UP TGT Mathematics Mock Test - 2 - UPTET MCQ

The Himachal Pradesh Cricket Association (HPCA) stadium, Himachal Pradesh becomes the first BCCI-accredited venue with a groundbreaking ‘hybrid pitch’ by Netherlands-based SISGrass. This innovation combines natural turf with 5% polymer fiber, endorsed by the ICC for T20 and 50-over matches. The pitch promises resilience, durability, and consistent bounce, easing maintenance burdens. The Universal machine, employed in Dharamshala, will replicate this technology in Ahmedabad and Mumbai, following successful installations at iconic English grounds like Lord’s and The Oval.

Fish was the royal emblem of Pandya Kingdom. The Pandyas were also mentioned by Megasthanese who said that then kingdom was famous for pearls.

The correct answer is 'Kuka Movement'.

Key Points

• The Kuka movement is also known as the Namdhari mission.
• The Kukas were also referred Namdharis, who were a sect within Sikhism.
• They started as a group for religious purification in Sikhism.
• The aim of the movement was to restore the Sikh rule in Punjab and oust foreign powers.
• The Kukas wore only white, hand-woven clothes and boycotted British education, products, and laws.
• Baba Ram Singh started the Kuka Movement.

Additional Information

• Khilafat Movement
  • ​The Khilafat Movement (1919–24) was a pan-Islamist political protest campaign launched by Muslims in British India.
  • Its aim was to restore the caliph of the Ottoman Caliphate, promote Muslim interests, and bring the Muslims into the national struggle.
• Radhaswami Movement
  • ​Radhaswami Movement was founded in 1861 by Tulsi Ram.
  • Tulsi Ram was also popularly known as Shiv Dayal Saheb or Swamiji Maharaj.
  • The Radhaswamis believe in one Supreme Being, the supremacy of the Guru, a “company of the pious people” (Satsang), and simple social life.

CONCEPT:

As we know that the equation of motion for the t^th second is written as;

S_t = u + g(2t-1)

Here S is the distance, u is the initial velocity, a is the acceleration and t is the time.

CALCULATION:

According to the t^th second, the equation of motion is 

St = u + g(2t-1)   -----(1)

Here the initial velocity of the freely falling body is equal to zero, i.e. u = 0 

The equation (1) becomes,

St = 0 + g(2t-1)

⇒ S_t =  g(2t-1)   -----(2)

Now for t = 1 second, the equation (2) becomes,

 S₁ = g(2-1) 

⇒ S₁ =  g    -----(3)

 For t = 2 seconds, the equation (2) becomes,

 S₂ = g(4-1) 

⇒ S₂ =  g    -----(4)

Now for t = 3 seconds, the equation (2) becomes,

 S₃ = g(6-1) 

⇒ S₃ =  g    -----(5)

 For t = 4 seconds, the equation (2) becomes,

 S₃ = g(8-1) 

⇒ S₃ =  g    -----(6)

Now to calculate the ratio of the distances traveled by a freely falling body in the 1^st, 2nd, 3rd, and 4th second we divide the equations (3), (4), (5), and (6) we get;

Hence, option 4) is the correct answer.

Concept:

Projectile motion: 

• When a particle is projected obliquely near the earth surface, it moves simultaneously in horizontal and vertical directions. This type of motion is called projectile motion.

Where, u = projected speed

θ = angle at which an object is thrown from the ground.

g = acceleration due to gravity = 9.8 m/s2

Calculation:

Formula used:

• 2 sin2x = 1 - cos2x + C
• ∫ sin x dx = - cos x + C
• ∫cos x dx = sin x + C

Calculation:

Given that,

⇒  = sin x cosec x - 2 sin²x

⇒  = 1 - 2 sin2x

⇒  = 1 - 2 sin2x

Integrating both sides with respect to x, we get,

⇒  = ∫( 1 - 2 sin2x)

Explanations:

Group Homorphisms :

Let (G,o) & (G’,o’) be 2 groups, a mapping “f ” from a group (G,o) to a group (G’,o’) is said to be a homomorphism

if –f(a o b) = f(a) o' f(b) for all a, b ∈ G

∴ option 1 is correct 

Concept:

Identities:

