - 1

⇒ -2 ≤ x ≤ 0

∴ x ∈ [-2, 0]

Concept:

Greatest Integer Function (GIF) gives the greatest integer less than or equal to the given value of x.

Formula:

If I  ≤ x ≤ (I + 1), then [x] = I where I is an integer

Calculation:

Concept:

Let second-degree equation be,

ax2 + by2 + 2hxy + 2gx + 2fy + c = 0

Discriminant is calculated as:

Δ = 

Δ = abc + 2fgh - af2 - bg2 - ch2

Case: 1

If discriminant (Δ) of this equation doesn’t equal to zero (Δ ≠ 0)

• If h2 – ab > 0, it represents a hyperbola and a rectangular hyperbola (a + b = 0).
• If h2 – ab = 0, it represents a parabola.
• If h2 – ab < 0, it represents an ellipse. (a ≠ b)
• If h2 – ab < 0, it represents a circle. (a = b)

Case: 2

If discriminant (Δ) of this equation is equal to zero (Δ = 0)

• If h2 – ab > 0, it represents two distinct real lines or pair of perpendicular straight lines
• If h2 – ab = 0, it represents a parallel lines.
• If h2 – ab < 0, it represents non-real lines.

Calculation:

Given: Second degree equation, 12 x2 - 10 xy + 2y2 + 11 x - 5y + λ = 0

Compare with second-degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

So, a = 12, h = - 5, b = 2, g = 5.5, f = -2.5, c = λ

The given equation represents the pair of straight lines so;

Δ = 0

abc + 2fgh - af2 - bg2 - ch2 = 0

24λ + 137.5 - 75 - 60.5 - 25λ = 0

λ = 2

Given:

sin2 β - sin 30° = 0

Concept used:

⇒ sin 30° = 1/2

⇒ sin 45° = 1/√2

Calculations:

⇒ sin2β - sin 30° = 0

⇒ sin2 β = sin 30° 

⇒ sin2 β = 1/2

Square rooting both sides,

⇒ sinβ = 1/√ 2

⇒ sinβ = sin 45°

∴ The value of β is 45°.

Consider the parallelogram ABCD as in figure. The diagonal AC is common to both the triangles ABC and ADC.

Given $AB=\overrightarrow{a}andBC=\overrightarrow{b}.$ In a parallelogram, the opposite sites are equal and parallel and hence,

According to Triangle law, in ∆ABC, the diagonal

Also for the ∆ADC,

Concept used:

Integral test:

Let f(x) be positive decreasing continuous function such that f(n) = a_n

then series a_n is Convergent iff f(x) dx converges

Calculations:

(n) (say);

ϕ (n) is +ve and decreases as n increases. So let us apply the integral test.

 dt {t = x², dt = 2xdx}

Given:

Formula Used:

2cos^-1x = cos^-1(2x² - 1)

Concept:

Length of latus rectum of an ellipse  is given by 

Distance between the foci of the ellipse is given by 2ae.

Calculation:

Given that distance between the foci = 4√2

⇒ 2ae = 4√2

Given the length of latus rectum is 4.

Substituting the value of ,

We know eccentricity of the ellipse is given by,

Concept:

Arithmetic mean

Consider any two numbers, say m and n

And P is the arithmetic mean between two numbers.

The sequence will be m, P, and n in A.P.

P – m = n – P

 = (Sum of the numbers)/(number of terms)

Geometric mean

If a, b and c are three quantities in GP, then and b is the geometric mean of a and c.

This can be written as b² = ac or b =√ac.

Harmonic Mean

Harmonic mean is calculated as the reciprocal of the arithmetic mean of the reciprocals.

The formula to calculate the harmonic mean is given by:

Harmonic Mean = n[(1a)+(1b)+(1c)+(1d)+...]

