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

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

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

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.

Distributive law is given by : 

S₁ = 0 and S₁ = S₁₁

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

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

Internally touch ∴ common tangent is one.

 Let 2 tan^-1⁡x = y

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. 

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 (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)

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

 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.

None

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

Apply , C1→C1 - C3, C₂→C₂-C₃

= 10 - 12 = -2

(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]
(x_²+y²)² = (a²+b²)/(c²+d²)

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)

 replacing x and y both by 1, we get

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

Centre (0, 4) and r = 1

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

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°