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

C₁C₂ < r₁+ r₂

Between n & n², numbers divisible by n are:
2n, 3n, 4n, ….... (n – 1)n
No. of numbers = (n – 1) – 2 + 1 = n – 2

3sin^-1(x) = sin^-1(3x - 4x^3) when -1/2<=x<=1/2
Definitely 0.4 comes in this range of x and so

3sin^-1(0.4) = sin^-1[3*0.4 - 4*0.4^3]
3sin^-1(0.4) = sin^-1[1.2 - 4*0.064]
3sin^-1(0.4) = sin^-1[1.2 - 0.256]
3sin^-1(0.4) = sin^-1[0.944]

Finally , sin(3sin^-1(0.4)) = sin{sin^-1(0.944)} = 0.944

Sum of first n terms of AP, S_n = 2n² + 5n
 
Now choose n =1 and put in the above formula,
First term = 2+5 = 7
 
Now put n=2 to get the sum of first two terms = 2x4 + 5x 2= 8 + 10 = 18
 
This means 
 
first term + second term = 18
 
but first term =7 as calculated above
 
so, second term = 18-7 = 11
 
So common difference becomes, 11-7 = 4
 
So the AP becomes, 7, 11, 15, ....
nth term = a + (n - 1)d = 7 + (n-1)4 = 7 + 4n - 4 = 3 + 4n

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

All the elements in set-A are presented in set-B. So "A" is a subset of "B".

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


∴ z = 0

(1 + x)ⁿ > 1 + nx
For n = 1, (1 + x)ⁿ = (1 + x)¹ = 1 + x
1 + nx = 1 + x
(1 + x)ⁿ = 1 + nx        
For n = 2, (1 + x)ⁿ = (1 + x)² = 1 + x² + 2x
1 + nx = 1 + 2x
(1 + x)ⁿ ≥ 1 + nx (Equality when x = 0)
So, (1 + x) ⁿ > 1 + nx for n ≥ 2 

A 3-digit no.: _ _ _
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 7, 8, 9
In the first place there are 9 possible digits (0 cannot be the 1st digit)
In the second and third place there are 10 possible digits.
Total no. of 3-digit numbers = 9 *10 *10 = 900 
For 3-digit no. without digit 7,
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 8, 9
In the first place there are 8 possible digits (0 cannot be the 1st digit)
In the second and third place there are 9 possible digits.
Total no. of 3-digit numbers = 8 *9 *9 = 648 
No. of 3-digit numbers with at least one digit 7 = 900 – 648 = 252

Geometric mean of two numbers a and b is √ab
As two numbers are 14 and 56
Geometric mean is √14×56
= ± √2×7×2×2×2×7
= ±(2×2×7)
= ±28

f(x + y)=f(x) f(y) wrt x we get:
f′(x+y) (1 + y′) = f(x) f′(y) (y′) + f′(x) f(y)
Now on substituting x = 0 and y = 3, we get :
f′(3) (1 + 0)=f(0) f′(3) 0+ f(3) f′(0)
f′(3) = 3 × 11 = 33

And y has a relative minimum at x = 0 if f ‘(0)+a = 0 . a =0.

Probability of not getting a head = 1/2
Probability of getting at least one head in n tosses is > 0.8 (given)
⇒ 1 -P (getting no head) > 0.8


Putting values for n, we see that the least value of n satisfying (1) is n=3.

Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

For n=1, we have 7² +16 -1 = 64 which is divisible by 64
 
For n=k, we assume that 7^2k + 16k-1 is divisible by 64
 
We must check for n=k+1, is the statement true
7^2(k+1) +16(k+1) -1 = 49 * 7^2k + 16k +15
 
we adjust this expression by adding and subtracting terms, that is
we add 49*16k, -49 but we have to subtract 49*16k and -49
 
we have the following
49*7^2k +16k +15 = 49*7^2k + 49*16k -49 -49*16k -49 +16k +15 =
49(7^2k +16k -1) - 49(16k-1) +16k +15 =
49(7^k +16k -1) - 48*16k +64
all these terms are divisible by 64

In case of negative or fractional power, expansion (1+x)^n is valid only when |x| < 1
(6 - 3x)^-1/2
= (6^-1/2 (1 - x/2)^-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2

Period of sin x is 2π.
Period of sin 5x is 2π/5

Let
If α, β,and γ are the three roots then we must have
ax³ + bx² + cx + d = a(x−α)(x−β)(x−γ)
Comparing coefficients of various power of x on both sides leads to
α + β + γ = −b/a −−−−(1)
αβ + βγ + γα = c/a −−−−(2)
αβγ= −d/a−−−(3)
On substituting β = αr and γ = αr2 we get
α(1+r+r²) = −b/a−−−(4)
α²(r+r² + r³) = c/a−−(5)
α³/r³ = -d/a−−−−−(6)
Now, dividing (5) by (4) to αr=− c/b and substituting this in(6)weget
(−c/b)³ = -d/a
⇒ c³/b³ = -d/a
∴ c³a = b³d

The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5, 6}

Number of possible outcomes of a dice = 6

Prime numbers on a dice are 2, 3, and 5.

Composite numbers on dice = 4, 6

Total prime numbers on a dice = 3

Total composite numbers on a dice  = 2

So, total (prime + composite) = 5

Probability of getting a number that is either prime or composite = 5/6

Let ABCD is a square circumscribing a circle of radius a cm.
The side of square ABCD = Diameter of circle
⇒AB=2a
Therefore, perimeter of square AB=4×AB=4×2a=8 cm

Let the angle be x
Then the complement angle will be 90^∘ −x

Given x is 14^∘ more than 90^∘ −x
⟹ x = 14^∘ + 90^∘ −x
⟹ 2x = 104^∘
⟹ x = 52^∘
The measure of the angle is 52^∘

x³+y³+15xy-125
=x³ + y³ +3 xy ×5 - 125
=x³ + y³ +3xy(x+y) - (5)³
= (x+y) ³ - (5)³
=(5)³ - (5)³
=0

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