UP PGT Math Mock Test - 4 - UPTET MCQ

Laws are trigonometry can be applied to all type of triangles. These rules are generalised.

n³ + (n+1)³ + (n−1)³
=3(n)³ + 6n
=3n(n² + 2)
=3n(n² − 1 + 3)
=3(n(n−1)(n+1)+3n)
n(n−1)(n+1) and 3n are both divisible by 3. Hence their sum is also divisible by 3. 
Therefore the whole number is divisible by 9.

Let A be the set of students who have passed in Mathematics, and B be the set of students who have passed in Physics.
We are given that |A| = 55 and |B| = 67, where |A| and |B| represent the number of students in sets A and B, respectively.
We are also given that there are 100 students in total. We need to find the number of students who have passed in Physics only, which means we need to find |B - A|.
First, let's find the number of students who have passed in both Mathematics and Physics, which can be represented by |A ∩ B|.
Using the principle of inclusion-exclusion, we have:
|A ∪ B| = |A| + |B| - |A ∩ B|
Since there are 100 students in total, we can say that |A ∪ B| = 100.
Now, we can find |A ∩ B|:
100 = 55 + 67 - |A ∩ B|
100 = 122 - |A ∩ B|
|A ∩ B| = 22
Now, we can find the number of students who have passed in Physics only, which is |B - A|:
|B - A| = |B| - |A ∩ B|
|B - A| = 67 - 22
|B - A| = 45
so, the correct answer is 45.

 replacing x and y both by 1, we get

Other two vertices will make two right angled triangles with AB as the common hypotenuse. So they must lie on the circle with AB as the diameter and O as the centre. Radius of that circle will be 5 units.
There will be 5 such pairs in which both the coordinates are integers.
[(5, 0), (–5, 0), [(4, 3), (4, – 3)],
[(–3, 4), (3, –4)] [(–3, –4), (3, 4)] and [(0, 5), (0, –5)]

Correct Answer :- C

Explanation : lt(x->a-) -|x-a|/x-a

= - |-a-a|/(-a-a)

= -|-2a|/(-2a)

= -2a/(-2a)

= 1

cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1

Let  E₁ : Head appears
E₂ : Tail appears
A : A reports that head appears

∴ Rwquired probability = 
By Bayes’ Theorem

n(A) = 4
B = {5, 6}
n(B) = 2
n(A x B) = n(A)*n(B) = 4*2 = 8

Let P(x₁,y₁) be a point on the line 3x + 4y = 12
Equation of variable chord of contact of P(x₁,y1₎ w.r.t circle x2 + y2 = 4 
xx₁ + yy₁ − 4 = 0   ...(1)
Also 3x₁ + 4y1 − 12 = 0
⇒ x₁ + 4/3y₁ − 4 = 0   ...(2)
Comparing (1) and (2), we get
x = 1; y = 4/3
∴ Variable chord of contact always passes through (1, 4/3)

Consider the words starting with A.
A−−−−
Number of ways of filling the 2nd place is 4
Number of ways of filling the 3rd place is 3
Number of ways of filling the 4th place is 2
Number of ways of filling the 5th place is 1
Hence in total 4×3×2×1=24
Similar cases for the words starting with I.=24
Hence 2×24 =48
Consider the words ending with A−−−−A
Number of ways of filling the 1st place is 4
Number of ways of filling the 2nd place is 3
Number of ways of filling the 3rd place is 2
Number of ways of filling the 4th place is 1
Hence in total 24 ways.
Similarly for I- 24 ways.
Hence total number of words 4×24=96
But this includes the words which start and end with Vowels.
Hence A−−−I
Number of such words is 3!.
Similarly I−−−A.
Number of such words is 3!.
Hence total number of words beginning with or ending with Vowels is
= 96−3!−3!=96−12
= 84

A = {(5x+2, 2x, 2x) (5x+2, x+2, 2x) (5x+2, 2x, x+2)}
= (5x+2) {(1, 2x, 2x) (1, x+2, 2x) (1, 2x, x+2)}
= (5x+2) [(1(x+2)^2 - 4x^2) - 2x(x+2 - 2x) + 2x(2x - x - 2)]
= (5x+2) [(x+2 - 2x)(x+2+2x)-2x(2 - x) + 2x(x-2)]
= (5x+2) [(2-x)(3x+2) -2x(2-x) -2x(2-x)]
= (5x+2) (2-x)[3x+2-2x-2x]
= (5x+2) (2-x)²

