HPSC PGT Mathematics Mock Test - 4 - HPSC TGT/PGT MCQ

Trapezoid, Rhombus, and Parallelogram are the 2 dimensional figure.

While;

Prism is a 3-dimensional figure.

Hence, "Prism" is the correct answer.

Important Points

"Use of formula for area-measurement should be taught before estimation of areas" is NOT correct in the context of teaching area- measurement of plane figures.

Measurement of plane figures: A plane, or flat, surface has two dimensions—length and width. The perimeter of a plane figure is the distance around it.

In the context of teaching area- measurement of plane figures correct statements:

• Students should be given a chance to 'discover' the various formulae.
• Grid paper can be a useful device for introducing and implementing the formal units of measurement.
• Tessellations with regular shapes can be useful for teaching area measurement.

In Cartesian coordinate system Shortest distance between the lines

Because, the element b in the domain A has no image in the co-domain B.

Let, sin^-1(0.6) = A…………(1)
  ( Since, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) )

A³ = I ⇒  Pre - multiplying both sides by A⁻¹,A⁻¹, A³ = A⁻¹ I ⇒ A² = A⁻¹

Centre of the ellipse is the intersection point of 
x+y−1=0.........(1) 
x−y=0............(2)
Substituting x from equation 2 in equation 1 two equations, we get,
2y=2,   y=1 
Replacing, we get x=1
⇒(1,1) is the centre

Distributive law is given by : 

The given matrix is a skew – symmetric matrix.,therefore , A = - A’.

The x-axis touches at A(1, 0) and x = y touches at B(1, 1). Hence the equation to the curve through these points is given by y(y – x) + k(x – 1)² = 0. For this to represent a parabola, 4k = 1. The equation is x² – 4xy + 4y² – 2x + 1 = 0. Vertex  focus

T_(r+1) =  ⁿC_r x^n-r a^r
T_(r+1) =  ^n-5Cr xn-5-r (1/x⁴)^r
= ^n-5C_r x^n-5-r (1/x^4r)
= ^n-5C_r x^n-5-5r 
=> n - 5 - 5r = 2r
=> n - 2r = 5(r + 1)

Since g is continuous at a , therefore , if g (a) > 0 , then there is a nhd.of a, say (a-e , a+ e) in which g (x) is positive .This means that f ‘ (x)>0 in this nhd of a and hence f (x) is increasing at a.

, which does not exist at x = 2 . However , we find that  , at x = 2 . Hence , there is a vertical tangent to the given curve at x = 2 .The point on the curve corresponding to x = 2 is (2 , 0). Hence , the equation of the tangent at x = 2 is x = 2

Δ = 4(a² + c² - ac) = 2(a2 + c² + (a - c²)) > 0

 Let 2 tan^-1⁡x = y

Let  E₁ : Head appears
E₂ : Tail appears
A : A reports that head appears

∴ Rwquired probability = 
By Bayes’ Theorem

2a/2be^y = 2b/2ce^y = 2c/2de^y
= a/be^y = b/ce^y = c/de^y = k
a/b = b/c = c/d = ke^y = k'
a=bk', b=ck',  c = dk'
b = (dk')k'
b = d(k')²
a = bk' = d(k')² k'
= d(k')³
a,b,c,d ⇒ d(k')³, d(k')², d(k'), d
G.P. sequence

Let (x₁ y₁) be point on 
chord to 

Area formed by the lines = 4ab

Second quadrant is from 90° to 180°
As we know that cos90° = 0, cos180° = -1
So the value of cos is decreasing from 0 to -1

LHD = limh→0 ((−h)² sin(−1/h) − 0)/−h
limh→0 −h² sin(1/h)/−h
= limh→0 h × sin(1/h)
= 0
RHD = limh→0 (h² sin(1/h) − 0)/h
= limh→0 h × sin(1/h)
= 0
RHD=LHD
∴ f(x) is differentiable at x= 0

 Radius of circle is given by -
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)²+(y-k)²=r²
equation is (x-2)² + (y-1)² = (√37)²
x² + 4 - 4x + y² + 1 - 2y = 37
x² + y² - 4x - 2y - 32 = 0

Let A be the set of students who have passed in Mathematics, and B be the set of students who have passed in Physics.
We are given that |A| = 55 and |B| = 67, where |A| and |B| represent the number of students in sets A and B, respectively.
We are also given that there are 100 students in total. We need to find the number of students who have passed in Physics only, which means we need to find |B - A|.
First, let's find the number of students who have passed in both Mathematics and Physics, which can be represented by |A ∩ B|.
Using the principle of inclusion-exclusion, we have:
|A ∪ B| = |A| + |B| - |A ∩ B|
Since there are 100 students in total, we can say that |A ∪ B| = 100.
Now, we can find |A ∩ B|:
100 = 55 + 67 - |A ∩ B|
100 = 122 - |A ∩ B|
|A ∩ B| = 22
Now, we can find the number of students who have passed in Physics only, which is |B - A|:
|B - A| = |B| - |A ∩ B|
|B - A| = 67 - 22
|B - A| = 45
so, the correct answer is 45.

a, b are roots of x² – 5x + 8 = 0 
a + b = 5, ab = 8:

∴ required equation is
x² – (9/8)x + 1 = 0  
⇒  8x² – 9x + 8 = 0 

Let (h,k) be the midpoint of the chord 7 x +y -1 = 0

Represents same straight line 

⇒ Equation of the line joining (0,0) and (h,k)  is y - x = 0.

 Let α,β be roots of ax³+bx+c=0
γ,δ be roots of px²+2qx+r=0
α/β = γ/δ …………(1) and  
β/α = δ/γ ………….(2)
(1) + (2)
⇒ α/β + β/α = γ/δ + δ/γ
= [(2²+β²)/αβ + 2]= [γ²+δ²]/γδ + 2
⇒ [(α)²+(β)²+2αβ]/αβ ​= [γ²+δ²+2γδ]/γδ
​= [(α+β)²]/αβ = [(γ+δ)²]/γδ
⇒ (4b²/a²)/(c/a) = (4q²/p²)/(r/p)
⇒ b²/ac = q²/pr.

A = {1, 2} and B = {5, 6, 7} 
A x B = {(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)} 

Number of words beginning with A = 4! = 4 × 3 × 2 = 24
Number of words beginning with N = 4! = 4 × 3 × 2 = 24
Number of words beginning with R = 4! = 4 × 3 × 2 = 24
Number of words beginning with U = 4! = 4 × 3 × 2 = 24
So, in total 96 words will be formed while beginning with letter A, N, R and U.
Now, 97th word according to dictionary order will be VANRU
Now, 98th word according to dictionary order will be VANUR
Now, 99th word according to dictionary order will be VARNU
Finally, 100th word according to dictionary order will be VARUN.
Hence, option (c) is the correct answer.