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

The correct answer is 10.

• The Common Services Centers (CSC), also called as Atal Seva Kendras in Haryana.
• There are ten Atal Seva Kendra in Bhiwani district in the urban region while 300 rural Atal Seva Kendra is also there.
• Common Services Centers (CSC) scheme is a mission mode projects under the Digital India Programme.

Notes:

• Important demographics of Bhiwani district:

The correct answer is Bansilal.

• Bansi Lal Legha founded Haryana Vikas Party.
• He became the third Chief Minister of Haryana.
• He took an oath of Haryana chief minister for the first time on 31 May 1968.
• He became Chief Minister four times in 1968, 1972, 1986 and 1996.
• He was prominently known as the architect of modern Haryana.

Additional Information

• Om Prakash Chautala is a former Chief Minister of Haryana.
  • He served as the Chief Minister of Haryana from 1989 to 2005 at four different times.
• Bhupinder Singh Hooda is an Indian National Congress politician.
  • He also served as the Chief Minister of Haryana from 2005 to 2014.
  • He served for 9 years as chief minister.

The correct answer is Jahaj Kothi.

Key Points

• Jahaj Kothi is an Ancient Ship shaped building located in Firoz Shah Complex. It was originally Jain Temple which later became the residence of George Thomas (1756–1802) and James Skinner (1778–1841).
• George Thomas, a native of Ireland and the uncrowned ruler of the territory between Sirsa and Rohtak built this building as his residence.

Additional Information

• Gujri mahal: The Gujari Mahal was built by Firoz Shah Tughlaq for his beloved Gujri Rani, a native of Hisar with whom he fell in love during the course of one of his hunting expeditions. 
• Devi Bhawan: One of the most important religious places to visit in Hisar is Shri Devi  Bhawan Mandir. This temple was constructed on the branch of West Jamuna canal in 1770 by Maharaja Amar Singh of Patiala.

According to the sequence in the dictionary: 

The correct order of the given words is:

2. Arrange

1. Arrest

3. Arrogant

4. Arrow

Hence, the correct answer is "2, 1, 3, 4".

Additional Information

1. First of all, check the common letter in all.

2. Arrange them according to the alphabet they come first. 

3. When common letters are arranged, then go for the rest.

The family chart is as follows:

i) A is brother of B
ii) B is brother of C
iii) C is husband of D
iv) E is father of A
Drawing the family tree as per the given information:

Thus we can see D is the Daughter - in - law of E.
Hence, the correct answer is "Daughter - in - law".

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).

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 

Two vectors A and B are said to be collinear , if they are parallel to the same line irrespective of their magnitudes and directions.

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.

If θ is the acute angle between

then cosine of the angle between
these two lines is given by :


Here, 


Then, 

x² = 16 ⇒ x = ±4 and 2x = 6 ⇒x = 3

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

Option (a) :- for x = nπ
let n=1
so, x= 1 x π =π
therefore , sinx + icos2x
sinπ + icos 2π
0 + i= i
Now,
cosx - isin2x
cosπ - isin2π
or, -1 -0 = -1
hence no conjugate
Option (b) :- x = 0
sin0 + icos0
0+i= i
now ,
cos0 - isin0
= 1
hence no conjugate.
option (c) :- x = (n+1/2)π
if n= 1
x=3π/2
sin3π/2 + icos3π
-1 - i
Now,
cos3π/2 - isin3π
0-0= 0
hence no conjugate
Thus option D is correct

Sin 7π = Sin 7*180 = Sin 2π * 7  = 0

Total number of students =10
There are 7 boys and 3 girls.
Seven boys can be arrange in a row  7P7 = 7!ways.
So, 8 places in which we can arrange 3 girls are 8P3 = 8!/(8−3)! = 336ways
The number of arrangement is 7!×336 = k * 8!
= 42

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

f(x)={(a²−ax+x²)−(a²+ax+x²)^1/2}/{(a+x -(a-x)^1/2)^1/2}
on rationalising, we get
lom(x → 0) a² - ax + x² -(a² +ax + x²) * (a+x)^1/2 + (a-x)^1/2]/[(a+x)-(a-x) * (a² - ax + x²)^1/2 + (a² + ax + x²)]
lim(x → 0) [-2ax * [(a)^1/2 + (a)^1/2]]/[2x * (a²)^1/2 * (a²)^1/2]
-a/(a)^1/2 = -(a)^1/2​

A = {1, 2, 3, 4}, B = {3, 4} and C = {2, 3} 
(A ∩ B) = {3, 4} C = {2, 3}
(A ∩ B x C) = {3,4} x {2,3}
⇒ {(3,2) (3,3) (4,2) (4,3)}

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

Put

sin^-1 x = θ ⇒ x = sin θ

f(x) = 3x + 1,  g(x) = x² + 2 
f-g = (3x+1) - (x² + 2)
= 3x + 1 - x² - 2
= 3x - x² -1

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



Hence no solution

Matrix C is given in Transpose form. So, Matrix C is 3x1 order.

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

Correct Answer :- a

Explanation : Semi latus rectum of ellipse = one half the last rectum

b²/a = 1/3*2a

b² = 2a²/3

b = (2a/3)^1/2...........(1)

So, b²/a = a(1-e²)

b² = a²(1-e²)

Substituting from (1)

2a²/3 = a²(1-e²)

e² = 1-2/3

e² = 1/(3)^1/2