AWES TGT Maths Mock Test - 1 - AWES TGT/PGT MCQ

King Charles III is the monarch depicted on the new UK banknotes, which will co-circulate with those featuring Queen Elizabeth II's portrait starting from June 5, 2024.

IIT Bombay is ranked 45th globally in engineering according to the QS World University Rankings 2024, highlighting its excellence in the field and its leading position in South Asia.

Key Points 

• It is well knowledge that young infants acquire significant concepts more quickly in their mother tongue or home language.
• The mother tongue or the language used by the neighborhood is typically the same as the home language.
• However, there may occasionally be a home language used by other family members in multilingual households that is distinct from the mother tongue or the local vernacular.
• National Education Policy 2020 proposes that wherever feasible, the home language, mother tongue, local language, or regional language will be used as the medium of teaching until at least Grade 5, but preferably until Grade 8 and beyond.

Hence, At the primary level, National Education Policy 2020 proposes mother tongue/Home language as the medium of instruction across the nation.

P(x) when divided by x−1, x-2 and x−3 leaves remainder 3,7 and 13 respectively.
From polynomial-remainder theorem, P(1) = 3 and P(2) = 7 P(3) = 13
If the polynomial is divided by (x−1)(x−2)(x-3) then remainder must be of the form ax+b (degree of remainder is less than that of divisor)
 
⇒P(x) = Q(x)(x−1)(x−2)(x-3)+(ax² + bx + c), where Q(x) is some polynomial. 
Substituting for x=1, x=2 and x=3:
P(1) = 3 = a+b+c
P(2) = 7 = 4a+2b+c
P(3) = 13 = 9a+3b+c
=> Subtract P(2) from P(1)
-3a - b = -4.............(1)
Subtract P(3) from P(2)
5a + b = 6..............(2)
From (1) and (2) Solving for a and b, we get 
a = 1 and b = 1, c = 1
⇒Remainder = x² + x + 1

If 2 tan^-1 (cos x) = tan ^-1(2 cosec x),

2tan^-1(cos x) = tan^-1 (2 cosec x)

= tan^-1(2 cosec x) 

= cot x cosec x = cosec x = x = π/4

(acos x + bsin x)² + (asin x – bcos x)² = a² + b²
⇒ c² + (asin x – bcos x)² = a² + b²
⇒ (asin x – bcos x)² = a² + b² – c²

The nth term of an AP is given by T_n = S_n − S_n−1.
First compute S₄ = 2×4×4 + 5×4 = 32 + 20 = 52.
Then compute S₃ = 2×3×3 + 5×3 = 18 + 15 = 33.
So T₄ = S₄ − S₃ 
= 52 − 33
= 19.

f (x + 2y) = f (x) + f (2 y) + 4xy for x, y ∈ R putting x = y = 0, we get f (0) = 0

In a ΔABC, we know that
tanA + tanB + tanC = tanA tanB tanC
∴ tanB + tanC = tanA(tanBtanC – 1) 


⇒ tan² B - (K - 1) tan B + K = 0
For real values of tan B, Disc.
(K – 1)² – 4K > 0 
K² – 6K + 1 > 0 

A (1,1,1), B (2,3,5), C(-1,0,2) direction ratios of AB are < 1,2,4 >
Therefore, direction ratios of normal to plane ABC are < 2, -3,1 >
As a result, equation of the required plane is 2x – 3y +z = k then 

Hence, equation of the required plane is 

When a = b = 1, c = 2, it gives minimum value (since a,b,c not all equal)

The equation y = 2x − x² i.e. y – 1 = - (x - 1)² represents a downward parabola with vertex at (1, 1) which meets x – axis where y = 0 .i .e . where x = 0 , 2. Also , the line y = - x meets this parabola where – x = 2x − x² i.e. where x = 0 , 3. 
Therefore , required area is :

As β ≠ 1 is afourth root of unity,
β4 = 1 ⇒ (1 - β) (1 + β + β² + β³) = 0


∴ z = 0

The lady can select one cotton saree out of 15 cotton varieties in 15 ways since
any of 15 varieties can be selected. Corresponding to each selection of a cotton saree, she can
choose a polyester saree in 13 ways. Hence the two sarees (one cotton and one polyester), by
multiplication principle of counting, can be selected in 15 x 13= 195 ways

n(A × B) = n(A) × n(B)

lim xsecx    x=0
lim x(1/cosx) = lim (x/cosx)
Put x = 0
= 0/cos0
= 0/1 = 0

Equation of the tangent at (h, k) is 

Equation of auxiallary circle x² + y² = a²

She has 6 friends and he wants to invite one or more. That is the same as saying he wants to invite at least 1 of his friends.
 
So, the number of ways he could do this is:
Invite only one friend
Invite any two friends
Invite any three friends
Invite any four friends
Invite any five friends
Invite all six friends
This can be thought of in terms of combinations. Inviting  r  friends out of  n  is same as choosing  r  friends out of  n . So, we can write the possibilities as:
⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ 
= 6 + 15 + 20 + 15 + 6 + 1 
= 63

(x+a)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^(n-1) a¹ + ⁿC₂ x^(n-2) a² + ..........+ ⁿC_(n-1) xa^(n-1) + ⁿC_n  aⁿ
Now, ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n-1,    ⁿC₂ = ⁿC_n-2,........
therefore, ⁿC_r = ⁿC_n-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

Number of tickets selected from first station =20
from second =19
.... for last station =0
We have to select 2 consecutive stations
so total number of possible tickets = P(20,2)

Option d is correct, because it is the property of definite integral
 ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

As the opinions of different person is different about the intelligent student. So, we can't exactly have the same student's name in a set, hence it is not a well-defined collection.

As the angle increases from 270° to 360°, the sine increases from -1 to 0. As the angle increases from 270° to 360°, and the value of cos 270° = 0
and cos 360° = 1
So, the cosine increases from 0 to +1.

a = p² + 1
Since series is of consecutive integer
Sum of (2p+1) terms = n/2 (2a+(n−1)d)
= [(2p+1)/2][2(p² + 1)+(2p + 1 − 1)1]
= [(2p + 1)/2][2p² + 2 + 2p]
=(2p + 1)(p² + p + 1)
= 2p³ + 2p² + 2p + p² + p + 1
= p³ +p³ + 1 + 3p² +3p
= p³ + (p+1)³

Consider the given function:
c1/c0 + 2c2/c1 +  3c3/c2 + 4c4/c3 +......+ncn/cn−1
=n/1 + [2n(n−1)]/2! . 1/n + [3n(n−1)(n−2)]/3! . 1/[n(n−1)/2!] +......+n.1/n
=n+(n−1)+(n−2)+......1
 =1+2+3+......+n
 = n(n+1)/2

The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends; one radian is just under 57.3 degrees.

Two coins are simultaneously tossed.
So sample space={HH,HT,TH,TT}
No. of favourable outcomes=getting at least one head={HH,HT,TH}
=3
Total number of outcomes=4
Probability of getting at least one head=No. of favourable outcomes/Total number of outcomes
=3/4