Mathematics Test 137

Result

Mathematics Test 137

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Question 1
4 / -1

The equation represents the equation of a circle with ω, ω² as extremities of a diameter, then λ is , (where ω, ω² are cube roots of unity)

A
√2

B
2

C
3

D
4

Solution

If P(z) lies on a circle whose diametrically opposite points are A(z₁) & B(z₂) then
Question 2
4 / -1

If sin^–1x + sin^–1y + sin^–1z = π/2 then x² + y² + z² + 2xyz is equal to -

A
-1

B
0

C
1

D
1/2

Solution

Given sin^–1x + sin^–1y + sin^–1z = π/2

⇒ (1-x²) (1-y²) = (xy + z)²
⇒ x² + y² + z² + 2xyz = 1
Question 3
4 / -1

If b² > 4ac for the equation 4x⁴ + bx² + c = 0, then all the roots of the equation will be real if:

A

B

C

D

Solution

Clearly, if all the roots of are real, then the equation must have both the roots positive. But
which is a quadratic with positive coefficient of t². Hence for both the roots to be positive:

⇒ a & care of same sign and a & b are of opposite sign
Question 4
4 / -1

The value of the definite integral for 0 < α < π, equal to

A
sin α

B
tan^-1(sinα)

C
α sin α

D

Solution
Question 5
4 / -1

Coordinates of the orthocentre of the triangle whose sides are x = 3, y = 4 and 3x + 4y = 6 will be:

A
(0, 0)

B
(3, 0)

C
(0, 4)

D
(3, 4)

Solution

Sides of a triangle ABC are given by x = 3, y = 4, 3x + 4y = 6 It forms a right angle triangle ABC with B(3, 4)as right angle.

Hence B is the orthocentre as perpendiculars drawn from A and C meet at B.
Question 6
4 / -1

An equilateral triangle is inscribed in the parabola y²= 4ax whose vertex is at the vertex of the parabola. The length of its side is

A
2a√3

B
4a√3

C
6a√3

D
8a√3

Solution
Question 7
4 / -1

The projection of the line joining the points (3,4,5) and (4,6,3) on the line joining the points (-1,2,,4) and (1,0,5) is

A
4/3

B
2/3

C
1/3

D
1/2

Solution

The direction ratios of the line joining are proportional to 2,-2, 1
∴ Its direction cosines are
Thus, the projection of the line joining and on PQ is given by
Question 8
4 / -1

Consider the following statements regarding the events E, A, B and C .

(i) Event E can take place due to the occurrence of any of the events A, B, C.

(ii) Events A, B and C are equiprobable, mutually exclusive and exhaustive.

(iii) Probability of occurrence of event E is 5/12.

Considering above data the value of P(E/C) is:

A
3/4

B
3/8

C
4/5

D
5/8

Solution
Question 9
4 / -1

(P→q) ↔ (q V ~ P) is –

A
equivalent to

B
Tautology

C
Fallacy

D
Neither tautology nor fallacy.

Solution

Truth table

From truth table
(P→q) ↔ (q V ~ P) is always true so it is a tautology
Question 10
4 / -1

The equation of the hyperbola whose foci are (6, 5), (- 4, 5) and eccentricity 5/4 is

A

B

C

D

Solution

Center of the hyperbola is the mid-point of the line segment joining two foci. Therefore, coordinates of the center are (1, 5). Also since the foci have same y coordinate therefore the line joining foci is parallel to x-axis. Hence the transverse axis is parallel to x-axis. Clearly the equation of the hyperbola is of form