• i² = -1
• i³ = -i
• i⁴ = 1
• i⁴ⁿ = 1
• i⁴ⁿ⁺¹ = i
• i⁴ⁿ⁺² = -1
• i⁴ⁿ⁺³ = -i

Alternate Method

Concept:

Euler's formula states that for any real number x,

Where i = √-1

The differential equation is:

The above equation can be written as:

Then solution is

x.e^y = ∫ ye^y dy + c

x.e^y = (y - 1) e^y + c

so x = (y - 1) + ce^-y

Curve passing through (0, 0) so,

0 = 0 – 1 + c ⇒ c = 1

Now, x = (y - 1) + e^-y

If curve passes through (α, 1), then

Given:

In the given figure, A tangent AC touches the circle (with centre at O) at point B. If ∠DBC = 30∘, then find the value of ∠BOD.

Concept used:

Alternate segment theorem:

As per the above figure, the angle (θ) between a tangent and a chord is equal to the angle (θ) in the alternate segment.

The angle subtended by an arc at the centre is twice to that of the angle subtended by it at any point on the remaining part of the circumference of the circle.

Calculation:

According to the question, the required figure is:

By using the Alternate segment theorem:

∠DEB = ∠DBC = 30^∘ 

Since the angle subtended by an arc at the centre is twice to that of the angle subtended by it at any point on the remaining part of the circumference of the circle,

Therefore, 

⇒ ∠DOB = 2∠DEB

⇒ ∠DOB = 2 × 30∘ 

⇒ ∠DOB = 60∘ 

∴ The required angle is 60^∘,

Concept:

The geometric mean is defined as the nth root of the product of n numbers.

The geometric mean of a data set  is given by: 

Calculation:

To Find: Geometric mean of 6, 8, 16 and 27

Here n = 4

Now,

Given:

∫  = gof(x) + C

Solution:

Putting x3 = (x3/2)2 and then putting x3/2 = t, √xdx = dt

⇒ I = ∫  

After substitution -

I = ∫ 

⇒ I = sin-1(t) + C

⇒ I = sin-1(x3/2) + C

On comparing with gof(x) + C

⇒ g(x) = sin-1(x) and f(x) = x3/2

Given:

x² + y² = π²

y = sin x

Calculation:

We have,

⇒ x² + y² = π² is a circle of radius π and center at the origin.

Required area = Area of the circle (1st quadrant) - 

⇒ Required Area =   

⇒ Required Area = 

⇒ Required Area = 

⇒ Required Area = 

⇒ Required Area =  sq. units

∴ The required area is sq. units.

Given:

PQ = 25 cm, PA = 17 cm and PB = 12 cm

Concept used:

Tangents drawn from the same external point are equal in length.

Calculation:

PQ = PR = 25 cm

Hence, AQ = 25 - AP = 25 - 17 = 8 cm

Now, BR = 25 - BP = 25 - 12 = 13 cm

B being an external point, RB = BM = 13 cm

A being an external point, AQ = AM = 8 cm

Hence, AB = AM + BM = 8 + 13 = 21 cm

∴ The length of AB is 21 cm.

Concept:

Calculation:

Given:

y = e3x + e-5x 

At x = 0

we have

Concept:

The equation of parabola having their vertices at the origin and foci on the x-axis is 

Calculations:

The equation of parabola having their vertices at the origin and foci on the x-axis is

               ....(1)

Differentiating on both side, we get

⇒ 

⇒ 

Put this value in equation (1), we get

⇒ 

⇒ y(y - 2xy') = 0, Here y' = 

∴ y-2xy' = 0

Hence,  the differential equation to the family of parabolas having their vertices at the origin and foci on the x-axis is y - 2xy' = 0

Concept:

A function f(x, y) is said to be homogeneous of degree n in x and y, it can be written in the form

f(λx, λy) = λn f(x, y)

Euler’s theorem:

If f(x, y) is a homogeneous function of degree n in x and y and has continuous first and second-order partial derivatives, then

If z is homogeneous function of x & y of degree n and z = f(u), then

The correct answer is: nu

Given:

D, E and F finish the puzzle in a competition with the probability of ¾, 4/9 and 2/5 respectively.

Concepts used:

Probability that at least one of them will finish the competition = 1 - P (No one will finish the competition)

Calculation:

D, E and F finish the puzzle in a competition with the probability of ¾, 4/9 and 2/5 respectively.