We know that,

lim_x→3 (xⁿ – 3ⁿ)/(x – 3) = n(3)^n-1

Thus, n(3)n-1 = 108 {from the given}

n(3)^n-1 = 4(27) = 4(3³) = 4(3)^4-1

Therefore, n = 4

(1 + x)ⁿ  = 1 + ⁿC₁x¹  + ⁿC₂x²+..............................+ⁿCnxⁿ
Three consecutive terms coefficients
ⁿCₐ  : ⁿCₐ₊₁  :  ⁿCₐ₊₂  :::  6 : 33 :  110
⇒  ⁿCₐ  = 6K  =>  n!/(a!)(n-a)!  = 6K  => n! = 6K (a!)(n-a)!
ⁿCₐ₊₁   = 33K   => n!/(a+1)!(n-a-1)! = 33K  => n!  = 33K (a+1)!(n-a-1)!
ⁿCₐ₊₂ = 110K  => n!/(a + 2)!(n-a-2)! = 110K  => n! = 110K (a + 2)!(n-a-2)!
6K (a!)(n-a)!  = 33K (a+1)!(n-a-1)!
⇒ 2  (a!)(n-a)(n-a - 1)! = 11 (a + 1)a! (n-a-1)!
⇒ 2(n-a) = 11(a + 1)
⇒ 2n - 2a = 11a + 11
⇒ 2n = 13a + 11
⇒ 13a = 2n - 11
33K (a+1)!(n-a-1)!  = 110K (a + 2)!(n-a-2)!
⇒ 3 (a+1)!(n-a-1)(n-a-2)!  = 10 (a + 2)(a + 1)!(n-a-2)!
⇒ 3 (n - a - 1) = 10(a + 2)
⇒ 3n - 3a - 3 = 10a + 20
⇒  3n = 13a + 23
⇒ 13a = 3n - 23
2n - 11 = 3n - 23
⇒ n = 12

On comparing the given equations with: 
, we get: 

Correct Answer :- d

Explanation:- 3(6-6) -2(6-9) +3(4-6)

= 3(0) + 6 - 6

= 0

Since a, b, c are in H.P.

Now log (a + c) + log (a – 2b + c)
= log [(a + c){(a + c) – 2b}]
= log [(a + c)² – 2b (a + c)]
= log [(a + c)² – 2 × 2ac]
= log (a – c)² or log (c – a)²
= 2 log (a – c) or 2 log (c – a)
∴ log (a + c) + log (a – 2b + c) = 2 log (c – a)

2a/2be^y = 2b/2ce^y = 2c/2de^y
= a/be^y = b/ce^y = c/de^y = k
a/b = b/c = c/d = ke^y = k'
a=bk', b=ck',  c = dk'
b = (dk')k'
b = d(k')²
a = bk' = d(k')² k'
= d(k')³
a,b,c,d ⇒ d(k')³, d(k')², d(k'), d
G.P. sequence

⁸C_r = ⁸C_p = ⁸C^8-p     
So, either r = p or r = 8 – p
i.e. r = p or r + p = 8

^n+r-1C_r-1

Since g is continuous at a , therefore , if g (a) > 0 , then there is a nhd.of a, say (a-e , a+ e) in which g (x) is positive .This means that f ‘ (x)>0 in this nhd of a and hence f (x) is increasing at a.

Given that first term =a
S= sum to infinity of a G.P= a/(1−r)
r=1−a/S
S_n=a(1−rⁿ)/(1-r)
S_n=S(1−rⁿ)
S_n=S[1−(1−a/S)ⁿ]

The sine function has no domain restrictions. That means that the domain is (−∞,+∞)
However, the range of a since function is restricted as such : [−1,+1].

Number of words beginning with A = 4! = 4 × 3 × 2 = 24
Number of words beginning with N = 4! = 4 × 3 × 2 = 24
Number of words beginning with R = 4! = 4 × 3 × 2 = 24
Number of words beginning with U = 4! = 4 × 3 × 2 = 24
So, in total 96 words will be formed while beginning with letter A, N, R and U.
Now, 97th word according to dictionary order will be VANRU
Now, 98th word according to dictionary order will be VANUR
Now, 99th word according to dictionary order will be VARNU
Finally, 100th word according to dictionary order will be VARUN.
Hence, option (c) is the correct answer.

 r =7 m, θ = 1°
Length subtended by Arc = (2πr/360o)×θ
= (2×22×7×1)/(7×360)
​= 0.12m