HPSC PGT Mathematics Mock Test - 5 - HPSC TGT/PGT MCQ

The correct answer is Jind.

Important Points

• Pratap Singh revolted against the British in Jind, Haryana.
• Raja Bhag Singh suffered a severe paralytic attack in March 1813. Unfit to run the administration of his state, the ailing chief wished to appoint Prince Pratap Singh the ablest and wisest of all his sons as his regent to do his work.
• But the British government to whom the anti-British bearing of the prince was known stood in his way and got Rani Sobrahi appointed in place of the price in 1814. This was unbearable for Pratap Singh and he raised the standard of revolt on June 23, 1814.
• Pratap being a popular figure the state forces also revolted and joined him forthwith. With their help, the prince lost no time in occupying the Jind fort and established his government after putting the Rani the puppet of the British government to the sword.

Additional Information

• Kaithal revolt was started in the year 1843 and its leader was Gulab Singh and Suraj Kaur.

The correct answer is ​Sirajudaula.

Key Points

Battle of Plassey -

• It is a battle fought between the East India Company force headed by Robert Clive and Siraj-Ud-Daulah (Nawab of Bengal).
• The rampant misuse by EIC officials of trade privileges annoyed Siraj.
• The continuing misconduct by EIC against Siraj-Ud-Daulah led to the battle of Plassey in 1757.
• In 1757, when Sirajudaula was defeated in the Battle of Plassey, East India Company decided to build a new fort, one that could not be easily attacked.
• Hence the correct answer is option 3.
• Calcutta had grown from three villages called Sutanati, Kolkata and Govindapur.
• The Company cleared a site in the southernmost village of Govindapur and the traders and weavers living there were asked to move out.
• Around the new Fort William, they left a vast open space which came to be locally known as the Maidan or garer-math.

Additional Information

•  
• Mir Qasim Nawab of Bengal was defeated in the Battle of Buxar.
• Mir Jafar became the Nawab of Bengal after the Battle of Plassey, he gave the British to become the Nawab.

Logic: The alphabet that comes first alphabetically, comes first in the dictionary.

So, the correct arrangement should be:

2. Subtle.

4. Sucres.

3. Sudoku.

1. Sugary.

5. Sullen.

Hence, the correct answer is "2, 4, 3, 1, 5".

The first figure is a triangle, then there are 3 line segments inside it.

Similarly,

Circle, then there should be 3 line segments inside it.

The figure is shown below:-

Hence, Option (2) is the correct answer.

The diagonal elements of a skew-symmetric is zero.

Write a³x² + abcx + c³ = 0  as 

Given,  x²/5 + y²/3= 1
We know S₁F₁ × S₂F₂ = b²
∴ S₁F₁ × S₂F₂ = 3

Hence (C) is the correct answer.

If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then , the directions cosines of the line are given by :

Correct Answer :- a

Explanation : The formula to find the coefficient of a^pb^qc^rd^s in (xa+yb+zc+vd)ⁿ is (n!x^p y^q z^r v^s)/p!q!r!s!

So, coefficient of a⁴b³c²d in (a−b+c−d)¹⁰

= [10!(1)⁴(-1)³(1)²(-1)1]/(4!3!2!1!)

= 12600

(fof) (x) = f[f(x)] = f(x² - 3x + 2)² - 3(x² – 3x + 2)
= y² – 3y + 2 = (x² – 3x + 2)² – 3(x² – 3x + 2) + 2 =  (x⁴ – 6x³ + 10x² – 3x)

Point should be on y axis so x=0
y²+8y+5 = 0
y²+8y+16 =5+16
(y+4)² = 21
y+4 = (21)^½
y = -4 +- (21)^½
A = (0, 4+(21)^½)       B=(0, 4-(21)^½)
AB= [(0-0)² + {(4+(21)^½)  - (4-(21)^½)}]
= (2(21)^½)
= 2(21)^½

⇒ |r₂ - r₂| is minimum when x is minimum.
|r₂ - r₂| = 1

sec^-1x + cosec^-1x =​ π/2 ; x belongs to [ -1 , 1 ]

4x²−9y²−8x=32
⇒4(x²−2x)−9y2 = 32
⇒4(x²−2x+1)−9y² = 32 + 4 = 36
⇒(x−1)²]/9 − [y2]/4 = 1
⇒a²=9, b²=4
∴e=[1+b²/a²]^1/2 
= [(13)^1/2]/3

Because, f(- x) = f(x) is the necessary condition for a function to be an even function, which is only satisfied by x²+ sin²x .

2001 = 3 x 23 x 29 and (3+23+29)
= 55 ⇒ a = 3, b = 23, c = 29

The given series is a geometric series in which a=2,r=3,l=4374.
Therefore,
Required sum = (lr−a)/(r−1)
​= (4374×3−2)/(3−1)
​= 6560

• We know that, 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25
• Therefore the set A = {1, 4, 9, 16, 25...} can be written in set builder form as: 
  A = {x: x is the square of a natural number}

Equation of tangent to parabola y = mx- am² .......(1) equation of chord of hyperbola whose midpoint is (h, k) is  hx - ky = h² - k² ...... (2) form (1) and  (2) 

The given curves are : (i) y = x – 1 , x > 1 . (ii) y = - (x – 1) , x < 1. (iii) y = 1 these three lines enclose a triangle whose area is : 1/2 .base.height = 1/2 .2 .1 = 1 sq. unit.

∫tan(log x) dx
log x = t
x = e^t
dx = e^t dt
f(t) = ∫tan t dt
f’(t)= sec² t
= sec²(log x)