Mathematics Test 105

Result

Mathematics Test 105

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Weekly Quiz Competition

Question 1
4 / -1

The series of natural numbers is divided into groups as (1), (2, 3, 4), (3, 4, 5, 6, 7), (4, 5, 6, 7, 8, 9, 10) .......... . If sum of elements in the 20^th group is ℓ, then ℓ is equal to

A
1368

B
(38)²

C
(39)²

D
38 ×39

Solution
Question 2
4 / -1

The lines L₁ : x = y = z, L₂ : x = y/2 = z/3 and a line L₃ is passing through (1, 1, 1) form a triangle of area √6 units, (1, 1, 1) being one of the vertices of the triangle. Then the point of intersection of the L₃ with L₂ is

A
(1, 2, 3)

B
(2, 4, 6)

C
(4/3, 8/3, 4)

D
(1, 5, 7)

Solution
Question 3
4 / -1

The combined equation of 2 altitudes of an equilateral triangle is x² – 3y² – 4x + 6√3 y–5 = 0. The third altitude has equation.

A
x + 2 = 0

B
y = √3

C
x = 2

D
None of these

Solution
Question 4
4 / -1

The mean marks of students of a school is 56. The mean marks of girls is 60 and that of boys is 50. If number of boys and girls are n and m respectively, then 9n / m equals

A
9

B
2

C
3

D
6

Solution
Question 5
4 / -1

C₁ and C₂ are two circles whose equations are given as x² + y² = 25 and x² + y² + 10x + 6y + 1 = 0. Now C₃ is a variable circle which cuts C₁ and C₂ orthogonally. Tangents are drawn from the center of C₃ to C₁, if the locus of the mid point of the chord of contact of tangents is (where a, b ∈ Z⁺, and a and b are relative prime), then b/a is

A
1

B
2

C
4

D
5

Solution

Centre of C₃ will lie on the radical axis of C₁ and C₂ which is 10x + 6y + 26 = 0.

Let center of C₃ is (h, k).

Equation of chord of contact through (h, k) to the C₁ may be given as hx + ky = 25 ... (I)

Let the mid point of the chord is (x₁, y₁) the equation of the chord with the help of mid point may be given as
Question 6
4 / -1

A line is drawn from a point P(x, y) on curve y = f(x), making an angle in anti-clockwise with the +ve x-axis which is supplementary to the one made by the tangent to the curve at P(x, y). The line meets the x-axis at A. Another line perpendicular to the first, is drawn from P(x, y) meeting the y-axis at B. If OA = OB, where O is the origin, then the curve which passes through (1, 1).

A
x² – y² + 4xy = 4

B
x² – y² – 2xy + 2 = 0

C
x² – y² + 2xy = 2

D
x² – y² – 4xy + 4 = 0

Solution

The equation of the line through P(x, y) making an angle with the x-axis which is suplementary to the angle made by the tangent at P(x, y) is
Question 7
4 / -1

Let p : Sindhu plays to win

q : Sindhu gets Bharath Ratna

Then the contrapositive of "~(~q^p)" is

A
If Sindhu plays to win then she gets Bharath Ratna

B
If Sindhu gets Bharat Ratna then she plays to win

C
If Sindhu does not get Bharath Ratna then she does not plays to win

D
option 1 and option 3 are correct

Solution
Question 8
4 / -1

Let A and B are non-singular matrices of order 3 such that |A| = 5 and A^–1B² + AB = 0, then A² |A²| – adj (adjB) is equal to

A
null matrix

B
25A² – 5B

C
25A²

D
50A²

Solution
Question 9
4 / -1

Number of points of non-differentiability of f(x) = | | x | – 1 | + | cosπx |; –2 < x < 2 is

A
7

B
6

C
5

D
4

Solution
Question 10
4 / -1

If C₀, C₁, ....., C₂₀₁₂ are binomial coefficients in the expansion of (1 + x)²⁰¹² and a₀, a₁, ....., a₂₀₁₂ are real numbers in arithmetic progression then value of a₀C₀ – a₁C₁ + a₂C₂ – a₃C₃ + ..... + a₂₀₁₂C₂₀₁₂ is a

A
Even Number

B
Odd number

C
Natural number

D
Prime number

Solution

a₀C₀ – a₁C₁ + a₂C₂ – a₃C₃ + ..... + a₂₀₁₂c₂₀₁₂

= a₀ (C₀ – C₁ + C₂ + ..... + C₂₀₁₂)

– d (C₁ – 2C₂ + 3C₃ – 4C₄ + ..... – ²⁰¹²C₂₀₁₂)

= a₀ (0) – d (C₁ – 2C₂ + 3C₃ – 4C₄

+ ..... – ²⁰¹²C₂₀₁₂)

= a₀(0) – d (0)

= 0