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AWES TGT Maths Mock Test - 3 - AWES TGT/PGT MCQ


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30 Questions MCQ Test AWES TGT Mock Test Series 2024 - AWES TGT Maths Mock Test - 3

AWES TGT Maths Mock Test - 3 for AWES TGT/PGT 2024 is part of AWES TGT Mock Test Series 2024 preparation. The AWES TGT Maths Mock Test - 3 questions and answers have been prepared according to the AWES TGT/PGT exam syllabus.The AWES TGT Maths Mock Test - 3 MCQs are made for AWES TGT/PGT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for AWES TGT Maths Mock Test - 3 below.
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AWES TGT Maths Mock Test - 3 - Question 1

The diagram given below shows that 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 1

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

AWES TGT Maths Mock Test - 3 - Question 2

In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals:

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 2

cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1

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AWES TGT Maths Mock Test - 3 - Question 3

The number of ways in which 8 different flowers can be strung to form a garland so that 4 particular flowers are never separated is

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 3

It is asked to find out the number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated.

 

According to the question, there are 8 flowers, but it is given that four particular flowers are never separated. So four particular flowers will always remain adjacent to each other.

To solve this problem we should know the following cases of circular permutations:

  • If clockwise and anti-clockwise orders are different, the total number of circular permutations is: (n-1)!.
  • If anti-clockwise and clockwise orders are taken as the same, then the total number of circular permutations is given by (n-1)!/2!

It is important to know which formula is applicable in this case.

When we turn a garland it forms a different arrangement. But it still is the same garland.
For example, consider a set of numbers like ‘1234’. When turned around it looks like ‘4321’ but it still is the same permutation of numbers.
In a similar way, the garland when turned might have a different permutation but it still is the same garland.
Therefore we’ll use the 2nd formula.
For n distinct objects in which clockwise and anticlockwise orders are taken as the same, the total number of circular permutations is: (n-1)!/2
Here in this case, given that four flowers must be together in all cases.
So we’ll club together all four particular flowers and consider this as one object.
Now we have 5 different objects - four flowers and four particular flowers clubbed as one object. Now we have to arrange these five objects in five circular seats.
In that case, the number of permutations is (5-1)!
And in every permutation, the clubbed flowers can be arranged in 4! ways.
So there are a total of (5-1)! * 4! Permutations.
But we have seen that the garland is the same in anticlockwise and clockwise order.
Therefore for every permutation, we have included its rotated permutation.
So the total number of circular permutations of garlands is : (5-1)!*4! /2

(5-1)! * 4! /2 = (4*3*2*1)*(4*3*2*1)/2
= 24 * 12 = 288

Thus the total number of ways in which 8 different flowers can be seated to form a garland so that four particular flowers are never separated is 288.

AWES TGT Maths Mock Test - 3 - Question 4

Correct form of distributive law is

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 4

Distributive law is given by : 

AWES TGT Maths Mock Test - 3 - Question 5

If the function  is continuous at x = 0 then a = 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 5

AWES TGT Maths Mock Test - 3 - Question 6

If the tangent at (3, –4) to the circle x2 + y2 - 4x + 2y - 5 = 0 cuts the circle x2 + y2 + 16x + 2y + 10 = 0 in A and B then the mid point of AB is

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 6

S1 = 0 and S1 = S11

AWES TGT Maths Mock Test - 3 - Question 7

The curve described parametrically by x = t2 + t + 1, y = t2 - t + 1 represents

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 7

x = t2 +t +1andy = t2 -t +1
⇒ x + y - 2 = 2t2 and x - y = 2t
⇒ 2(x + y -2) = (x - y)2 ⇒ x2 + y2 -2xy -2x -2y + 4 = 0
Comparing with the equation
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, 
∴ abc + 2fgh - af2 - bg2 - ch2 ≠ and h2 = ab

AWES TGT Maths Mock Test - 3 - Question 8

How many three digit odd numbers can be formed by using the digits 1,2,3,4,5,6 if the repetition of digit is not allowed:

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 8

There can be only 3 digits at the last digit (1,3,5) the middle digit should have 5 No's. out of 6, because one digit is used in last digit similarly 1st digit will have 4 No's.

Therefore : 4×5×3 => 60

AWES TGT Maths Mock Test - 3 - Question 9

The number of common tangents of the circles x2 + y2 – 2x – 1 = 0 and x2 + y2 – 2y – 7 = 0

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 9



Internally touch ∴ common tangent is one.

AWES TGT Maths Mock Test - 3 - Question 10

What is the value of 2 tan-1⁡x?

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 10

 Let 2 tan-1⁡x = y

AWES TGT Maths Mock Test - 3 - Question 11

Equation of chord AB of circle x2 + y2 = 2 passing through P(2, 2) such that PB/PA = 3, is given by

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 11

Any line passing through (2, 2)  will be  of  the  form 
When this line  cuts  the  circle  x2+y2=2 , (rcosθ+2)2 +(rsinθ+2)2 = 2 
⇒ r2 + 4(sinθ + cosθ)r+6 = 0
now if r1 = α, r2 = 3α then 4α
=  - 4(sinθ + cosθ), 3α2 = 6 ⇒ sin2θ = 1⇒ θ = π/4 .
So required chord will be y – 2  = 1 (x –2) ⇒ y = x.
Alternative solution 
PA.PN = PT2 = 22 + 22 - 2 = 6
PB/PA = 3
From (1) and (2), we have PA = √2, PB = 3 √​2
⇒ AB = 2 √​2
Now diameter of the circle is 2 √​2(as radius is √​2) 
Hence line passes through the centre ⇒ y = x .
Hence (B) is the correct answer. 

