Mathematics Test 154

Result

Mathematics Test 154

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Weekly Quiz Competition

Question 1
4 / -1

The existence of unique solution of the system of equations,

x + y + z = β,

5x – y + αz = 10,

2x + 3y – z = 6

depends on

A
α only

B
β only

C
α and β both

D
neither α nor β

Solution

Here, Δ ≠ 0

As, Δ = (23 – α) ≠ 0
Question 2
4 / -1

The value of

A
⁵⁰C₁₀

B
⁵⁰C₁₀ – ³⁵C₁₀

C
⁴⁰C₁₀

D
(⁴⁰C₁₀ – 1)

Solution

= ¹⁵C₁ × ³⁵C₉ × ¹⁵C₂ × ³⁵C₈ +...+¹⁵C₁₀ × ³⁵C₀

= ⁵⁰C₁₀ – ³⁵C₁₀
Question 3
4 / -1

The total number of shortest path from A to B is:-

A
1326

B
1428

C
1632

D
1248
Question 4
4 / -1

If |z| = 1 then is equal to ?

A
40

B
30

C
20

D
35

Solution
Question 5
4 / -1

A ten digit number is formed (without repetition), the probability that the difference of the digits at equal distances from the begining and the end is always 1 is ?

A
17 / 1994

B
4 / 27

C
1 / 945

D
34 / 243

Solution

dis joint pair are (0,1), (2,3), (4,5), (6,7), (8,9) so required probability =
Question 6
4 / -1

The coordinates of the foot of the perpendiculars from the vertices of a triangle on the opposite sides are (20,25), (8,16) and (8,9). The orthocentre of the triangle lies at the point-

A
(5,10)

B
(15,30)

C
(10,15)

D
(50,–5)

Solution
Question 7
4 / -1

The co-ordinates of the orthocentre of the triangle formed by the lines 2x² – 2y² + 3xy + 3x + y + 1 = 0 and 3x + 2y + 1 = 0 are :-

A

B

C

D

Solution

2x² – 2y² + 3xy + 3x + y + 1

⇒ 2x² + 3(y + 1) x – (2y² – y – 1) = 0

⇒ (x + 2y + 1) (2x – y + 1) = 0

∴ The sides of the triangle are

x + 2y + 1 = 0        ....(1)

2x – y + 1 = 0        ....(2)

3x + 2y + 1 = 0       ....(3)

∵ Lines (1) and (2) are perpendicular

∴ Orthocentre H lies on their point of intersection

∴ H ≡
Question 8
4 / -1

The point A on the parabola y² = 4ax for which |AC – BC| is maximum, whose B(0, a) and C(–a, 0) is :-

A
(a, 2a)

B
(4a, 4a)

C
(a, –2a)

D
None of these

Solution
Question 9
4 / -1

The distance of the point having position vectorfrom the straight line passing through the point (2, 3, –4) and parallel to the vector,

A
7

B
4 √3

C
2 √13

D
6

Solution
Question 10
4 / -1

If the distance of the point P(1,–2,1) from the plane x + 2y – 2z = a, where a > 0, is 5, then the foot of the perpendicular from P to the plane is-

A

B

C

D

Solution