Mathematics Test 141

Result

Mathematics Test 141

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Weekly Quiz Competition

Question 1
4 / -1

The number of terms in the expansion of which are integers, is given by

A
501

B
500

C
250

D
251

Solution

⇒ R = 0,2,4,........,250
∴ Number of integers terms = 251.
Question 2
4 / -1

In the Argand plane if O, P, Q represent respectively the origin, the complex number z and z + iz, then angle OPQ is

A
π/4

B
π/3

C
π/2

D
2π/3

Solution
Question 3
4 / -1

There are unlimited number of identical balls of four different colours. How many arrangements of at most 8 balls in row can be made by using them?

A
97380

B
87380

C
87370

D
87480

Solution

The number of arrangements of one ball = 4,
because there are only four different balls.
The number of arrangements of two balls = 4 x 4 = 4² , etc.
∴ the required number of arrangements
Question 4
4 / -1

The value of iⁱ is :

A
ω

B
-ω²

C
π/2

D
e^-π/2

Solution
Question 5
4 / -1

For what value of 'a' is the area bounded by the curve y = a²x² + ax + 1 and the straight line y = 0, x = 0 & x = 1 the least ?

A

B

C

D
None of these

Solution
Question 6
4 / -1

If a₁, a₂, a₃,....a₂₀ are AMs between 13 and 67, then the maximum value of product a₁ a₂ a₃ ....a₂₀ is.

A
(20)²⁰

B
(60)²⁰

C
(80)²⁰

D
(40)²⁰

Solution
Question 7
4 / -1

The area bounded by the curvesy = (x - 1)² ,  y = (x+1) ² and y = 1/4 is

A
1/3  sq unit

B
2/3  sq unit

C
1/4  sq unit

D
1/5  sq unit

Solution

The curves y = (x - 1)², y = (x + 1)² and y = 1/4 are shown as

where point of intersection are
Question 8
4 / -1

Let z and ω be two non-zero complex numbers such that z = ω and arg z + arg ω = π, then z equals

A
ω

B
- ω

C

D

Solution
Question 9
4 / -1

If two distinct chords, drawn from the point (p, q) on the circle x² + y² = px + qy (where pq ≠ 0) are bisected by the x-axis, then

A
p² = q²

B
p² = 8q²

C
p² < 8q²

D
p² > 8q²

Solution

Let (h,0) be the mid-point of chord then its equation is
T = S,.
Question 10
4 / -1

If two vertices of a triangle are (5, -1), (-2, 3) and the orthocentre of the triangle lies at the origin, then the third vertex is

A
(4, 7)

B
(-4, -7)

C
(4, -7)

D
(-4, 7)

Solution

Let the third vertex C be (x₁. y₁) since 0(0, 0) is the orthocenter