AWES TGT Maths Mock Test - 3 - Question 12

The equation of the line passing through the centre and bisecting the chord 7x +y -1 = 0 of the ellipse 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 12

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.

AWES TGT Maths Mock Test - 3 - Question 13

The value of 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 13




AWES TGT Maths Mock Test - 3 - Question 14

Tangents are drawn from the points on the line x – y + 3 = 0 to parabola y2 = 8x. Then the variable chords of contact pass through a fixed point whose coordinates are

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 14

Let (k,k+3) be the point on the line x−y+3=0
Equation of chord of contact is S1=0
⇒yy1=4(x+x1)
⇒y(k+3)=4(x+k)
⇒4x−3y−k(y−4)=0
Therefore, straight line passes through fixed point (3,4)

AWES TGT Maths Mock Test - 3 - Question 15

tan 2x =

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 15

tan2x = tan(x+x)
(tanx + tanx)/(1 - tan2x)
= 2tanx/(1 - tan2x)

AWES TGT Maths Mock Test - 3 - Question 16

If the ratio of the roots of ax2 + 2bx + c = 0 is same as the ratio of the roots of px2 + 2qx + r = 0 then

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 16

 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
= [(222)/αβ + 2]= [γ22]/γδ + 2
⇒ [(α)2+(β)2+2αβ]/αβ ​= [γ22+2γδ]/γδ
​= [(α+β)2]/αβ = [(γ+δ)2]/γδ
⇒ (4b2/a2)/(c/a) = (4q2/p2)/(r/p)
⇒ b2/ac = q2/pr.

AWES TGT Maths Mock Test - 3 - Question 17

Find the projection of the vector 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 17

AWES TGT Maths Mock Test - 3 - Question 18

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 18

AWES TGT Maths Mock Test - 3 - Question 19

Evaluate: 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 19

None

AWES TGT Maths Mock Test - 3 - Question 20

If S is the sum of infinity of a G.P. whose first term is `a', then the sum of the first n terms is

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 20

Given that first term =a
S= sum to infinity of a G.P= a/(1−r)
r=1−a/S
Sn=a(1−rn)/(1-r)
Sn=S(1−rn)
Sn=S[1−(1−a/S)n]

AWES TGT Maths Mock Test - 3 - Question 21

The function 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 21

AWES TGT Maths Mock Test - 3 - Question 22

 is equal to 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 22

Apply , C1→C1 - C3, C2→C2-C3

= 10 - 12 = -2

AWES TGT Maths Mock Test - 3 - Question 23

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 23

(x+iy)(x−iy) = (a+ib)(a−ib)/(c+id)(c−id) 
⇒x2−iy2 = √[(a2−i2b2)/(c2−i2d2)]
⇒x2+y2 = √[(a2+b2)/(c2+d2)]  [1i2 = -1]
(x2+y2)2 = (a2+b2)/(c2+d2)

AWES TGT Maths Mock Test - 3 - Question 24

If z1 + z2 + z3 = 0 and |z1| = |z2| = |z3| = 1, then value of equals

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 24





AWES TGT Maths Mock Test - 3 - Question 25

If three positive real numbers a, b and c (c > a) are in Harmonic Progression, then log (a + c) + log (a – 2b + c) is equal to:

(2010)

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 25

Since a, b, c are in H.P.

Now log (a + c) + log (a – 2b + c)
= log [(a + c){(a + c) – 2b}]
= log [(a + c)2 – 2b (a + c)]
= log [(a + c)2 – 2 × 2ac]
= log (a – c)2 or log (c – a)2
= 2 log (a – c) or 2 log (c – a)
∴ log (a + c) + log (a – 2b + c) = 2 log (c – a)

AWES TGT Maths Mock Test - 3 - Question 26

Let f be a differentiable function satisfying the condition for all 
, then f ' (x) is equal to

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 26

 replacing x and y both by 1, we get


AWES TGT Maths Mock Test - 3 - Question 27

The mean deviation of the following data 14, 15, 16, 17, 13 is:

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 27

Here N= 5 , sigma x = 75
so mean = 15
now taking deviation from mean,( By ignoring signs)
we get sigma deviation from mean = 6
Now applying the formula of mean deviation
M.D.= SIGMA deviation from mean/ n
so M.D.= 6/5
= 1.2

AWES TGT Maths Mock Test - 3 - Question 28

Circle with centre (0, 4) and passing through the projection of (2, 4) on x-axis is

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 28

Centre (0, 4) and r = 1

AWES TGT Maths Mock Test - 3 - Question 29

If cosec A - cot A = 4/5, then cosec A = 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 29

cosec A - cot A = 4/5  ........(i)
Also cosec2 A - cot2 A = 1
⇒ (cosec A - cot A) (cosec A - cot A) = 1

⇒ cosec A + cot A = 4/5   ......(ii)
From (i) and (ii), cosec A = 41/40

AWES TGT Maths Mock Test - 3 - Question 30

In the adjoining figure, the bisectors of ∠CBD and ∠BCE meet at the point O. If ∠BAC = 700, then ∠BOC is equal to :-

 

Detailed Solution for AWES TGT Maths Mock Test - 3 - Question 30



If the outer bisectors of the angle B and angle C meet at a point O outer side of the triangle, then ∠BOC = 90° - ∠A/2 = 90° - 70°/2 = 55°

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