AWES PGT Math Mock Test - 1 - AWES TGT/PGT MCQ

• The Archaeological Survey of India (ASI) have stated excavations in Benisagar – which is a village in Jharkhand’s West Singhbhum district. The excavation has yielded hundreds of artifacts indicating continuous human habitation from the fifth century A.D. until the 16-17 century A.D.
• Benisagar has been recognized by the ASI and listed among the top 100 significant archaeological monuments in the country.

In education process aim, content and methods are evaluated.

Evaluation is concerned with assessing the effectiveness of teaching, teaching strategies, methods and techniques. It provides feedback to the teachers about their teaching and the learners about their learning.

Learning disabilities in Mathematics can be assessed most appropriately by the diagnostic test. 
A diagnostic test is a test that is used to measure the level of understanding a student possess regarding any subject area. Teachers conduct diagnostic tests for reading and maths skills and use these results to provide remedial instructions or classes to students as per their level.
Hence, the correct answer is, 'Diagnostic Test.'

Key Points 

• The curricular and pedagogical structure and the curricular framework for school education will therefore be guided by a 5 + 3 + 3 + 4 design, consisting of the stages:
  • Foundational (3 years of preschool + Grades 1-2):- The Foundational Stage will consist of play/activity-based learning. It will also include a focus on good behavior, courtesy, ethics, personal and public hygiene/cleanliness, teamwork, cooperation, etc. 
  • Preparatory (Grades 3- 5):-  The Preparatory Stage will transition gradually from play-based learning to more formal but interactive classroom learning, with the introduction of some (light) textbooks, in order to lay a solid groundwork across subjects, including reading, writing, speaking, physical education, art, languages, science, and mathematics.
  • Middle (Grades 6-8):- The Middle Stage will see the introduction of subject teachers for learning/discussion of the more abstract concepts in each subject that students will be ready for at this stage, across the sciences, mathematics, arts, social sciences, and humanities.
  • High school (Grades 9-12 in two phases, i.e. 9 and 10 in the first and 11 and 12 in the second) stages respectively, with an option of exiting at Class 10 and re-entering in the next phase:- The High School (or Secondary) Stage will comprise of four years of multidisciplinary study, building on the subject-oriented pedagogical and curricular style of the Middle Stage, but with greater depth, greater critical thinking, greater attention to life aspirations, and greater flexibility and student choice.

Thus, it is concluded that Primary classes (Classes I to V) will consist of  Preparatory and  Foundational Stages according to NEP, 2020.

Schools act as an agent of socialization through the hidden curriculum. According to sociologists, the hidden curriculum is an unintended result of education. It is the process by which the norms and values of a society are taught to children, alongside the expectations of their particular society.

Childhood refers to the period of the human lifespan, ranging from birth to puberty. It consists of many developmental stages and 'adolescence' is one of them.

• 'Adolescence' comes from the Latin word 'Adolescere' which means 'to grow to become mature'. It a stage which lies between the age of '12 to 19 years'.
• Emotion is a mental state associated with fear, anger, love, etc. Emotional states of an individual are the combination of physiological arousal, psychological appraisal, and behavioral changes.
• Emotionally disturbed children can be defined in terms of certain observable characteristics such as withdrawn behavior, failure to achieve a level reasonably commensurate with ability or tendency towards fighting, and other aggressive behavior.

​Key Points

Emotional disturbances of adolescents:

• They show the same emotions as they used to show in their childhood  i.e., fear, disgust, but the emotions of adolescence are different in their patterns.
• The emotions at this stage are more uncontrolled, intensive, and irrational.  As they gradually grow, their emotional behavior improves and become mature.
• Children who suffer from emotional disturbances generally do nail-biting, thumb sucking, depression, pulling or stretching of hairs, fear of specific objects or things, etc.
• A child wearing a well-fitted dress is not a symptom of emotional disturbance but wearing proper clothes that fit them well makes them confident.

Hence, it could be concluded that wearing a well-fitted dress is not found in the symptoms of emotional disturbances of adolescents.

Apply , C₁ → C₁ - C₂, C₂ → C₂ - C₃,

Because here row 1 and 2 are identical

2a=10,a=5
2b=8,b=4
b²=a²(1−e²)
e²=1−16/25
e=3/5
Focus of ellipse (±3,0)
Taking (3,0) as centre, circle is drawn,
Radius of circle will be =5−3 = 2 units.

