Mathematics Test 140

Result

Mathematics Test 140

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Question 1
4 / -1

A

B

C

D

Solution
Question 2
4 / -1

ABC is an acute - angled triangle with circumcentre 'O' and orthocentre H. If AO = AH, then angle A is

A
π/2

B
π/6

C
π/8

D
π/3

Solution

As we know in triangle, the distance of the ortho-centre from vertex A is 2R cos A
In the question it is given distance of the circum-centre and ortho-centre from the vertex A is equal.
Question 3
4 / -1

The locus of the mid point of the line segment joining the focus to a moving point on the parabola y² = 4ax is another parabola with directrix

A
x = -a

B
x = -a/2

C
x = 0

D
x = a/2

Solution

Let p (h, k) be the mid point of the line segment joining the focus (a, 0) and a general point Q (x, y) on the parabola. Then
Question 4
4 / -1

The eccentricity of the ellipse which meets the straight line on the axis of x and the straight line on the axis of y and whose axes lie along the axes of coordinates, is :

A

B

C

D

Solution
Question 5
4 / -1

A circle is described whose centre is the vertex and whose diameter is three-quarters of the latus rectum of a parabola y² = 4ax. The common chord of the circle and parabola is

A
x = a/2

B
x = −a/2

C
x = a/4

D
None of these

Solution

⇒ x = a/2 common chord of circle and parabola.
Question 6
4 / -1

A (3, 2, 0), B (5, 3, 2) and C (-9, 6, -3) are three points forming a triangle and AD is bisector of the angle AD meets BC at the point :

A

B

C

D
None of these

Solution

The bisector of ∠BAC i.e., AD divides the side BC in the ratio AB : AC
Question 7
4 / -1

If p, q ∈{1,2,3,4} the number of equations of the form px²+ qx + 1 = 0 having real roots is

A
15

B
9

C
7

D
8

Solution

For real roots, we must have
Question 8
4 / -1

A

B

C

D
None of these

Solution
Question 9
4 / -1

The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is

A
6/5

B
5/3

C
10/3

D
3/5

Solution

Let the equation of the circle is (x - 1)² + (y - k)² = k²
It passes through (2, 3) ⇒ (2 - 1)² + (3 - k)² = k²
Question 10
4 / -1

If f(x) = a - (x - 3)^8/9, then maximum value of f(x) is:

A
3

B
a - 3

C
a

D
None of these

Solution

At x = 3: f' (x) is not defined,
hence sign scheme of f' (x) is

∴ f(x) is maximum at x = 3
Hence maximum value of f (x) is equal to "a"