Mathematics Test 224

Result

Mathematics Test 224

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Weekly Quiz Competition

Question 1
4 / -1

A
1/6

B
1/7

C

D
-1/6

Solution
Question 2
4 / -1

The value of denotes the greatest integer function) is

A
0

B
1

C
2

D
None of these

Solution
Question 3
4 / -1

If R = {(x, y)∣x, y ∈ Z, x²+ y²≤ 4} is a relation in Z , then domain of R is

A
{0, 1, 2}

B
{0, − 1, − 2}

C
{− 2, − 1, 0, 1, 2}

D
None of these

Solution

∵ R = {(x, y)∣x, y ∈ Z, x²+ y²≤ 4}

R = {(−2, 0),(−1, 0),(0, −1),(−1, 1),(1, −1),(0, −1)(0, 1),(0, 2),(0,−2),(1, 0),(0, 1),(1, 1),(−1, −1),(2, 0),(0, 0)}

Hence, domain of R = {−2, −1, 0, 1, 2}
Question 4
4 / -1

If y = 4x − 5 is tangent to the curve y²= px³+ q at (2,3) then (p,q) is

A
(2,7)

B
(−2, 7)

C
(−2, −7)

D
(2,−7)2,-7

Solution
Question 5
4 / -1

The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a man and a woman is

A
1960

B
15!

C
(15!)²

D
14!

Solution

First team can be selected in (¹⁵C₁.¹⁵C₁) way, second in (¹⁴C₁.¹⁴C₁) way and so on. So, number of ways of selecting 15 teams
Question 6
4 / -1

A
0

B

C

D
π/2

Solution
Question 7
4 / -1

The area bounded by the curves y = (x − 1)², y = (x + 1)² and y = 1/4 is

A
1/3 sq.units

B
2/3 sq.units

C
1/4 sq.units

D
1/5 sq.units

Solution

The curves y = (x − 1)², y = (x + 1)²and y = 1/4 are shown as
Question 8
4 / -1

The equation of the pair of straight lines parallel to the y−y-axis and which are tangents to the circle x² + y² − 6x−4y−12 = 0 is

A
x² − 4x − 21 = 0

B
x² − 5x + 6 = 0

C
x² − 6x − 16 = 0

D
None of these

Solution

Given circle is x²+ y²− 6x − 4y − 12 = 0

⇒ (x − 3)²+ (y − 2)²= 25 = 5²

Centre = (3, 2) radius = 5

If a line parallel to y-axis, x = λ touches the given circle, then

Pair of lines (x + 2)(x − 8) = 0

⇒ x²−6x −16 = 0.
Question 9
4 / -1

If y = log₁₀x + log_x10 + log_x x + log₁₀10, then dy/dx is equal to

A

B

C

D
None of these

Solution
Question 10
4 / -1

If a,b,c,d,e,f are in arithmetic progression. Then e − c is equal to

A
2(c − a)

B
2(d − c)

C
2(f − d)

D
d − c

Solution

a, b, c, d, e are in A.P.

Let common difference = D

∴ e − c = (a + 4D) − (a + 2D) = 2D = 2(d − c)