He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.
1 friend can be selected out of 6 in 6C1 = 6 ways
2 friends can be selected out of 6 in 6C2 = 15 ways
3 friends can be selected out of 6 in 6C3 = 20 ways
4 friends can be selected out of 6 in 6C4 = 15 ways
5 friends can be selected out of 6 in 6C5 = 6 ways
6 friends can be selected out of 6 in 6C6 = 1 ways
Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63

If AB = BA = I , then A and B are inverse of each other. i.e. A is invers of B and B is inverse of A.

f (x + 2y) = f (x) + f (2 y) + 4xy for x, y ∈ R putting x = y = 0, we get f (0) = 0

a = cos α + i sin α, z = iy, y ∈ R
-(cos α + i sin α) y² + iy + 1 = 0
⇒ y² cos α = 1, sin α = y
y sin α = 1, sin cey ≠ 0
∴ sin² α =  cos α, cos² α+ cos α - 1 = 0

Lets first place the men (M). '*' here indicates the linker of round table
 
* M -M - M - M - M *
which is in (5-1)! ways
So we have to place the women in between the men which is on the 5 empty seats ( 4 -'s and 1 linker i.e * )
So 5 women can sit on 5 seats in (5)! ways or
1st seat in 5 ways
2nd seat 4
3rd seat 3
4th seat 2
5th seat 1
i.e 5*4*3*2*1 ways
So the answer is 5! * 4! = 2880

(1/1² + 1/2² + 1/3² + ........∞) = π²/6
⇒ (1/1² + 1/3² + 1/5² + ........∞) + (1/2² + 1/4² + 1/6² + ........∞) = π²/6
=> (1/1² + 1/3² + 1/5² + ........∞) + 1/2²(1/1² + 1/2² + 1/3² + ........∞) = π²/6
=> (1/1² + 1/3² + 1/5² + ........∞) + 1/2²(π²/6) = π2/6
=> (1/1² + 1/3² + 1/5² + ........∞) + π²/24 = π²/6
= π²/8

And y has a relative minimum at x = 0 if f ‘(0)+a = 0 . a =0.

n (P) = 5, n(Q) = 12 and n(PUQ) = 14
n(PUQ) = n(P) + n(Q) - n(P∩Q) 
14 = 5 + 12 - n(P∩Q)
n(P∩Q) = 3

The lady can select one cotton saree out of 15 cotton varieties in 15 ways since
any of 15 varieties can be selected. Corresponding to each selection of a cotton saree, she can
choose a polyester saree in 13 ways. Hence the two sarees (one cotton and one polyester), by
multiplication principle of counting, can be selected in 15 x 13= 195 ways

sin²x = 1 - cos²x
∫sinx(sin²x - 3cos²x + 15)dx
Put cos²x = t
 ∫sinx(1 - cos²x - 3cos²x + 15)dx
=  ∫sinx (16 - 4cos²x)dx
Put t = cosx, differentiate with respect to x, we get 
dt/dx = -sinx
= -  ∫ [(16 - 4t²)]^1/2dt
= -2 ∫ [(2)² - (t)²]^½
= -2{[(2)² - (t)²]^½ + 2sin^-1(t/2)} + c
= - cosx {[4 - (cos)²x]^½ - 4sin^-1(cosx/2)} + c

lim xsecx    x=0
lim x(1/cosx) = lim (x/cosx)
Put x = 0
= 0/cos0
= 0/1 = 0

Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

Lower triangular matrix is given by : 
 ,

here , a_ij = 0
if i is less than j.and a_ij ≠ 0, if i is greater than j.

She has 6 friends and he wants to invite one or more. That is the same as saying he wants to invite at least 1 of his friends.
 
So, the number of ways he could do this is:
Invite only one friend
Invite any two friends
Invite any three friends
Invite any four friends
Invite any five friends
Invite all six friends
This can be thought of in terms of combinations. Inviting  r  friends out of  n  is same as choosing  r  friends out of  n . So, we can write the possibilities as:
⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ 
= 6 + 15 + 20 + 15 + 6 + 1 
= 63

ar⁶/ar² = 16
r⁴ = 16
r⁴ = (2)⁴
r = 2
S₉ = a(rⁿ - 1)
2555 = a((2)⁹ - 1)
2555 = a(512-1)
2555/511 = a
a = 5

(x+a)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^(n-1) a¹ + ⁿC₂ x^(n-2) a² + ..........+ ⁿC_(n-1) xa^(n-1) + ⁿC_n  aⁿ
Now, ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n-1,    ⁿC₂ = ⁿC_n-2,........
therefore, ⁿC_r = ⁿC_n-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

Option d is correct, because it is the property of definite integral
 ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

 has a local maximum value = - 2 at x = - 1 and a local minimum value = 2 at x = 1.

As x R y if x + 2y = 8, therefore, domain of the relation R is given by x = 8 – 2y ∈ N.
When y = 1, 
⇒ x = 6 ,when y = 2, 
⇒ x = 4, when y = 3, 
⇒ x = 2.
therefore domain is {2, 4, 6}.

2.6.10.14... to n factors = (2n)!/n! 
= 2(1.3.5.7………………25)
2.6.10.14... to 50 factors = (2*2*25)!/25! 
 = 100!/25!

By formula 

Cos 2A = Cos²A-Sin²A

Cos²A-Sin²A=Cos²A-(1-Cos²A) =Cos²A -1 +Cos²A=2 Cos²A - 1

so, Cos 2A = 2 Cos²A - 1