AWES PGT Math Mock Test - 5 - AWES TGT/PGT MCQ

The ‘Asia-Pacific Leaders’ Conclave on Malaria Elimination’ will be organized in New Delhi by the Union Ministry of Health and Family Welfare in partnership with the Asia Pacific Leaders Malaria Alliance (APLMA) as a precursor to World Malaria Day.
Its goal is to reaffirm political commitment to achieve malaria elimination by 2030.

The Education of children with special needs should be provided along with children with non-special needs. To promote the education of special needs children along with non-special needs children concept of inclusive education was proposed. In Inclusive Education, all students are welcomed, they have been given equal opportunity irrespective of caste, creed, religion, special-needs, non-special needs.
Hence, the correct answer is, 'Along with other normal children.'

'Wrong' answers of students should be corrected by the teachers during class discussions as children should be encouraged to express themselves without fear of being incorrect.

Primary level students learn more by observing pictures and specimens. Thus, they are most effective in teaching primary level students. Pictures and specimens are important visual tools which facilitate learning.

The given inequality is 5x– 3 < 7
=> 5x – 3 + 3 < 7 + 3                             [3 is added both sides]
=> 5x < 10
=> x < 10/5
=> x < 2
When x is a real number, the solutions of the given inequality are given by x < 2, i.e., all real numbers x which are less than 2.

Determinant = [(a² + b²).1 - (-2ab)]
= (a+b)²

Correct Answer :- a

Explanation : The formula to find the coefficient of a^pb^qc^rd^s in (xa+yb+zc+vd)ⁿ is (n!x^p y^q z^r v^s)/p!q!r!s!

So, coefficient of a⁴b³c²d in (a−b+c−d)¹⁰

= [10!(1)⁴(-1)³(1)²(-1)1]/(4!3!2!1!)

= 12600

(fof) (x) = f[f(x)] = f(x² - 3x + 2)² - 3(x² – 3x + 2)
= y² – 3y + 2 = (x² – 3x + 2)² – 3(x² – 3x + 2) + 2 =  (x⁴ – 6x³ + 10x² – 3x)

sec^-1x + cosec^-1x =​ π/2 ; x belongs to [ -1 , 1 ]

As we know that, cosA = (b² + c² - a²)/2bc
a = 6, b=4, c=8
cosA = (16+64-36)/2(4)(8)
44/64 
= 0.687
A = cos-1(0.687)
which is approx equals to 46°36’2″

L.C.M of 4 and 6 is 12.

Given, A = {x:x is a multiple of 4}
              = {4,8,12,16,20,…}

and     B = {x:x is a multiple of 6}
              = {6,12,18,24,…}

∴A ∩ B = {12,24,…}
             = {x:x is amultiple of 12}

The odd digits which can be considered are (1,3,5,7 and 9).
If we start from left, the first digit can be any number out of the five digits. The second place can be filled with any four digits except the digit at the first place.
Similarly, four ways for the third digit until we reach the tenth place.
Therefore, the total numbers formed are 5 × 49

Equation of tangent to parabola y = mx- am² .......(1) equation of chord of hyperbola whose midpoint is (h, k) is  hx - ky = h² - k² ...... (2) form (1) and  (2) 

The smallest number in the series is 1000, a 4-digit number.  

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left-most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

So, The left-most digits have values 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

So, the next 3 digits have values 5.

Hence, there are numbers.

or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

Transpose displays row data in a column or column data in a row.

Let
A – Getting 15 as the sum in a throw of three dice.
B – Getting 4 on the first die.
Required probability 

(As total number of ways = 1×6×6=36. There are two favourable events (4,6,5) (4,5,6))

Total number of jackets=3
Total number of shirts=10
Total number of pair of slacks=5
Ways of selecting one jacket =3C1=3
Ways of selecting one shirt =10C1=10
Ways of selecting one pair of slack = 5C1=5
Number of different outfits that can be made=3×10×5=150
Number of different outfits that can be made=150

lim x→0 (1−cosx)/(xlog(1+x))
= lim x→0 {2sin² x/2}/[xlog(1+x)]
= lim x→0 {{2sin²x/2/(x/2)²}*/1/4}/[1/xlog(1+x)]
= lim x→0 [log(1+x)/x]1/2
= 2​

Using 

and substitute (7, 1)

By verification f ' (2 -) ≠ f ' (2 +)
∴ f(x) is not differentiable at x = 2

 lt x->-2 (1/x + 1/2) / x+2
lt x->-2 (2+x)/2x(x+2)
lt x->-2 (1/2x)
= 1/2(-2) = -¼