UP PGT Math Mock Test - 6 - UPTET MCQ

There are 8 letters in the word TRIANGLE.
2 alphabets are fixed, remaining are 6 alphabets
So, number of arrangements = ⁶P₆
= 6!
= 720

Solving x + y = k + 1 and y = x(1 – x) ⇒ x² – 2x + k + 1 = 0  and Δ = 0 

sin³10° + sin³50° - sin³70° 
= 1/4 [(3 sin 10° - sin 30°) + (3 sin 50° - sin 150°) - (3 sin 70° - sin 210°)]

The ratio of the coefficients of x^p and x^q in the expansion of (1+x)^p+q is

Let S = {Persons who drive scooter}
C = {Persons who drive car}
Given, n(S ∪ C) = 80, n(S) = 35, n(C) = 60
Therefore, n(S ∪ C) = n(S) + n(C) - n(S ∩ C)
80 = 35 + 60 - n(S ∩ C)
80 = 95 - n(S ∩ C)
Therefore, n(S ∩ C) = 95 – 80 = 15
Therefore, 15 persons drive both
scooter and car.
Therefore, the number of persons who
drive a scooter only = n(S) - n(S ∩ C)
= 35 – 15
= 20
Also, the number of persons who drive car only = n(C) - n(S ∩ C)
= 60 - 15
= 45

In dy ‘a’ is the called as lower limit and ‘b’ is called the upper limit of the integral. The function f in is called the integrand. The letter ‘y’ is a dummy symbol and can be replaced by any other symbol.

Required area :
= (1/4) area of ellipse – area of right angled triangle AOB.   (π - 2)

Here, matrix P is of order 2 × 3 and matrix Q is of order 2 × 2 , then , the product PQ is defined only when : no. of columns in P = no. of rows in Q. And the order of resulting matrix is given by : rows in P x columns in Q.

The function is injective (one-to-one) if every element of the codomain is mapped to by at most one element of the domain

The function is surjective (onto) if every element of the codomain is mapped to by at least one element of the domain

f(x)=2x 

Domain of f is N 

Codomain of f is N

Every element of N(codomain) is mapped to only one element in N(domain)

i.e. f(x)=f(y)

⟹2x=2y⟹x=y

Hence f is one one function

Every even element of N(codomain) has corresponding value in N(domain).

But, for any odd number in N(codomain) has no corresponding value in Ndomain.

Hence, f is not onto function.

- lim x-->k+ = x-{x}
Lim x→k+   k-{k}
                  = 0           

⇒ αS = α + 2α² + 3α³ + ...+(n -1)α^n-1 +nαⁿ

On subtracting, we get

Since, α is nth root of unity, therfore
αⁿ = 1 

Clearly f is one one and onto, so invertible

 By L-Hospital rule

If a line has the direction ratios – 18, 12, – 4, then its direction cosines are given by:

Divide the numerator and denominator by x, so that given function becomes
f(x)=1+sinx/x/(1+cosx/x)
Now as x→∞.sinx/x→0 , because sin x would oscillate between +1 and -1, which in either case divided by ∞ would be 0. Thus the limit of the numerator would be 1. Like wise the limit of the denominator would also be 1.
Thus limit as a whole would be 1

d = 100
S_n = 21700
n = 14
S_n = n/2[2a + (n-1)d]
21700 = 14/2[2a + (14-1)100]
= 21700 = 7[2a + 1300]
= 3100 = 2a + 1300
⇒ 3100 - 1300 = 2a
⇒ 1800 = 2a
⇒ a = 900

Formula

Let X – No. of defective items in a sample of 10 items drawn successively.
Trial of drawing defective items with replacement are Bernoulli trials.
Probability of success = p = 5/100 = 1/20
∴ Probability of failure = 

    (1)
P/not more than 1 defective

Given: y = sin 3x . sin³

= 3 sin²x.sin 4x

Equation of line perpendicular to x−y=0 is given by
y=−x+c
Also this line is tangent to the hyperbola x²−2y²=18
So we have m=−1, a²=18, b²=9
Thus Using condition of tangency c² = a²m²−b²
= 18−9=9
⇒ c = ±3
Hence required equation of tangent is x+y = ±3