⇒ P (No one will finish the competition) = {1 – (3/4)} × {1 – (4/9)} × {1 – (2/5)} = 1/4 × 5/9 × 3/5 = 1/12

Probability that at least one of them will finish the competition = 1 - P (No one finishes the competition)

⇒ Probability that at least one of them finishes the competition = 1 – (1/12) = 11/12

∴ Probability that at least one of them will finish the competition is 11/12.

Concept:

L’Hospital’s Rule:- 

L’Hospital’s Rule tells us that if we have an indeterminate form  or  all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

Given:

Concept used:

Trace of the Matrix: Sum of the diagonal Elements of the Matrix.

Calculations:

$A=\left[\begin{array}{cc}2& 4\\ 3& 2\end{array}\right]$

The diagonal elements are 2, 2 

⇒ Sum of the diagonal elements is 2 + 2 = 4 

∴ Trace = 4 

Given:

If G is a Prime order group of p

Concept used:

Lagranges theorem:

The order of subgroups divides the Order of Group but if some m divide the order of the group that doesn't mean that group has the subgroup of order m 

Proper subgroup :  H ≠ e and H ≠ G is the subgroup of G

Calculations:

G is a Prime order Group of p 

so, it has only 2 divisors 1 and itself that divides the order of Group 

⇒ Only 2 subgroups are generated by 1 and p

⇒ <1> = G and <p> = {e} where e is the identity 

∴ option 1 is correct, No proper Subgroup 

Given :

  and  

Formula used :

The harmonic mean of p & q         

Calculations :

Let   and 

⇒  and 

Using the above concept, the harmonic mean of x & y is

Given:

PQ = 24 cm and RS = 10 cm

MN = 7 cm

(ON - OM) = 7 cm

Concept used:

The perpendicular drawn from the center on a chord bisects it in two equal parts.

A² - B² = (A + B) (A - B)

Radius2 = (chord/2)2 + (Distance between chord and center)2

Calculation:

Hence, PM = 12 cm and RN = 5 cm

OP = OR (Radii of the circle)

Considering ΔPOM,

OP² = OM² + 12²      ....(1)

OR² = ON² + 5²      ....(2)

From (1) & (2),

OM2 + 122 = ON2 + 52

⇒ ON2 - OM2 = 119

⇒ (ON + OM) (ON - OM) = 119

⇒ (ON + OM) = 17      ....(3)

and (ON - OM) = 7      ....(4)

From (3) & (4),

ON = 12

Hence, OR = √(12² + 5²) = 13 cm

∴ The length of the radius of the circle is 13 cm.

Concept:

The length of the tangent from an external point P (x₁, y₁) to the circle represented by the equation: x² + y² + 2gx + 2fy + c = 0 is given by:

Calculation:

Given: x2 + y2 - 4x + 2y + 3 = 0 and Let P = (1, - 3)

By substituting x = 1 and y = - 3 in the expression x2 + y2 - 4x + 2y + 3 , we get

⇒ x2 + y2 - 4x + 2y + 3 = 1² + (-3)² - 4 - 6 + 3 = 3 > 0.

So, the point P lies outside of the given circle.

As we know that, the length of the tangent from an external point P (x1, y1) to the circle represented by the equation: x2 + y2 + 2gx + 2fy + c = 0 is given by: 

Here, x₁ = 1, y₁ = - 3, g = - 2, f = 1 and c = 3.

So, the length of the tangent from the point P to the given circle: 

Concept:

Any number proportional to direction cosines is called direction ratio.

The angle between two two lines is given by

→ ,

where, (a1, b1, c1) = direction ratios of line 1

and (a2, b2, c2) = direction ratios of line 2

The equation for the angle bisector for ax2 + 2hxy + by2 = 0 is given by

Given:

x2 + 2pxy - y2 = 0 and x2 - 2qxy - y2 = 0 bisect angles between each other.

Calculation:

Given :

If f = {(1, 2)} and g = {(2, 3)} 

Concept used:

Mapping Diagram consists of two columns in which one denotes the domain of a function f whereas the other column denotes the Range.

Example:  f = {(2, -5), (3, 4)} means 2 maps to -5 and 3 maps to 4 

Calculations:

gof(1) 

⇒ g(2) because f(1) = 2

⇒ g(2) = 3

∴ gof(1) = 3 

∴ option 2 is correct.