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

Uttaramerur inscription was mentioned by Prime Minister Modi recently. This inscription was created during the reign of Parantaka Chola I (907-953 AD)
The walls of the Vaikunda Perumal Temple in Uttaramerur are inscribed with details revealing systems of local governance from the 10th century CE.

IIM Mumbai collaborated with Starburst to fortify India's Aerospace, New Space, and Defence (ASD) startup ecosystem. This partnership aims to nurture and empower ASD startups across India, fostering innovation within the sector and providing unparalleled support for their growth and contribution to India’s ASD industry.

The Indian Navy concluded the bi-annual 'Prasthan' exercise off the Mumbai coast, working with defence, state, and civil agencies to enhance preparedness and coordination in addressing contingencies related to oil production platforms.

Pavan Davuluri has been named as the new head of Microsoft's combined Windows and Surface teams, succeeding the previous leader, Panos Panay. Davuluri, a 23-year veteran at Microsoft, brings extensive experience in PC, Xbox, Surface, and Windows leadership roles, including efforts with Arm-based processors.

Useful strategies for teaching students with mental retardation include the following:

• Teach one concept or activity component at a time
• Teach one step at a time to help support memorisation and sequencing
• Teach students in small groups, or one-on-one if possible
• Always provide multiple opportunities to practice skills in a number of different settings
• Use physical and verbal prompting to guide correct responses, and provide specific verbal praise to reinforce these responses

"Use of formula for area-measurement should be taught before estimation of areas" is NOT correct in the context of teaching area- measurement of plane figures.

Measurement of plane figures: A plane, or flat, surface has two dimensions—length and width. The perimeter of a plane figure is the distance around it.

In the context of teaching area- measurement of plane figures correct statements:

• Students should be given a chance to 'discover' the various formulae.
• Grid paper can be a useful device for introducing and implementing the formal units of measurement.
• Tessellations with regular shapes can be useful for teaching area measurement.

Learning disabilities are neurologically-based processing problems. These processing problems can interfere with learning basic skills such as reading, writing and/or math. They can also interfere with higher level skills such as organisation, time planning, abstract reasoning, long or short term memory and attention. They are not caused due to the method of teaching used by the teacher. They may be present due to improper brain development, or infection in brain, etc. Constant fights amongst parents also lead to mental strain and anxiety in students, which may lead to learning disabilities.

Informal education is a form of education that is learnt by the student outside the school. Family plays a major role in the informal education of a child. 

f (x) = x² -4x +5 = (x -2)² +1
x ∈ [2,∞) ⇒ x > 2 ⇒ x - 2 > 0 ⇒ (x - 2)² +1 > 1 ⇒ B = [1,∞). 

In Cartesian coordinate system Shortest distance between the lines

The given matrix is a skew – symmetric matrix.,therefore , A = - A’.

R is reflexive
∴ (3, 3), (6,6), (9, 9), (12, 12) ∈ R 
again ∴ (6, 12) ∈ R but (12, 6) ∉ R ⇒ R is not symmetric
R is transitive
[∴ (3, 6) ∈ R, (6, 12) ∈ R, (3, 12) ∈ R others are clear 

Other two vertices will make two right angled triangles with AB as the common hypotenuse. So they must lie on the circle with AB as the diameter and O as the centre. Radius of that circle will be 5 units.
There will be 5 such pairs in which both the coordinates are integers.
[(5, 0), (–5, 0), [(4, 3), (4, – 3)],
[(–3, 4), (3, –4)] [(–3, –4), (3, 4)] and [(0, 5), (0, –5)]

2a/2be^y = 2b/2ce^y = 2c/2de^y
= a/be^y = b/ce^y = c/de^y = k
a/b = b/c = c/d = ke^y = k'
a=bk', b=ck',  c = dk'
b = (dk')k'
b = d(k')²
a = bk' = d(k')² k'
= d(k')³
a,b,c,d ⇒ d(k')³, d(k')², d(k'), d
G.P. sequence

If we substitute the value of x=0 in numerator and denominator, we get the indeterminate form 0/0 . We should use L'hopital's rule.
lim x→ 0 (√1+2x-√1-2x)/sinx
Differentiate it we get
lim x→ 0 (1/√1+2x) + (1/√1-2x)/cosx
= (1/√1+2(0)) + (1/√1-2(0))/cos 0
=2

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)

(1/2)^{(x2 - 2x)} < (1/4) 
(1/2)^{(x2 - 2x)} < (1/2)²
x² − 2x > 2......(as after multiplicative inverse sign of inequality changes)
x² − 2x − 2 > 0
x²  -  2x + 1 - 3  >
(x-1)²  - 3  > 0
(x-1)²  > 3 
So for the above to hold good both the expression must be positive or both must be negative. After finding the solution the range of the solution will be
either x > 3
(3,¥)

Even if f(x) is differentiable everywhere, |f(x)| need not be differentiable everywhere. An example being where the function changes its sign from negative to positive. f(x)=x at x=0B. 
∣f∣= {f if f>0                −f if f<0}
​So,∣f∣² = {f² if f>0        −f² if f<0}
​Differentiating ∣f∣²,
{2ff′ if f>0     −2ff′ if f<0}
​At f=0, LHD=RHD=∣f(0)∣²=0, so function is differentiable.

(1 + x)ⁿ  = 1 + ⁿC₁x¹  + ⁿC₂x²+..............................+ⁿCnxⁿ
Three consecutive terms coefficients
ⁿCₐ  : ⁿCₐ₊₁  :  ⁿCₐ₊₂  :::  6 : 33 :  110
⇒  ⁿCₐ  = 6K  =>  n!/(a!)(n-a)!  = 6K  => n! = 6K (a!)(n-a)!
ⁿCₐ₊₁   = 33K   => n!/(a+1)!(n-a-1)! = 33K  => n!  = 33K (a+1)!(n-a-1)!
ⁿCₐ₊₂ = 110K  => n!/(a + 2)!(n-a-2)! = 110K  => n! = 110K (a + 2)!(n-a-2)!
6K (a!)(n-a)!  = 33K (a+1)!(n-a-1)!
⇒ 2  (a!)(n-a)(n-a - 1)! = 11 (a + 1)a! (n-a-1)!
⇒ 2(n-a) = 11(a + 1)
⇒ 2n - 2a = 11a + 11
⇒ 2n = 13a + 11
⇒ 13a = 2n - 11
33K (a+1)!(n-a-1)!  = 110K (a + 2)!(n-a-2)!
⇒ 3 (a+1)!(n-a-1)(n-a-2)!  = 10 (a + 2)(a + 1)!(n-a-2)!
⇒ 3 (n - a - 1) = 10(a + 2)
⇒ 3n - 3a - 3 = 10a + 20
⇒  3n = 13a + 23
⇒ 13a = 3n - 23
2n - 11 = 3n - 23
⇒ n = 12

The sign of the sinA and tanA in third quadrant is negative and positive.

We have , 

Therefore, range of f(x) is {-1}.