UP PGT Math Mock Test - 8 - UPTET MCQ

lim (x-> 0) (1-cos2x) / x

= lim (x->0) {1-(1–2sin² x)}/x

= lim (x->0) 2sin^x/x

When we put the limit…the answer is 0/0 and it is undetermined.

So we have to apply LHR rule to determine it,

= lim(x->0) 2 * 2 sin x cos x / 1

= lim(x->0) 2 sin 2x

Now put the limit, we get

2 * sin 0 = 0

Any square matrix when multiplied with its inverse gives the identity matrix. Note that non square matrices are not invertible.

The graphs of the two functions are shown below: If a > 0 in any parabolic function then parabola open up side

From the above figure, it is obvious that the graphs of the two functions intersect at three points.

(2^1/5 +3^1/10)⁵⁵ is
Total term = 55 + 1 = 56

Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;  
Number of irrational terms = 56 - 6 = 50

Order = 3 , degree not defined ,because the function y’ present in exponential form.

|x| is not differentiable at x = 0
|x| is continuous at x = 0

The orthogonal projection on x- axis is given by : 

To find area the curves y = √x and x = 2y + 3 and x – axis in the first quadrant., We have ; y²−− 2y −− 3 = 0 (y – 3) (y + 1) = 0 . y = 3 , - 1 . In first quadrant , y = 3 and x = 9.
Therefore , required area is ;

Taking the new axes as  we see that the parabola can be  with the required condition |X| <  200

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

Correct option is D)

A={1,2,4}, B={2,4,5}, C={2,5}

A−C={1,4}

B−C={4}

(A−C)×(B−C)={1,4}×{4}

={1,4},{4,4}.

• This method involves specifying a rule or condition which can be used to decide whether an object can belong to the set.
• This rule is written inside a pair of curly braces and can be written either as a statement or expressed symbolically or written using a combination of statements and symbols.
• {x :  x ≠ x}
  This implies a null set.

The normal line to circle is →y² - 2 y + 4 x -2 xy=0
→ y(y-2) - 2x(y-2)=0
→ (y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)² + (2-1)²]^½ 
=(1+1)^½ =(2)^½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→ (x-1)²+(y-2)²= 2
x²+y² -2x-4y+3 = 0

Coefficient of x⁵ in (1 + x)¹⁰ 
Coefficient of x²⁵ in (1 + x)³⁰

No. of proper subsets = 2ⁿ-1

sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A

f ‘ (x) = 2x – 2 = 2 (x - 1) < 0 if x < 1 i.e. x x∈ (−∞,1) x∈ (−∞,1). Hence f is strict decreasing in left decreasing in (−∞,1)

To solve this problem, we'll use the tangent addition formula:

• tan(A + B) = (tanA + tanB) / (1 - tanA * tanB)

We can rewrite 45° as 22° + 23°. So,

• tan(45°) = tan(22° + 23°)

Using the tangent addition formula:

• 1 = (tan22° + tan23°) / (1 - tan22° * tan23°)

Cross-multiplying, we get:

• 1 - tan22° * tan23° = tan22° + tan23°

Rearranging the terms, we get:

• tan22° + tan23° + tan22°.tan23° = 1

Therefore, the value of tan22° + tan23° + tan22°.tan23° is 1.

use 

Split into two integrals

Concept:
If a number ω = a + ib is purely imaginary, then

• a = 0 
• and ω =

Calculation:
Given, z is a complex number such that is purely imaginary,
Let = a + ib, then a = 0
And

Given sequence : 2, 2√2, 4….
First term a₁ = a = 2 and 2nd term a₂ = 2√2, then
Common ratio r = a₂/a = (2√2)/2
Let a_n = 64
∴ ar^(n-1) = 64
⇒ 2.(√2)^(n-1) = 32
⇒ (2)^(n-1)/2 = 32
∴ (2)^(n-1)/2 = (2)⁵
⇒ (n − 1)/2 = 5,
⇒ n = 11

Correct option is D.
To find A×B we take one element from set A and one from set B. 

Given that 99 elements are common to both set A and set B.

Suppose these common elements are N_1​,N₂​,N₃​,...N₉₉​. 

Select an ordered pair for A×B such that both are selected out of these common elements. Examples: (N₁​,N_2​),(N₃​,N_5​)
All these will also be elements of B×A. Hence number of elements common to A×B and B×A is 99×99=99² ( first element in ordered pair can be selected in 99 ways; second element can also be selected in 99 ways)

n[(A×B)∩(B×A)]=n[(A∩B)∩(B∩A)]=(99)(99)=99²