Mathematics Test 256

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Mathematics Test 256

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Question 1
4 / -1

If the standard deviation of x₁, x₂, …, x_n is 3.5, then the standard deviation of – 2x₁ – 3, – 2x₂ – 3, …, – 2x_n – 3 is:

A
– 7

B
– 4

C
7

D
1.75

Solution

We are given:

The standard deviation (SD) of the data set x₁, x₂, ..., xₙ is 3.5

A new data set is formed: -2x₁ - 3, -2x₂ - 3, ..., -2xₙ - 3

If we transform a data set by:

yᵢ = a·xᵢ + b, then:

Standard Deviation changes only by the scale factor a

It does not change with a shift (b)

Also, standard deviation is always non-negative, so we take |a|

Given:

Original SD = 3.5

Transformation: each xᵢ → -2xᵢ - 3

So:

Multiply by a = -2

Add b = -3 (this does not affect SD)

So new standard deviation = |a| × original SD = |−2| × 3.5 = 2 × 3.5 = 7
Question 2
4 / -1

Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?

A
b² = 3(a² + c²) + 9d²

B
b² = a² + c² + 3d²

C
b² = 3(a² + c² + d²)

D
b² = 3(a² + c²) - 9d²

Solution
Question 3
4 / -1

In how many different ways can the letters of the word 'GEOGRAPHY' be arranged such that the vowels must always come together?

A
2520

B
2530

C
15130

D
15120

Solution

Given:

The given number is 'GEOGRAPHY'

Calculation:

The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA).

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,

The number of ways to arrange these letters = 7!/2!

⇒ 7 × 6 × 5 × 4 × 3 = 2520

In the 3 vowels(EOA), all vowels are different

The number of ways to arrange these vowels = 3!

⇒ 3 × 2 × 1 = 6

Now,

The required number of ways = 2520 × 6

⇒ 15120

∴ The required number of ways is 15120.
Question 4
4 / -1

The equations ax + 9y = 1 and 9y - x - 1 = 0 represent the same line if a =

A
1

B
-1

C
9

D
None of these

Solution

Given:

Equation₁ = ax + 9y = 1

Equation₂ = 9y - x - 1 = 0

Concept used:

If linear equations are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. Here, the equations have an infinite number of solutions, if

a₁/a₂ = b₁/b₂ = c₁/c₂

Calculation:

We have equations,

ax + 9y = 1

⇒ ax + 9y - 1 = 0

and, 9y - x - 1 = 0

⇒ x - 9y + 1 = 0

Here, a₁ = a, b₁ = 9, c₁ = -1

and, a₂ = 1, b₂ = -9, c₂ = 1

As we know that

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ a/1 = 9/-9 = -1/1

⇒ a = -1 = -1

⇒ a = -1

∴ The value of a is -1.
Question 5
4 / -1

If y = y(x) is the solution of the differential equation such that y(0) = 0, then y(1) is equal to:

A
log_e 2

B
2e

C
2 + log_e2

D
1 + log_e2

Solution
Question 6
4 / -1

The mean and variance of a binomial distribution are 4 and 2, respectively. What is the probability of two successes?

A
37/256

B
219/256

C
1/2

D
7/64

Solution
Question 7
4 / -1

The solutions of the equation cos²x + cos²2x + cos²3x = 1 are given by

A
x = nπ ± π/2 ; x = 2mπ ± π/4 ; x = kπ ± π/6 ; where n, m, k ∈ Z

B
x = 2nπ ± π/2 ; x = mπ ± π/4 ; x = 2kπ ± π/6 ; where n, m, k ∈ Z

C
x = nπ ± π/2 ; x = mπ ± π/4 ; x = kπ ± π/6; where n, m, k ∈ Z

D
x = (2n − 1)π ± π/2 ; x = mπ ± π/4 ; x = kπ ± π/6; where n, m, k ∈ Z

Solution

cos²x + (2cos²x − 1)²+ (4cos³x − 3 cos x)²= 1

⇒16 cos⁶x − 20 cos⁴x + 6cos²x = 0 ⇒ 2cos²x(2cos²x − 1)(4cos²x − 3) = 0

⇒ cos²x = 0 ; cos²x = 1/2 = cos²π/4 ; cos²x = 3/4 = cos²π/6

⇒ x = nπ ± π/2 ; x = mπ ± π/4 ; x = kn ± π/6, n, m, k ∈ Z
Question 8
4 / -1

If a plane bisects the line segment joining the points (1, 2, 3) and (-3, 4, 5) at right angles, then this plane also passes through the point

A
(1, 2, -3)

B
(-3, 2, 1)

C
(-1, 2, 3)

D
(3, 2, 1)

Solution
Question 9
4 / -1

The value of k for which the points A (1, 0, 3), B (– 1, 3, 4), C (1, 2, 1) and D (k, 2, 5) are coplanar, is

A
1

B
2

C
0

D
-1

Solution
Question 10
4 / -1

A unit tangent vector at t = 2 on the curve x = t² + 2, y = 4t³– 5, z = 2t² – 6t is

A

B

C

D
None of these

Solution

The position vector of any point at t is r = (2 + t²) i + (4t³ - 5) j + (2t² - 6t) k