UP PGT Math Mock Test - 7 - UPTET MCQ

lt h->0 [sin(x+h)^½ - sin(x)^½]/h
Differentiate it with ‘h’
lt h->0 {cos(x+h)^½ * ½(x+h)^1/2] - 0}/1
lt h->0 {cos(x+h)^½ * ½(x+h)^1/2]
lt h->0 cos(x)^½ / 2(x)^½

Only non-singular matrices have inverses.
Suppose that A is invertible. This means that we have the inverse matrix A⁻¹ of A.
Consider the equation Ax=0. We show that this equation has only zero solution.
Multiplying it by　A⁻¹ on the left, we obtain
A⁻¹ Ax = A^-1 0 
⇒ x = 0

It lies on y = 0.
∴ 

We have product of the roots = 2e^{4 ln k} - 1 = 31 (given) 

⇒ k⁴ = 16 ⇒ k⁴ - 16 = 0
⇒ (k – 2)(k + 2)(k² + 4) = 0  ⇒ k = 2, -2
as k > 0, we get k = 2.
Sum of the roots = 5k = 10 

Let α, α² be the roots of 3x² + px + 3 = 0. Then 

As α³ = 1 ⇒ α = 1, ω, ω²
We take α = ω so that α² = ω²
Thus - 1 = ω + ω² = α + α² 

In vector form:
Vector equation of a plane that passes through the intersection of planes
 expressed in terms of a non – zero constant λ is given by :

Clearly , (M−1)⁻¹ = (M⁻¹)¹ is not true.

General term T_r+1 of (x+y)ⁿ is given by 
T_r+1 = ⁿC_r x^n-r y^r
T₂ = ⁿC₂ x^n-2 y = 240
T₃ = ⁿC₃ x^n-3 y² = 720
T₄ = ⁿC₄ x^n-4 y³ = 1080
T₃/T₂ = [(n-1)/2] * [y/x] = 3......(1)
T₄/T₂ = {[(n-1)(n-2)]/(3*2)} * x²/y² = 9/2
T₄/T₃ = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n

Total 3 digit numbers = 900
There are only 5 odd numbers (1,3,5,7,9)
So 3 digit numbers having containing only odd digits = 5*5*5 = 125
So, 3 digit numbers containing at least one even digit = 900–125 = 775

There are 2 rows and 5 columns in the matrix.
Therefore, order of the matrix: 2 x 5.

Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)

= CosA*Cos B - Sin A*Sin B
= Cos (A+B)
= cos(π/4-x+π/4-y)
= cos(π/2-x-y)
= cos{π/2 - (x+y)}
= sin(x+y)

(i) xRx ⇔ 2x² - 3x.x + x²  

∴ R is reflexive
(ii) For x = 1, y= 2; 2x² - 3xy + y² = 0
∴ 1 R 2 but 2.2² - 3.2.1 + 1² = 3 ≠ 0.
So, 2 is not R-related to 1.
∴ R is not symmetric. 

Length of tangents from origin

Out of integers from 1 to 11, there are 5 even and 6 odd integers.
Let A – both the numbers chosen are odd
B –sum of the numbers chosen is even. 

By definition , The angle θ between the planes A₁x + B₁y + C₁z + D₁ = 0 and A₂ x + B₂ y + C₂ z + D₂ = 0 is given by :

We have x² + px + 1 = (x – a)(x – b)
Thus, E = (c – a)(c – b)(-d - a)(-d - b)
= (c² + pc + 1)[(-d²) - pd + 1] = (c² + pc + 1)(d² - pd + 1)       [∴ a + b = -p]
But c² + qc + 1 = 0 and d² + qd + 1 = 0
∴ E = (-qc + pc)(-qd - pd) = cd(q - p)(q + p) = cd(q² - p²) = q² - p² [∴ cd = 1]

P(A∪B) = P(A) = P(A) + P(B) - P(A∩B)
P(B) = 2/3
 = P(B) - P(A∩B) = 

When P = 1000 and t = 0 ., then ,
c = 1000, therefore, we have :

z = x + iy
|z| = (x² + y²)^1/2
x² + y² = 1
Equation of circle with centre (0,0) and radius 1 unit.

i) There are 5 places to fill

2 places of alphabet(out of 26) can be filled in =26∗25 ways

3 places of digits(out of 10) can be filled in = 10∗9∗8 ways

No of different plates = 26∗25∗10∗9∗8=468000

ii) We have total four place to cover with two alphabets and two digits.

For the first place:

Out of 26 alphabets we need one  

∴ combinations for first place = ²⁶C₁ = 26

For the second place:

We are left with 25 alphabets now as no repetitions are allowed.

∴combinations for second place= ²⁵C₁ = 25

For the third place:

Out of 10 digits we need to select one

∴combinations for third place = ¹⁰C₁ = 10

For the fourth place:

We are left with 9 digits now as no repetitions are allowed

∴combinations for fourth place= ⁹C₁ = 9

∴Total Combinations of number plates = 26 * 25 * 10 * 9 = 58,500

∑a(b - c) = 0, then both roots of equation is 1. 
Product of roots 


Hence, roots are in H.P.