UP PGT Math Mock Test - 3 - UPTET MCQ

Hence , the given equation has only one solution.

Δ = 1/2 [2(1-8) -7(1-10) +1(8-10)]
= 47/2

Let D – Lost card is diamond
D – Lost card is not diamond.

When one diamond card is lost, the number of diamond cards are 12 out of 51.
No. of ways of choosing 2 from 12 cards ¹²C₂ .
Similarity, choosing 2 diamond cards out of 51 are ⁵¹C₂.
 probability of getting two cards, when one diamond is lost

When the lost and is not diamond, number of diamond cards are 13 out of 51.
No. of ways of choosing 2 cards when one card is lost which is not diamond =

The two curves parabola and the line meet where,

Required area ;

We know that a number between 100 and 1000 has three digits. 
So we have to find all the three digit numbers with distinct digits.

We cannot have 0 at the hundred’s place for 3-digit number.  So the hundred’s place can be filled with any of the 9 digits.  1,2,3,…..9.  So, there are 9 ways of filling the hundred’s place.

Now, 9 digits are left including 0.  So ten’s place can be filled with any of the remaining 9 digits in 9 was.  Now the unit’s place can be filled with any of the remaining 8 digits.  So, there are 8 ways of filling the unit’s place.

Hence, the total number of required numbers = 9 x 9 x 8= 648.

In the vector form, equation of a plane which is at a distance d from the origin, and  is the unit vector normal to the plane through the origin is given by : 

There’s a result in mathematics used for this. It says that a power set B of any set A is a set of all the subsets of A and the number of elements of B will be 2^n where n is the number of elements of A.
So taking your question as an example;
A = {1,2,3}
B : set of all subsets of A
List out all the subsets of A - {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}
Number of elements in A (n) = 3 
so 23 = 8
So, B = {{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}} 
and the number of elements are 8.

R = {(a, b): a = b − 2, b > 6}
Now, since b > 6, (2, 4) ∉ R
Also, as 3 ≠ 8 − 2, (3, 8) ∉ R
And, as 8 ≠ 7 − 2
∴ (8, 7) ∉ R
Now, consider (6, 8).
We have 8 > 6 and also, 6 = 8 − 2.
∴ (6, 8) ∈ R

Probability that first critic favour is P(ε₁) = 5/7 
Similarly, P(ε₂) = 4/7  P(ε₃) = 3/7 
Majority are in favour if at least two favour 

Differentiating xy=16, (xdy)/x+y=0
⇒ dy/dx=−y/x=−1/16= Slope of tangent
⇒  Its (tangent's) equation : y=−1=−1/16(x−16)
⇒ 16y−16=−x+16 ⇒ 16y=−x−32
⇒ 16y+x+32=0
It will cut x−axis at A(−32, 0)
& y−axis at B(0, −2)
⇒  Area of △OAB= 1/2×2×32 
⇒ 32

In linear algebra, the identity matrix, or sometimes ambiguously called a unit matrix, of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. It is denoted by I_n, or simply by I if the size is immaterial or can be trivially determined by the context

Since equation of required circles is S + λS ' = 0
As it passes through (1, 1) the value of 
If 7 + 2p = 0, it becomes the second circle,
Therefore it is true for all values of p

Let a and r the first term and common ratio, respectively.

We are drawing two cards without replacement from a deck of 52 cards.

No. of ways = ⁵²C₁ x ⁵¹C₁ = 52 x 51 = 2652

Sum of the roots = 0

Since A ⊆ B
∴ A ∪ B = B
So, n(A ∪ B) = n(B) = 6

f(g(2)) compare f(gx) x=2
on comparing g(x)=x, g(2)=2
f(g(x)) = f(2) = x² = 2²= 4

xy - 3x - 2y + λ = 0.
Then abc + 2fgh − af² − bg² − ch² = 0
⇒ 3/2 − λ/4 = 0
⇒ λ = 6
∴ Equation of asymptotes is xy-3x-2y+6=0
⇒ (x-2)(y-3)=0
⇒x - 2 = 0 and y - 3 = 0

therefore , slopes of the tangents at (1,1) and (- 1 , -1)are equal. Hence, the two tangents in reference are parallel.

To be reflexive, a line must be perpendicular to itself, but which is not true. So, R is not reflexive
For symmetric, if  l₁ R l₂ ⇒ l₁ ⊥ l₂.
⇒  l₂ ⊥ l₁ ⇒ l₁ R l₂ hence symmetric
For transitive,  if l₁ R l₂ and l₂ R l₃
⇒ l₁ R l₂  and l₂ R l₃  does not imply that l₁ ⊥ l₃ hence not transitive.

α + β = -a,αβ = -b;γ+δ = -a,γδ = b
(α² + aα + b) (β² + aβ +b) = 4b²

f (x) is discontinuous when x² - 3|x| + 2 = 0
⇒ |x|² - 3|x| + 2 = 0 ⇒ |x| = 1, 2

Given, 
Number of visitors= 4
Number of hotels= 5
So, A can disperse among 5 hotels,
      B can disperse among 4 hotels,
      C can disperse among 3 hotels,
      D can disperse among 2 hotels.
∴ They can disperse in the hotels in 5 × 4 × 3 × 2 = 120 ways

By the property of transpose , (AB)’ = B’A’.

ar⁹ = 1/(2)⁸ -----------------------(1)
ar³ = 1/(2)² ------------------------------(2)
Comparing eq (1) & (2)
1/(2)⁸ r⁹ = 1/(2)^2 r⁴
1/r⁵ = (2)⁵
r⁵ = 1/(2)⁵
r = 1/2