UP PGT Math Mock Test - 1 - UPTET MCQ

Concept:

The null set is a subset of every set. (ϕ ⊆ A)

Every set is a subset of itself. (A ⊆ A)

The number of subsets of a set with n elements is 2ⁿ.

The number of proper subsets of a given set is 2n - 1

Calculation:

Statement I: If A = {x: x is an even natural number} and B = {y: y is a natural number}, A subset B.

A = {2, 4, 6, 8, 10, 12, ...} and A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...}.

It is clear that all the elements of set A are included in set B.

So, set A is the subset of set B.

Statement I is correct.

Statement II:  Number of subsets for the given set A = {5, 6, 7, 8) is 15.

Given: A = {5, 6, 7, 8}

The number of elements in the set is 4

We know that,

The formula to calculate the number of subsets of a given set is 2ⁿ

 = 2⁴ = 16

Number of subsets is 16

Statement II is incorrect.

Statement III: Number of proper subsets for the given set A = {5, 6, 7, 8) is 15.

The formula to calculate the number of proper subsets of a given set is 2ⁿ - 1

 = 2⁴ - 1

 = 16 - 1 = 15

The number of proper subsets is 15.

Statement III is correct.

∴ Statements I and III are correct.

9x² + 5y² - 30y = 0
9x² + 5(y−3)² = 45
We can write it as : [(x-0)²]/5 + [(y-3)²]/9 = 1
Compare it with x²/a² + y²/b² = 1
e = [(b² - a²)/b²]^½
e = [(9-5)/9]^1/2
e = (4/9)^½
e = ⅔

Geometric mean of two numbers a and b is √ab
As two numbers are 14 and 56
Geometric mean is √14×56
= ± √2×7×2×2×2×7
= ±(2×2×7)
= ±28

Area of standard ellipse is given by :πab.

Apply , C₁ → C₁ - C₂, C₂ → C₂ - C₃,

Because here row 1 and 2 are identical

z = -i
Comlex number z is of the form x + iy
x = 0, y = -1
arg(z) = π − tan−1|y/x|
⇒ π − tan−1|-1/0|
= π - (π/2)
⇒  π - π/2
⇒ π/2

When a = b = 1, c = 2, it gives minimum value (since a,b,c not all equal)

f ‘ (x) = 0 ⇒ f (x)is constant in (0 , 1)and also in (2, 4). But this does not mean that f (x) has the same value in both the intervals . However , if f (c) = f (d) , where c ∈ (0 , 1) and d ∈ (2, 4) then f (x) assumes the same value at all x ∈ (0 ,1) U (2, 4) and hence f is a constant function.

ANSWER :- b

Solution :- sin(A+B+C) = cos(A+B+C)/2 = 0

Hence, sinA/2 sinB/2 sinC/2 ≤ 1/8

Probability of not getting a head = 1/2
Probability of getting at least one head in n tosses is > 0.8 (given)
⇒ 1 -P (getting no head) > 0.8


Putting values for n, we see that the least value of n satisfying (1) is n=3.

Let the two numbers be α and β. Given that 

∴ Required equation is x² – 18x + 16 = 0 

She has 6 friends and he wants to invite one or more. That is the same as saying he wants to invite at least 1 of his friends.
 
So, the number of ways he could do this is:
Invite only one friend
Invite any two friends
Invite any three friends
Invite any four friends
Invite any five friends
Invite all six friends
This can be thought of in terms of combinations. Inviting  r  friends out of  n  is same as choosing  r  friends out of  n . So, we can write the possibilities as:
⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ 
= 6 + 15 + 20 + 15 + 6 + 1 
= 63

If AB = BA = I , then A and B are inverse of each other. i.e. A is invers of B and B is inverse of A.

Required probability 

Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

Let
If α, β,and γ are the three roots then we must have
ax³ + bx² + cx + d = a(x−α)(x−β)(x−γ)
Comparing coefficients of various power of x on both sides leads to
α + β + γ = −b/a −−−−(1)
αβ + βγ + γα = c/a −−−−(2)
αβγ= −d/a−−−(3)
On substituting β = αr and γ = αr2 we get
α(1+r+r²) = −b/a−−−(4)
α²(r+r² + r³) = c/a−−(5)
α³/r³ = -d/a−−−−−(6)
Now, dividing (5) by (4) to αr=− c/b and substituting this in(6)weget
(−c/b)³ = -d/a
⇒ c³/b³ = -d/a
∴ c³a = b³d

As the curve goes from c to d and the equation is x = f(y)
So the shaded area is ∫(c to d)f(y)dy

n (P) = 5, n(Q) = 12 and n(PUQ) = 14
n(PUQ) = n(P) + n(Q) - n(P∩Q) 
14 = 5 + 12 - n(P∩Q)
n(P∩Q) = 3

Hexagon has 6 vertices. Three vertices are chosen.
∴ n(S) = Total number of triangles formed


Let E : Triangle is equilateral
Out of these 20 triangles, only two triangles ACE and BDF are equilateral and each of their side is where ‘a’ is the side of regular hexagon.

(x+a)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^(n-1) a¹ + ⁿC₂ x^(n-2) a² + ..........+ ⁿC_(n-1) xa^(n-1) + ⁿC_n  aⁿ
Now, ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n-1,    ⁿC₂ = ⁿC_n-2,........
therefore, ⁿC_r = ⁿC_n-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x³.
Substiture x = 3y , so that, as x→0 , y→0.
∴ L = lim y→0 sin3y − 3y/(3y)³,
= lim y→0 (3siny − 4sin3y)− 3y)/27y³,
= lim y→0 {3(siny − y)/27y³) − (4sin³y/27y³)},
⇒ L = lim y→0 1/9 * (siny − y)/y³) − 4/27* (siny/y)³...(∗)
.Note that, here,
= lim y→0(siny − y)/y³)
= lim x→0 (sinx−x)/x³) = L.
Therefore, (∗) ⇒ L = 1/9*L − 4/27
or,  8/9L = −4/27
Hence, L = −4/27*9/8
=−1/6

Arranging the given data in ascending order,
we have 34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Here, Median M = (46+48)/2
=47
(∵ n = 10, median is the mean of 5th and 6th items)
∴ Mean deviation = ∑|xi−M|/n
=∑|xi−47|/10
= (13+9+5+3+1+1+7+8+16+23)/10
=8.6

As x R y if x + 2y = 8, therefore, domain of the relation R is given by x = 8 – 2y ∈ N.
When y = 1, 
⇒ x = 6 ,when y = 2, 
⇒ x = 4, when y = 3, 
⇒ x = 2.
therefore domain is {2, 4, 6}.

In a ΔABC, we know that
tanA + tanB + tanC = tanA tanB tanC
∴ tanB + tanC = tanA(tanBtanC – 1) 


⇒ tan² B - (K - 1) tan B + K = 0
For real values of tan B, Disc.
(K – 1)² – 4K > 0 
K² – 6K + 1 > 0