Mathematics Test 235

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Mathematics Test 235

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Weekly Quiz Competition

Question 1
4 / -1

Let A and B be two events such that where A̅ stands for the complement of the event A. Then the events A and B are:

A
mutually exclusive and independent.

B
equally likely but not independent.

C
independent but not equally likely

D
independent and equally likely.

Solution

Concept:

Events A and B are independent if P(A ∩ B) = P(A)P(B)

∴ The events A and B are independent but not equally likely.

The correct answer is Option 3.
Question 2
4 / -1

The value of : tan{sin^–1(cos(sin^–1 x))}. tan{cos^-1(sin(cos^–1x))} ; x ∈ (0, 1) is equal to ;

A
0

B
1

C
-1

D
none of these

Solution
Question 3
4 / -1

The area of the region described by A ={(x, y) : x² + y² ≤ 1 and y² ≤ 1 - x} is:

A

B

C

D

Solution
Question 4
4 / -1

A
0

B
∞

C
1/2

D
none of these

Solution
Question 5
4 / -1

One of the bisectors of the angle between the lines a(x – 1)² + 2h(x – 1)(y – 2) + b(y – 2)² = 0 is x + 2y – 5 = 0. The other bisector is

A
2x – y = 0

B
x + 2y = 0

C
2x – y + 4 = 0

D
2x – y + 3 = 0

Solution

Calculation:

Given, line a(x – 1)² + 2h(x – 1)(y – 2) + b(y – 2)² = 0

Angle bisector, x + 2y – 5 = 0

The bisectors between two lines will be perpendicular to each other.

Slope of the given bisector is − 1/2

The other bisector will have slope 2 and will pass through the point (1, 2)

⇒ Equation is y - y₁ = m(x - x₁)

⇒ y – 2 = 2(x – 1)

⇒ y – 2 = 2x – 2

⇒ 2x – y = 0

∴ The equation of other bisector is 2x – y = 0.

The correct answer is Option 1.
Question 6
4 / -1

The locus of the foot of perpendicular drawn from the centre of the ellipse x² + 3y² = 6 on any tangent to it is

A
(x² – y²)² = 6x² + 2y²

B
(x² – y²)²= 6x² – 2y²

C
(x² + y²)² = 6x² + 2y²

D
(x² + y²)² = 6x² – 2y²

Solution
Question 7
4 / -1

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is

A
3 - √2

B
3 + √2

C
2 - √3

D
2 + √3

Solution
Question 8
4 / -1

If the median of 21 observations is 40 and if the observations greater than the median are increased by 5 then the median of the new data will be

A
40

B
45

C
40 + 50/21

D
45 – 50/21

Solution

Concept:

The median of a set of data is the middle value when the data is arranged in ascending order.

Calculation:

For 21 observations, the median is the 11th value.

According to the question, the original median is 40.

Now, observations greater than the median are increased by 5.

Since only the values greater than the median are increased, the position of the median (the 11th observation) remains unchanged, and the value of the median (40) itself does not change.

⇒ The median of the new data will still be 40.

∴ The median of the new data remains 40.

The correct answer is Option 1.
Question 9
4 / -1

Two dice are thrown simultaneously to get the coordinates of a point on x – y plane. Then, the probability that this point lies on or inside the region bounded by |x| + |y| = 3 is

A
1/12

B
1/24

C
5/36

D
none of these

Solution

Calculation:

Given, 2 dice are thrown to get the coordinates of a point on x – y plane

⇒ Possible outcomes = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

= 36 cases

Now, point lies on or inside the region bounded by |x| + |y| = 3

⇒ Favourable outcomes = {(1, 2), (2, 1), (1, 1)} = 3 cases

∴ Required probability = 3/36 = 1/12

∴ The probability that the point lies on or inside the region bounded by |x| + |y| = 3 is 1/12.

The correct answer is Option 1.
Question 10
4 / -1

If x̅ & y̅ are two non-collinear vectors and a, b, c represent the sides of a ΔABC satisfying (a – b)x̅ + (b – c)y̅ + (c – a)(x̅ × y̅) = 0, then ΔABC is

A
An acute angle triangle

B
An obtuse angle triangle

C
A right angle triangle

D
A scalene triangle

Solution

Calculation:

Given, x̅ & y̅ are two non-collinear vectors

⇒ x̅, y̅ and (x̅ × y̅) are independent vectors.

Now, (a – b)x̅ + (b – c)y̅ + (c – a)(x̅ × y̅) = 0

Since, x̅, y̅ and (x̅ × y̅) are independent vectors.

⇒ a – b = b – c = c – a

⇒ a = b = c

⇒ All sides are equal

⇒ It is a equilateral triangle

∴ ΔABC is an acute angle triangle.

The correct answer is Option 1.