HPSC PGT Mathematics Mock Test - 10 - HPSC TGT/PGT MCQ

The correct answer is Martial art and Folk art.

Key Points

• The annual Kartik Cultural Festival:
  • The festival is the joint effort of Haryana Tourism, Ministry of Tourism and Department of Culture, Government of India, Department of Youth Affairs and Sports, Department of Cultural Affairs, Development Commissioners Handlooms and Handicrafts, North Zone Cultural Center, North Central Cultural Center, Nehru Yuva Kendra Sangathan and Ballabgarh Development and Beautification Society.
  • The main aim of the Kartik festival is to promote fort ambience, martial arts both Indian classical music and dance, and a rich variety of folk theatre. Hence, Option 2 is correct.

Additional Information

• • The Kartik Cultural Festival is the result of the continuous effort of the Haryana Government.
  • The motive was to keep alive the art and heritage of India.
  • The festival is organized at Nahar Mahal which was once the residence of Raja Nahar Singh (ascended throne in 1829) who ruled Ballabgarh Town under the reign of the last Mughal King Bahadur Shah Zafar.
  • It was built around 1739 AD by the forefathers of King Nahar.

For this question, we will use the given table-

The relation of NZ : MA  is given below-

Similarly,

We will apply the same relation to TV : ? as-

Hence, Option (3) is correct.

NEP 2020 emphasizes play-based learning for children aged 3-6 to provide a strong foundation for future learning.

A teacher makes an attempt to understand the potentialities of his/her students it is related to Educational Psychology. Educational Psychology is considered as the branch of psychology that deals with the scientific study of human learning. The study of learning processes takes place from both behavioural and cognitive perspectives, allows experimenters to know about the individual differences in cognitive development, intelligence.
Hence, the correct answer is, 'Educational Psychology.'

The National Education Policy 2020 proposed various reforms in school education as well as higher education including technical education that has to be adopted by all educational institutions.

Key Points The ECCE program needs to be determined by children’s developmental and contextual needs, providing for more need-based inputs and an enabling environment.

• Ensuring quality ECCE - early childhood care and education for all children between 3-6 years.
• It was believed that a common ‘curriculum’ would not be appropriate for all.
• It comprises details of the goals for different domains of development, i.e. physical, language, cognitive, socio-emotional, creative, and aesthetic appreciation, to be fostered to ensure holistic development of children under six years.​

Hence we can say that the full form of ECCE is Early Childhood Care and Education.

Additional Information Other reforms of NEP 2020 include -

• Equitable and inclusive education - Special emphasis is given to Socially and Economically Disadvantaged  Groups (SDGs).
• Ensuring Universal Access at All Levels of schooling from pre-primary school to Grade 12.
• New Curricular and Pedagogical Structure (5+3+3+4).
• Setting up of State School Standards Authority (SSSA).

National education policy 2020:

• The Union Cabinet has approved the new National Education Policy (NEP), 2020 with an aim to introduce several changes in the Indian education system.
• The main goal of the school education regulatory system is to improve educational outcomes so that children can learn to implement the acquired knowledge and skills in their practical life.

Key Points

Experiential learning:

• Experiential learning refers to the learning by gaining experience through active involvement. 
• In all stages, experiential learning will be adopted, including  arts-integrated and sports-integrated education, story-telling-based pedagogy, etc.
• Pedagogical approaches proposed in experimental learning under NEP-2020 includes:
  • Art-integrated education:
    •    It is a cross-curricular pedagogical approach that utilizes various aspects and forms of art and culture as the basis for learning of concepts across subjects.
    • It will be embedded in classroom transactions not only for creating joyful classrooms, but also for imbibing the Indian ethos through integration of Indian art and culture in the teaching and learning process at every level. 
  • Sports-integrated education:
    • It is another cross-curricular pedagogical approach that utilizes physical activities including indigenous sports, in pedagogical practices to help in developing skills such as collaboration, self-initiative, self-direction, self-discipline, teamwork, responsibility, citizenship, etc.
    • Sports-integrated learning will be undertaken in classroom transactions to help students adopt fitness as a lifelong attitude and to achieve the related life skills.
  • Story-telling-based pedagogy:
    • It refers to the art of narrating a story more interactively to people who are willing to listen. It is used as a strategy in teaching which broadens learner's reading choices.
      In a storytelling based pedagogy, the story chosen by the teacher to narrate is directly or indirectly related to the lesson to develop their fullest potential, both educationally and morally.