Calculation:

Given:

AM = A, GM = G, And HM = H

Let a and b be the two distinct positive real numbers

Then , G = √(ab),

Explanation:

Equilibrium of Concurring Forces

• Any number of forces acting upon a point or particle are said to be concurring forces. If the resultant of any system of concurring forces is zero, the forces are said to be in equilibrium.
• The effect of the resultant force is to cause acceleration of the particle. When the forces are in equilibrium, then, the particle has no acceleration, and if the particle is at rest it is said to be in static equilibrium.
• But if the resultant force is zero, all the concurring forces must evidently reduce to two equal and opposite forces, or, what is the same thing, any one of the forces must be equal and opposite to the resultant of all the others.
• Hence the algebraic sum of the components of all the forces in any three rectangular directions must be zero.
• We have then, for the conditions of equilibrium of concurring forces
• F_x = ΣF cos α = 0, F_y = ΣF cos β = 0, F_z = ΣF cos γ = 0.

We obtain, then, the following obvious results from the condition for equilibrium of concurring forces, which will be found useful in special cases :

1. If two concurring forces are in equilibrium, they must be equal in magnitude and opposite in direction.
2. If three concurring forces are in equilibrium, they must all act in the same plane. For the resultant of any two must act in their plane and be equal and opposite to the third.
3. If three concurring forces are represented in magnitude and direction by the sides of a triangle taken the same way round, the resultant is zero and the forces are in equilibrium.
4. Hence, if three concurring forces are in equilibrium, each one is proportional to the sine of the angle between the other two.
5. If three concurring forces are in equilibrium and their directions are represented by the sides of a triangle taken the same way round, their magnitudes will also be represented by the sides of that triangle, and rice versa.

Concept:

System of equations

a₁x + b₁y = c₁

a₂x + b₂y = c₂

For unique solution

For Infinite solution

For no solution

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Calculation:

Given:

5x + 7y = 2, 15x + 21y = μ

Here For no solution

$\frac{5}{15}=\frac{7}{21}\ne \frac{2}{\mu }$

Hence for no solution μ ≠ 6

Given:

A is a square matrix and B is non-singular matrix, so B is invertible and a square matrix

Calculations:

det(B^-1 AB)

= det(B^-1 BA)

= det(I_nA) (where I_n is an identity matrix of order n)

= det(A)

∴ option C is correct 

Concept:

Property of determinant of a matrix:

• Let A be a matrix of order n × n then det(kA) = kn det(A)

• If A and B are two square matrices then |AB| = |A||B|

• The determinant of a Matrix with two Identical rows or columns is equal to 0. 

Calculation:

Given:

(x + y)[-3x + 3y] - (y + z)[-3z + 3y] + (z + x)[-3z + 3x]

⇒ -3x² + 3xy - 3xy + 3y² + 3yz - 3y² + 3z² -3yz - 3z² +3xz - 3xz + 3x²

Δ = 0

Concept:

If two vectors are perpendicular, then their dot product is zero i.e. 

CALCULATION:

Given: The wo vectors  are perpendicular,

As we know that, if two vectors are perpendicular, then their dot product is zero i.e. 

⇒ (3i + 4j – 7k) . (pi – 6j + 3k) = 0

3p – 24 – 21 = 0

3p = 45

p = 15

Hence, option D is the correct answer.

Concept:

For two vectors  to be collinear,​  where λ is a scalar.

Concept:

For a square root function given by f(x) = √x to have real values, the radicand x must be positive

or equal to zero.

⇒ x ≥ 0

Solution:

By using above concept,

x - 1 ≥ 0 and 6 - x ≥ 0

⇒ x ≥ 1 and x ≤ 6

⇒ 1 ≤ x ≤ 6

∴ The domain of the given function is [1, 6].

Total number of balls = (5 + 4 + 11) = 20 balls

S = Sample space that one ball is drawn out of the box

∴ n(S) = 20

Let E = event that the ball drawn is neither black nor pink = event that the ball drawn is yellow

Thus, n(E) = 4

∴ Probability of occurrence of event;