UP PGT Math Mock Test - 10 - UPTET MCQ

 (1+x)ⁿ= nC0+ nC1x+ nC2x²+...+ nC0xⁿ ....(1)
(1−x)ⁿ= nC0 − nC1x+ nC2x²−...+ nC0 xⁿ....(2)
⇒(1+x)ⁿ−(1−x)ⁿ = nC0+ nC1x+ nC2x²+...+ nC0xⁿ− nC0+ nC1x− nC2x²+...+ nC0xⁿ
 =2(nC1x+nC3x³+....+xⁿ)

Given: n=60 and put x=1
⇒(1+1)⁶⁰−(1−1)⁶⁰
 =2⁶⁰
⇒2⁶⁰=2(nC1x+nC3x³+....+xⁿ)

∴Sum of odd powers of x in the expansion is = nC1x+nC3x³+....+xⁿ =2⁶⁰−1 =2⁵⁹

P(-x,-y) lies on the 3rd or 4th quadrant
So, according to the question, points lies in 3rd quadrant
i.e. 3(π/2)

• We have, A^c in X = The set of elements in X which are not in A = 0,1
• {0} ∈ A^c w.r.t. X is false, because {0} is not an element of A^c in X.
• ϕ ∈ A^c in X is false because ϕ is not an element of A^c in X.
• {0} ⊂ A^c in X is correct because the only element of {0} namely 0 also belong to A^c in X.
• 0 ⊂ A^c in X is false because 0 is not a set.

If A = {5, 10, 15}, B = ϕ
B - A will have those elements which are in B but not in A.
B - A = ϕ

none of these  : 7/25

• Every set has the empty set as a subset. So if a set has 1 element, like {0}, then it will have 2 subsets: itself and the empty set, which is denoted by { }.
• So, if a set has only one subset, then this set must be the empty set.

Here 2n is even integer, therefore,

term  will be  the middle term.
Now, (n + 1)^th term in 

 

The general form of an AP is  a,  a+d, a+2d,.....
where a is the first term and d is the common difference  
If we double the terms,  the new sequence would be
A , a+2d, a+4d,......
We can observe that this sequence is also an AP
First term is a
Common difference is 2d  nth term= 2a+(n-1)2d               
= 2[a+(n-1)d]

Let y has some fractional part
x = 6.5
y = 0.9
[x] = 6, [y] = 0, [x]+[y] = 6
[x + y] = [7.4] = 7
[x + y] ≠ [x]+[y]
Let x, y be integers
x = 6.5
y = 1
[x] = 6, [y] = 1, [x]+[y] = 7
[x + y] = [7.5] = 7
[x + y] = [x]+[y]
So, [x + y] = [x]+[y] holds for y ϵ I

Radius of circle =14cm
An angle = 135^o
Length of arc = ?
We know that, Length of arc = Radius × Angle
= 14 * 3π/4
= 14 * (3*22)/(7*4)
= 33cm

√x / √(16+√x) - 4

= (√x * √(16+√x) + 4) / ( √(16+√x) - 4) * (√(16+√x) + 4)

= (√x * √(16+√x) + 4) / 16+√x - 16

= √(16+√x) + 4

now at lim x->0

= √16 + 4 = 8

The greatest integer function ⌊x⌋ returns the largest integer less than or equal to x. For x², this means that ⌊x²⌋ will be the greatest integer less than or equal to x².

The integral becomes:

The function ⌊x²⌋ will be 0 on the interval (0,1)(0,1), 1 on the interval [1,√2​], and 4 on the interval [√2,2].

So, the integral is the sum of the areas of these intervals:

Evaluating these integrals:

So, the value of the integral is 4.

The number of non- empty subsets =
2ⁿ − 1 = 2⁴ − 1 = 16 − 1 = 15

The information is represented in the form of a 3 X 2 matrix as follows:

The entry in the third row and second column represents the number of women workers in factory III.

Given, lines can be written as

and 
∴ Requires SD =

Where 

∴ Requires SD 

x² = 16 ⇒ x = ±4 and 2x = 6 ⇒x = 3

There is no value of x which satisfies both the above equations. Thus, A = φ

a = 5, c = 3
angle c = 60^o
cos b = (a² + c² - b²)/2ca
= 1/2 = (25 + 9 - b²)/30
=> 15 = 34 - b²
=> b² = 19
=> b = (19)1/2
=> b = 4.35

The numbers should be divisible by 6. Thus, the number of favourable ways is ¹⁶C₃ (as there are 16 numbers in first 100 natural numbers, divisible by 6).
Required probability is 

The correct option is C 
a=c=1/4, b=1/2
y=x is a tangent
∴ slopes are equal.
dy/dx=2ax+b
⇒1=2a+b at (1,1)⋯(1)
Also, the parabola passes through (1,1)
⇒a+b+c=1⋯(2)
The parabola passes through (−1,0)
⇒0=a−b+c⋯(3)
Solving (1),(2),(3), we get -
∴a=c=1/4
and b=1/2

 are any three vectors then the correct expression for distributivity of scalar product over addition is :

As (1 – p) is root of the equation: x² + px + (1 – p) = 0

(1 – p)² + p(1 – p) + (1 – p) = 0

(1 – p)[1 – p + p + 1] = 0

(1 – p) = 0

p = 1

Therefore, given equation now becomes

x² + x = 0

x(x + 1) = 0

x = 0, -1

Radius of the given circle = |g|. For perpendicular tangents origin should lie on the director circle, i.e. distance of origin from the centre of the given circle  

z₁ = (√3/2 + i/2), z₂ = (√3 - i/2)
(z1)³ = (√3/2 + i/2)² 
= 3/4 - 1/4 + √3i/2 
= 1/2 + √3i/2
(z₃)³ = (z₁)² * z₁ 
= (1/2 + √3i/2) * (√3/2 + i/2)
= √3/4 + i/4 + 3i/4 - √3/4
= i
(z₂)² = (√3/2 - i/2)² 
= 3/4 - 1/4 - √3i/2 
= 1/2 - √3i/2
(z₂)³ = (z₂)² * z₂
= (1/2 - √3i/2) * (√3/2 - i/2)
= √3/4 - i/4 - 3i/4 - √3/4 
= -i
(z₂)⁵ = (z₂)³ * (z₂)² 
= (-i) * (1/2 - √3i/2)
= -i/2 + √3/2
z = (z₁)⁵ + (z₂)⁵
= i/2 - √3/2 - i/2 + √3/2
= 0
So Im(z) = 0

Sin 7π = Sin 7*180 = Sin 2π * 7  = 0