   ​Hence, it could be concluded that all of the above are the pedagogical approach proposed in experimental learning under NEP-2020.

Graph of y = x [x]

Equation of tangent at (x₁, y₁) on the parabola y² = 4x is
2x - y₁ + 2x₁ = 0      
length of perpendicular from centre of the circle (x - 3)² + y² = 9 is


Hence required equation 

Number of sides of a cube = 6
Number of colour of a cube = 6
Number of ways of colouring =?
Consider a cube in 3D space 
Each of the sides towards East West North South and remaning two up and down 
Now fixing up positions down is also fixed 
Total number of ways 
= 6!/(4×6)
​= 5×3×2 
= 30 Ways
Hence, the correct answer is 30 Ways

We want to find the indefinite integral of |x| with respect to x:

∫ |x| dx

1. Consider x ≥ 0:

   • Here, |x| = x.
   • So, ∫ |x| dx = ∫ x dx = (x²)/2 + C₁

2. Consider x < 0:

   • Here, |x| = -x.
   • So, ∫ |x| dx = ∫ (-x) dx = - (x²)/2 + C₂

To combine these results into a single expression, we note that for an indefinite integral, constants C₁ and C₂ can be absorbed into a single arbitrary constant C. A concise way to write this is:

∫ |x| dx = (x · |x|)/2 + C

• For x ≥ 0, this becomes x·x/2 = x²/2.
• For x < 0, this becomes x·(-x)/2 = -x²/2.
• Combining both the results , we get x|x|/2 + C

A = {2, 3, 4, 8, 10}
B = {3, 4, 5, 10, 12}
C = {4, 5, 6, 12, 14}
► (A ∩ B) = {3,4,10}
► (A ∩ C) = {4}
► (A ∩ B) U (A ∩ C) = {3,4,10}

A₁,A₂,......, An are inserted between a and b then the series will become 
a, A₁,A₂,A₃,......, An,b. Now a becomes the first term, A₁ will be second, A₂ will become third term
An will become A(n+1)th term 
therefore A(n-1) will become nth term.

x² = 16 ⇒ x = ±4 and 2x = 6 ⇒x = 3

There is no value of x which satisfies both the above equations. Thus, A = φ

tan^-1[(3+2x)/(2-3x)]
=> tan^-1[2(3/2 + x)/2(1-3/2x)]
=> tan^-1[(3/2 + x)/(1-3/2x)]
=> tan^-1(3/2) + tan^-1(x)

Given pth term of HP = qr
So pth term of AP = 1/qr
a+(p−1)d = 1/qr....(1)
and qth term of HP = pr
so qth term of AP = 1/pr
a + (q−1)d = 1/pr.....(2)
subtracting equation 1 and 2 we get, 
(p−q)d = (p−q)/pqr
d = 1/pqr
Now from equation 1, 
a = 1/qr − (p−1)/pqr
= (p−p+1)/pqr = 1/pqr
So rth term of AP = a+(r−1)d = 1/pqr + (r − 1)/pqr = 1/pq 
So, rth term of HP = pq

First we fix the alternate position of the girls. Five girls can be seated around the circle in (5−1)!=4! , 5 boys can be seated in five -vacant place by 5!
∴ Required number of ways =4!×5!

The given ellipse is 

Hence, the eccentricity  e1, of the hyperbola  is given by
1 = e₁ sinθ ⇒ e₁ = cosec θ ⇒ b² = sin²θ (cosec²θ - 1) = cos²θ

|kA| = kⁿ|A|
|3A| = (3)ⁿ |A|
= 27|A| = k|A|
k = 27

As (1 – p) is root of the equation: x² + px + (1 – p) = 0

(1 – p)² + p(1 – p) + (1 – p) = 0

(1 – p)[1 – p + p + 1] = 0

(1 – p) = 0

p = 1

Therefore, given equation now becomes

x² + x = 0

x(x + 1) = 0

x = 0, -1

Given, lines can be written as

and 
∴ Requires SD =

Where 

∴ Requires SD 

So, f has a local minima at 2 and a local maxima at - 2 .

sin² (θ - α) cos α = cos² (θ - α) sin α = m sin α cos α

The point (2 cosθ, 2 sinθ) lies on the circle x² + y2 = 4. From the figure, it  is  obvious that 



Hence no solution

Let the roots be α, β : β, γ and γ, α then
αβ = b, βγ = c and γα = a