UP PGT Math Mock Test - 2 - UPTET MCQ

Substitute a positive integer for ‘n’ and verify.

Out of 30 Boys,
No. of ways to select 1 boy = ³⁰C₁ = 30
Out of 18 Girls,
No. of ways to select 1 boy = ¹⁸C₁ = 18
No. of ways of selecting 1 boy and 1 girl =
30*18 = 540

Shortest distance between 2 curves occurs along the common normal. Let required point be (t, 2 -t). Normal at this will be y –(2 –t) = (x –t), it should passes through centre of circle so –1 –2 + t = 5 –t ⇒ t = 4, so point can be (4, -2).

cos105° + cos75°
= cos(90º + 15) + cos(90° - 15) 
= - sin15 + sin15 = 0

f (x) f (1/x) = f (x) + f (1/x) ⇒ f(x) = xⁿ +1. 
f (2) = 33 ⇒ 2ⁿ +1= 33 ⇒ 2ⁿ = 32 = 2⁵ ⇒ n = 5 ⇒ f (x) = x⁵ +1.

Consider the given function.
lim x→0 ((1-x)ⁿ −1)/x)
This is 0/0 form.
So, apply L-Hospital rule,
lim x→0 [nx-1x(1-x)_^n-1 −0)]/1
lim x→ 0(-n(1-x)^n-1)
=-n

= cos (log x) cos (cos y)   [cos(log(x/y))+cos(log(xy))] 
= cos (log x) cos (log y)  [cos(log x - log y)+cos(log x +log y)]
= cos (log x) cos (log y) [2 cos(log  x)cos(log y)] = 0

β = (1)^1/n ⇒ βn = 1 ........(1)
If S be the sum of the given series which is arithmetic geometric series, then 

Correct Answer :- b

Explanation : α₁,α₂,........,α₉₉

α₁, 1α₂,........,1α₉₉

α₁α₂, α₂α₃, α₃α₄................α₉₈α₉₉ 

Sum of α₁α₂, α₂α₃, α₃α₄................α₉₈α₉₉ = 0

Therefore, x - 1 = 0

x = 1

Middle term in the expansion of (1+x)²ⁿ; is 
=tn+1 = 2nCn.1⁽²ⁿ⁻ⁿ⁾.xⁿ
= {(2n)!.xⁿ}/(2n−n)!n!
= {{2n(2n−1)(2n−2)(2n−3)....4×3×2×1}/n! n!}/xⁿ
= {{2ⁿ[n(n−1)(n−2)....×2×1][(2n−1)(2n−3)....3×1]}/n! n!}xⁿ
= [[(2n−1)(2n−3)....3×1]/n!] 2ⁿxⁿ

For 5 lettered palindrome, we can fill the first 3 letters randomly and then the last 2 letters will become fixed.

there are 26 letters.
therefore the number of ways =26×26×26=26³
 = 17,576.

A = {(1,x) (0,1)}
A² = {(1,x) (0,1)} {(1,x) (0,1)}
= {(1+0, x+x) (0, 1)}
= {(1, 2x) (0,1)}
Similarly A¹⁰ = {(1, 10x) (0,1)}

For x²/(a² + b²) + y²/b² = 1
Equation for tangent y = mx +- [(a² + b²)m² + b²]^1/2
For x²/a² + y²/(a² + b²) = 1
Equation for tangent y = mx +- [a²m² + (a² + b²)}^1/2
For common tangent: [(a² + b²)m² + b²]^1/2 = [a²m² + (a² + b²)}^1/2
(a² + b²)m² + b² = a²m² + (a² + b²)
a²m² + b²m² + b² = a²m² + (a² + b²)
b²m² = a²
m = +- a/b
Equation for tangent : y = +- ax/b +-[(a² + b²)a²/b² + b²]^1/2
y = +- ax/b +-{[a⁴ + a²b² + b⁴]^1/2}/b
by = +- ax +-{[a⁴ + a²b² + b⁴]^1/2}

Due to the rule if a square matrix is of order n x n then, |kA| = kⁿ |A|.

cos35^o cos45^o
Multiply and divide numerator and denominator by 2
1/2{2cos35^o cos45^o}
= 1/2{cos(35^o+45^o) + cos(35^o-45^o)}
= 1/2{cos80^o + cos(-10^o)}    { cos(-x) = cosx 
= {cos80^o + cos10^o}/2

y = V^3/3
V = x² - 2x + 3
y = 1/3*v³
dy/dx = d/dx{1/3*(x²-2x+3)³} 
= 1/3*3*(x²-2x+3)²*(2x-2) 
{d/dx(xⁿ)=nx^n-1*dx/dx} 
= 2(x-1)*(x²-2x+3)²

The power set has 2ⁿ elements. For n = 11, size of power set is 2048.

so that the equation becomes 1/z = 1z²

where ω (≠1) is a cube root of unity.

The distance of a point whose position vector is  from the plane  given by :

Independent event but not equally likely

Since the exterior angle 140 degrees, The sum of the interior angles = (2n - 4)* right angles. So 140n = (2n - 4)* right angles, or
140n = (2n - 4)*90, or
140n = 180n - 360o, or
40n = 360°, or
n = 9 sides.

Let the other nos. be a and b
then (x+y+1+2+6)/5 = 4.4
x + y = 13 ---------------------(1)
Variance = 8.24


41.2 = 19.88 + (x2 + 19.36 – 8.8x) + (y2 + 19.36 – 8.8y) 
21.32 = x2 + y2 + 38.72 – 8.8(x + y) 
x2 + y2 + 38.72 – 8.8(13) – 21.32 = 0 
(using equation (1)) 
x2 + y2 – 97 = 0 …(2) 
Squaring equation (1) both the sides, 
we get (x + y)2 = (13)^2 
x2 + y2 + 2xy = 169 
97 + 2xy = 169 
(using equation (2)) 
xy = 36 or x = 36/y (1)
⇒ 36/y + y = 13 
y2 + 36 = 13y 
y2 – 13y + 36 = 0 
(y – 4)(y – 9) = 0 
Either (y – 4) = 0 or (y – 9) = 0 
⇒ y = 4 or y = 9 
For y = 4 x = 36/y 
= 36/4 = 12 
For y = 9 
x = 36/9 
x = 4 
Thus, remaining two observations are 4 and 9.