AWES TGT Maths Mock Test - 8 - AWES TGT/PGT MCQ

Chief Minister Prof Dr. Manik Saha inaugurated the 2nd State Level Shehri Samridhi Utsav in Agartala, organized by Tripura Urban Livelihood Mission. The event focuses on enhancing urban livelihood opportunities and empowering women in the region. Chief Minister Prof Dr. Manik Saha highlighted state government initiatives supporting women’s empowerment during the inauguration at Children’s Park, Agartala.

The Uttarakhand government plans to introduce the Uttarakhand Public and Private Property Damage Recovery Bill in the upcoming budget session. This legislation aims to hold rioters financially responsible for damages to both government and private property during protests or strikes. If enacted, Uttarakhand would be the third state, following Uttar Pradesh and Haryana, to implement such a law. The initiative aligns with the government’s agenda to uphold law and order and safeguard public and private assets from unlawful destruction.

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Playing games along with studies in school is very important for the physical growth and balance of the children. It also develops in them qualities like cooperation and spirit of teamwork to work with each other.

NEP 2020 recommends the incorporation of internships and projects to promote experiential learning.

NEP 2020 encourages private investment in school infrastructure development to improve facilities and resources.

 Let f(x) = 2tan x + 5x – 2 
Now, f(0) = – 2 < 0 
and f(π/4 ) = 2 + 5π/4 – 2 
= 5π/5 > 0, 
the line intersects the x-axis at x = ⅖ 
since 0 < ⅖ <  π/4
Thus, f(x) = 0 has atleast real solution root in ( 0, π/4)

⇒ x² = x²

∴ f is one-one

Let y∈R and let y=x³. then x=y^⅓ ∈ R

Thus for each y in the codomain R there exists y^1/3 in R such that 

f(y¹³)=(y¹³)³=y

∴ f is onto.

Hence f is one -one onto.

Taking the new axes as  we see that the parabola can be  with the required condition |X| <  200

The total number of ways is same as the number of arrangements of 4 fingers, taken 3 at a time.
So, required number of ways = ⁴P₃ 
= 4!/(4-3)!
= 4!/1!
= 4! => 24

Let √(5 – 12i) = x + iy
Squaring both sides, we get
5 – 12i = x² + 2ixy +(iy)² = x² – y² + 2xyi.
Comparing real and imaginary parts , we get
5 = x² – y² ———– (1) and xy = – 6 ———— (2)
Squaring (1), we get
25 = (x² – y²)2 = (x² + y²)² – 4x²y²
⇒ 25 = (x² + y²)² – 4(– 6)²
⇒ (x² + y²)² = 169
⇒ x² + y² = 13 ———- (3)
Adding (1) and (3) we get
2x² = 18
⇒ x = ± 3.
Subtracting (1) from (3) we get
2y² = 8
⇒ y = ± 2.
Hence, square root of √(5 – 12i) is (3 – 2i)
Similarly, √(5 + 12i) is (3 + 2i)
√(5 + 12i) + √(5 – 12i)
⇒ (3 + 2i) + (3 - 2i)
⇒ 6

sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A

• The set containing all objects or elements and of which all other sets are subsets is a Universal set.
• Here, option A has all the elements of set P, Q and R. So {1, 2, 3, 4, 5, 6, 7, 9} may be considered as universal set for all the three sets P, Q and R.

There are in total 6 numbers, 4,5,6,7,8.
Now consider the number 56000
Consider the numbers of the form
56−−−.
Considering no repetitions we get
3×2×1 = 6 numbers.
Similarly for 57−−− and 58−−−.
Hence 3×6 = 18 numbers.
Now consider the numbers starting with 6.
6−−−−
We get 4×3×2×1 = 24.
Similarly for the numbers starting with 7 and 8 we get in total 24 numbers each.
Hence total number of numbers greater than 56000 will be
= (24×3)+18
= 72+18 = 90

Now, dividing (1) by (2) we get,





Now, Dividing (2) by (3) we get,

y=sin4x,  and z=cos4x
So by using chain rule
df(x)/dx = dsin⁴x/dx + dcos⁴x/dx
=dy⁴/dy * dy/dx + dz⁴/dz * dzdx
=dy⁴/dy * dsinxdx + dz⁴/dz * dcosx/dx
=4y⁽⁴⁻¹⁾⋅cosx+4z(4−1)⋅(−sinx)
=4sin³xcosx − 4cos³xsinx
=4sinxcosx(sin²x − cos²x)
=2sin2x(−cos2x)
=−2sin2xcos2x
=−sin4x

Coefficient of x⁵ in (1 + x)¹⁰ 
Coefficient of x²⁵ in (1 + x)³⁰

Identity

Required area : 

S=(3 + 3/4 + 3/4² +....)+(d/4 + 2d/4² + 3d/4² + ....)
S = 3(1 + 1/4 + 1/4² + ..)+(s) (lets's say)
s = d/4 + 2d/4² + 3d/4³ + ..(1)
s/4 = d/4² + 2d/4³ + ...(2)
subtrating  (1) - (2)
3s/4 = d/4 + d/42 + d/43 + ....
s = d/3(1 + 1/4 + 1/4² + ..)
s= 4d/9
S=3(3/4) + 4d/9
S = 4 + 4d/9 = 8
d = 9

α + β = (a – 2), αβ = –a –1
α² + β² = (α + β)² - 2αβ = (a - 2)² + 2(a + 1)
α² + β² = α² - 2a + 6
ax² + bx+ c takes least value -D/4a, so, least value at of α² + β² is 5.

Correct option is D)

A={1,2,4}, B={2,4,5}, C={2,5}

A−C={1,4}

B−C={4}

(A−C)×(B−C)={1,4}×{4}

={1,4},{4,4}.

First person can sit on any of the 7 seats, second person can sit on other 6 vacant seats except one occupied by the first one. Similarly, the third can do sit on 5 so, total no. of ways = 7*6*5 = 210

A = Φ
B = {0, 1, 2, 3, 4, 5, 6}
A × B = Φ (Cartesian product with an empty set is an empty set)

Zeroes of f(x) = x² − 4x − 5 are obtained by solving:
x² − 4x − 5 = 0, which implies 
x² − 5x + x − 5 = 0 or, x(x − 5) + 1(x − 5) = 0 
which means x = 5 or x = −1 
Thus the assertion is correct. 
However, the reason is incorrect. 
The numbers given are not the zeroes  of x² − 4x + 7