Mathematics Test 211

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Mathematics Test 211

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Weekly Quiz Competition

Question 1
4 / -1

Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n, respectively, are

A
8 and 5

B
6 and 3

C
4 and 1

D
None of these

Solution

Total number of subsets of a set of m elements = 2^m

Total number of subsets of a set of n elements = 2ⁿ

According to the question,

2^m = 2ⁿ + 56

2^m - 2ⁿ = 56

2ⁿ(2^m - n - 1) = 56

2ⁿ(2^m - n - 1) = 8 × 7

2ⁿ(2^m - n - 1) = 2³(2³ - 1)

n = 3, m - n = 3

m = 6, n = 3
Question 2
4 / -1

For complex numbers z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂, (where i = √-1), we write z₁ ∩ z₂ for x₁ ≤ x₂ and y₁ ≤ y₂, then for all complex numbers z with 1 ∩ z, we have

A
0

B
1

C
2

D
3

Solution

As given for z₁= x₁ + iy₁ and z₂ = x₂ +iy₂, the symbol z₁ ∩ z₂ is used if x₁ ≤ x₂. Therefore for 1 ∩ z,we have 1 ≤ x and 0 ≤ y as 1 = 1 + 0.i and z = x + iy.
Question 3
4 / -1

In a plane, there are 37 straight lines, out of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through the same point, no line passes through both points A and B, and no two lines are parallel. The number of intersection points the lines have is equal to

A
535

B
601

C
728

D
963

Solution

The number of points of intersection of 37 lines is ³⁷C₂.

But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting ¹³C₂ points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B.

Therefore instead of getting ¹¹C₂ points, we get only one point B.

So, the number of intersection points = ³⁷C₂ - ¹³C₂ - ¹¹C₂ + 2 = 535 (2 is added because of points A and B)
Question 4
4 / -1

In a sequence of (4n + 1) terms, the first (2n + 1) terms are in AP, whose common difference is 2, and the last (2n + 1) terms are in GP, whose common ratio is 0.5. If the middle terms of the AP and GP are equal, then the middle term of the sequence is

A

B

C
n . 2ⁿ

D
None of these

Solution
Question 5
4 / -1

then the global maximum and local minimum values of f(x) for x ∈ [-2, 2], respectively, are

A
(4 + cos 2) and 1

B
11 and 1

C
11 and indeterminant

D
None of these

Solution

Since the given function is continuous but not differentiable at x = 0,

⇒ x = 0 is the point of local minima and f(0) = 1.

Also, there is no other critical point.

Now, f(-2) = 11, f(2) = 4 + cos 2

On comparing values of function at x = -2, 0, 2:

Global maximum occurs at x = -2 ⇒ f(-2) = 11

Hence, global maximum and local minimum values are 11 and 1, respectively.
Question 6
4 / -1

The area of the triangle formed by the tangent and the normal to the parabola y² = 4ax, both drawn at the same end of the latus rectum and the axis of the parabola, is

A
2√2a²

B
2a²

C
4a²

D
None of these

Solution

An end of the latus rectum = (a, 2a)

The equation of the tangent at (a, 2a) is: y(2a) = 2a(x + a), i.e. y = x + a

The normal at (a, 2a) is: y + x = 2a + a, i.e. y + x = 3a

Solving y = 0 and y = x + a, we get

x = -a, y = 0

Solving y = 0 and y + x = 3a, we get

x = 3a, y = 0

The area of the triangle with vertices (a, 2a), (-a, 0), (3a, 0) is: (1/2)(4a)(2a) = 4a²
Question 7
4 / -1

A

B

C

D
None of the above

Solution
Question 8
4 / -1

The lines are coplanar, if the value of λ is

A
2

B
13

C
-13

D
non-existent

Solution
Question 9
4 / -1

The mean of n items is If these n items are successively increased by 2, 2², 2³, …, 2ⁿ, then the new mean is

A

B

C

D
None of these

Solution
Question 10
4 / -1

A
0

B
1

C
2

D
3

Solution