AWES TGT Maths Mock Test - 6 - AWES TGT/PGT MCQ

An Se-young of South Korea won the gold medal in the women's singles event at the 2023 World Badminton Championships, defeating Carolina Marín of Spain in the final, 21-12, 21-10. An Se-young, aged 21, became the first woman from South Korea to win the world singles title. She had a dominant tournament, winning all of her matches in straight games, including victories over Olympic champion Chen Yufei and two-time world champion Akane Yamaguchi.

► n(Persons travelling by car) = 20 = n(C)
► n(Persons travelling by bus) = 50 = n(B)
► n(Persons travelling by both car and bus) = 10 = n(C ∩ B)
∴ n(C ∪ B) = n(C) + n(B) - n(C ∩ B)
= 20 + 50 - 10
= 60 

U = Set of all teachers in a school
B = Set of all Mathematics teachers in a school
B’ = U – B = Set of all Non-Mathematics teachers in a school

N = 3
No. of subsets = 2^N = 2³ = 8

Reflexive : A relation is reflexive if every element of set is paired with itself. Here none of the element of A is paired with themselves, so S is not reflexive.
Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive.

If R is a relation defined by xRy : ifx ⩽ y, then R is reflexive and transitive But, it is not symmetric. Hence, R is not an equivalence relation.

The function is injective (one-to-one) if every element of the codomain is mapped to by at most one element of the domain

The function is surjective (onto) if every element of the codomain is mapped to by at least one element of the domain

f(x)=2x 

Domain of f is N 

Codomain of f is N

Every element of N(codomain) is mapped to only one element in N(domain)

i.e. f(x)=f(y)

⟹2x=2y⟹x=y

Hence f is one one function

Every even element of N(codomain) has corresponding value in N(domain).

But, for any odd number in N(codomain) has no corresponding value in Ndomain.

Hence, f is not onto function.

Correct Answer :- c

Explanation : f(n-1)/2 ; n is odd        -n/2 ; n is even

f(1) = 0                f(3) = 1

f(2) = -1               f(4) = -2

f(5) = 2                f(6) = -3

f(n) + f(n+1) = -1    n is odd

if = 0     n is even

Range of f(n) = Z

f(n) is one-one, therefore f(n1) = f(n2)

(n₁-1)/2 = -n₂/2

=> 1 < (n1 + n2)/2 is not equal to 1/2

Therefore f(n) is onto.

4^2x = 4^-6

2x = -6

So,  x = -3

There are 8 letters in the word TRIANGLE.
2 alphabets are fixed, remaining are 6 alphabets
So, number of arrangements = ⁶P₆
= 6!
= 720

18P4/18C4
⇒ [18!/14!]/[18!/(4!*14!)]
⇒ 4!
⇒ 24

Substitute ‘n’ and verify the options.

d = 100
S_n = 21700
n = 14
S_n = n/2[2a + (n-1)d]
21700 = 14/2[2a + (14-1)100]
= 21700 = 7[2a + 1300]
= 3100 = 2a + 1300
⇒ 3100 - 1300 = 2a
⇒ 1800 = 2a
⇒ a = 900

The sequence is the Fibonacci series
1+1 = 0
1+2 = 3
2+3 = 5
3+5 = 8
5+8 = 13
8+13 = 21
13+21 = 34
21+34 = 55
34+55 = 89
The 11th term will be 89.

aα² + bα + c = 0 and bα² + cα + a = 0 ⇒ α = 1 ∴ a + b + c = 0

Let D₁ = b² – 4ac and  D₂ = d²  + 4ac
As ac ≠ 0, either ac < 0 or ac > 0
If ac < 0, then D₁ > 0
In this case P(x) = 0 has distinct two real roots 
If ac > 0, the D₂ > 0
In this case Q(x) = 0 has two distinct real roots.
Thus, in either case P(x)Q(x) = 0 has at least two distinct real roots. 

We can write the given equation as
(x – a)² = 3 – a
This shows that 
Both the roots of the given equation will be less than 3 if the larger of the two roots is less than 3, that is, if 


⇒ 3 - a > 1 or a < 2
Thus a > 3 and a < 2 ⇒ a < 2

Here, matrix P is of order 2 × 3 and matrix Q is of order 2 × 2 , then , the product PQ is defined only when : no. of columns in P = no. of rows in Q. And the order of resulting matrix is given by : rows in P x columns in Q.

In a diagonal matrix all elements except diagonal elements are zero.i.e. 

Operate,

Apply R₃→R₃- R₁, R₂→ R₂ -R₁,

⇒ -6(5x² - 20) +15(2x-4) = 0

⇒ (x- 2)(x+1) = 0⇒x=2, -1

⇒-6(5x² - 20) + 15(2x-4) = 0

⇒(x-2)(x+1) = 0 ⇒ x=2, -1

Let the centre be C (h, k). Since the circle passes through the origin,

Let x = c meets the circle in A and B.

From figure, CB² = CD² + BD²
⇒ OC² = CD² + b²  (∴CB=OC and BD=b)
⇒ h² + k² = (h - c)² + b²
⇒ k² + 2ch - c² - b² = 0
⇒ locus is y² + 2cx = b² + c².
Hence (C) is the correct answer.

Center of such circle in the case of parabolas  y² = 4ax, x² = 4ay is 

Combined equation of lines joining the points of intersection of the chord with origin is 



these lines are mutually perpendicular if 

thus chord becomes x cos θ + y sin θ = which is clearly a tangent to the circle having centre at origin and radius is equal to

Since sin⁻¹x is defined for |x|⩽1, and sec⁻¹x is defined for |x|⩾1,therefore,f(x) is defined only when|x|=1.so, D_f = {−1,1}.

 By L-Hospital rule