Mathematics Test 253

Result

Mathematics Test 253

Score
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Rank
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Weekly Quiz Competition

Question 1
4 / -1

A
3/4

B
3π/8

C
0

D
3/16π

Solution
Question 2
4 / -1

If λ ∈ R is such that the sum of the cubes of the roots of the equation x² + (2 - λ) x + (10 - λ) = 0 is minimum, then the magnitude of the difference between the roots of this equation is

A
4√2

B
20

C
2√5

D
2√7

Solution
Question 3
4 / -1

If tan A and tan B are the roots of the quadratic equation 3x² - 10x - 25 = 0, then the value of 3 sin² (A + B) - 10 sin (A + B) ×cos (A + B) - 25 cos²(A + B) is

A
-25

B
10

C
-10

D
25

Solution

Since tan A and tan B are roots of the equation 3x2 - 10x - 25 = 0, therefore tan A + tan B = 10/3.
Question 4
4 / -1

If the system of linear equations

x + ky + 3z = 0

3x + ky - 2z = 0

2x + 4y - 3z = 0

has a non-zero solution (x, y, z), then xz/y² is equal to

A
-10

B
10

C
-30

D
30

Solution
Question 5
4 / -1

A
Does not exist

B
Exists and is equal to -2

C
Exists and is equal to 0

D
Exists and is equal to 2

Solution
Question 6
4 / -1

If β is one of the angles between the normals to the ellipse x² + 3y² = 9 at the points (3 cos θ, √3 sin θ) and (-3 sin θ, √3 cos θ); θ ∈ (0, π/2), then 2 cot β/sin2θ is equal to

A
1/√3

B
√3/4

C
2/√3

D
√2

Solution
Question 7
4 / -1

A
Neither injective nor surjective

B
Invertible

C
Injective but not surjective

D
Surjective but not injective

Solution

∴ From the above diagram of f(x), f(x) is surjective but not injective.
Question 8
4 / -1

The number of four-lettered words that can be formed using the letters of the word BARRACK is:

A
270

B
144

C
264

D
120

Solution

B's - 1

A's - 2

R's - 2

C's - 1

K's - 1

The number of four-lettered words can be formed from (2 identical, 2 identical) or (2 identical, 2 different) or 4 different letters.
Question 9
4 / -1

A
a = 2n, b = 2ⁿ

B
a = 2ⁿ, b = 2n

C
a = 2ⁿ, b = n2^n-1

D
a = 2ⁿ, b = n2ⁿ

Solution
Question 10
4 / -1

A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws , is

A
5/6

B
1/6

C
5/11

D
6/11

Solution

Required Probability = 5/6 x 1/6 + (5/6)³x 1/6 + (5/6)⁵ x 1/6 + ....

= 1/6 x 5/6 / (1-25/36)

= 5/11