Mathematics Test 152

Result

Mathematics Test 152

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Weekly Quiz Competition

Question 1
4 / -1

If the line intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to :

A

B
√14

C
2 √7

D
14

Solution
Question 2
4 / -1

Let in equilateral ΔABC,

A(–1 + acosθ, 2 + asinθ),
B(–1 + acosα, 2 + asinα),
C(–1 + acosβ, 2 + asinβ)

and length of median through vertex A is 2b, then equation of circumcircle of triangle ABC is (where 'a' is consant) -

A
x² + y² + 18x – 36y + 5 – b² = 0

B
9x² + 9y² + 18x – 36y + 45 – 16b² = 0

C
9x² + 9y² + 18x – 36y + 45 – 4b² = 0

D
9x² + 9y² – 18x + 36y + 45 – 4b² = 0

Solution
Question 3
4 / -1

Length of the normal chord of the parabola y² = 4x, which makes an angle π/4 with x-axis is -

A
8

B

C
4

D

Solution

Normal is y + xt = 2t + t³

slope –t = 1 ⇒ t = –1

∴ P(1,–2)
Question 4
4 / -1

The equation of the perpendicular bisectors of the sides AB and AC of a triangle ABC are y = x and y = –x, respectively. If the point A is (1, 2), then the area of ΔABC is :-

A
6 sq. units

B
3 sq. units

C
9 sq. units

D
2 sq. units

Solution
Question 5
4 / -1

The sum of the series up to ∞ is :

A
3

B
2

C
1

D
1.5

Solution
Question 6
4 / -1

If all roots of the equation

f(x) = x⁶ – 12x⁵ + bx⁴ + cx³ + dx² + ex + 64 = 0

are positive, then which has greatest numerical (absolute) value :-

A
b

B
c

C
d

D
e

Solution

Let roots are x₁, x₂, x₃, x₄, x₅, x₆

AM ≥ GM

⇒ x₁ = x₂ = x₃ = x₄ = x₅ = x₆ = 2

(x–2)⁶ = x⁶ – 12x⁵ + bx⁴ + cx³ + dx² + ex + 64 = 0

b = ⁶C₂2², c = –⁶C₃2³, d = ⁶C₄2⁴, e = –⁶C₅2⁵

b = 60, c = –160, d = 210, e = –192

d = 210
Question 7
4 / -1

then value of f(25) is :-

A
2

B
0

C
8

D
25

Solution
Question 8
4 / -1

In the argand plane, all the complex numbers satisfying |z – 4i| + |z + 4i| = 10 lie on-

A
a straight line

B
a circle

C
an ellipse

D
a parabola

Solution

Z₁ (0, 4), Z₂(0, – 4), k = 10

|z₁ – z₂| = 8

k > |z₁ – z₂|

Ellipse
Question 9
4 / -1

Number of ways in which 4 people can be select out of 10 people sitting in a row such that exactly two are consecutive ?

A
105

B
75

C
11

D
91

Solution
Question 10
4 / -1

If  and det(A) = det(4I), where I is 3 × 3 identity matrix, then (a – b)³ + (b – c)³ + (c – a)³ can be equal to -

A
–10

B
–24

C
–18

D
–6

Solution

⇒   det(A) = (a – b)² (b – c)² (c – a)²

      & det (4I) = 64

⇒   (a – b)(b – c)(c – a) = ±8

∵   (a – b) + (b – c) + (c – a) = 0

∴   (a – b)³ + (b – c)³ + (c – a)³

       = 3(a – b)(b – c)(c – a) = ±24