Mathematics Test 226

Result

Mathematics Test 226

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Question 1
4 / -1

Let z₁ and z₂ be the nth roots of unity which are ends of a line segment that subtend a right angle at the origin. Then, n must be of the form

A
4k+1

B
4k+2

C
4k+3

D
4k

Solution
Question 2
4 / -1

A
−x + (1 + x²)cot⁻¹x + c

B
x − (1 + x²)cot⁻¹x + c

C
x − (1 + x²) tan⁻¹x + c

D
−x + (1 + x²)tan⁻¹x + c

Solution
Question 3
4 / -1

If A is a square matrix of order 3 with |A| = 2, then the value of |(A − A^T)⁵| + |(A^T− A)³| is

A
24

B
16

C
0

D
8

Solution

We know, (A^T − A) and (A − A^T) are skew symmetric matrices of (3 × 3) order.

Also, the value of the determinant of a skew symmetric matrix of order is zero.
Question 4
4 / -1

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by dP/dx = 100 − 12√x. If the firm employs 25 more workers, then the new level of production of items is

A
3500

B
4500

C
2500

D
3000

Solution
Question 5
4 / -1

AB is a vertical tower. The point A is on the ground and C is the middle point of AB. The part CB subtends an angle α at a point P on the ground. If AP = nAB, then the correct relation is

A
n = (n²+ 1) tan α

B
n = (2n²− 1) tan α

C
n²= (2n²+ 1) tan α

D
n = (2n²+ 1) tan α

Solution
Question 6
4 / -1

If x = −1 and x = 2 are points of extrema of f(x) = α log|x| + βx²+ x, then

A

B
α = 2, β = 1/2

C
α = −6, β = 1/2

D

Solution
Question 7
4 / -1

If a,b,c,d,e,f are in arithmetic progression. Then e − c is equal to

A
2(c − a)

B
2(d − c)

C
2(f − d)

D
d − c

Solution

a, b, c, d, e are in A.P.

Let common difference = D

∴ e − c = (a + 4D) − (a + 2D) = 2D = 2(d − c)
Question 8
4 / -1

The number of ways of arranging 8 men and 4 women around a circular table such that no two women can sit together, is

A
8!

B
4!

C
(8!)(4!)

D
7!(⁸P₄)

Solution

The number of ways of arranging 8 men = 7!

The number of ways of arranging 4 women such that no two women can sit together = ⁸P₄

∴ Required number of ways =7!(⁸P₄)
Question 9
4 / -1

Fifteen coupons are numbered 1, 2, 3,…..,15 respectively. Seven coupons are selected at random one at a time with replacement. The probability, that the largest number appearing on selected coupons is at most 9, is

A
(1/15)⁷

B
(3/5)⁷

C
(8/15)⁷

D
(2/5)⁷

Solution

Total coupons = 15

1≤1≤ selected coupon number ≤9

i.e., 1, 2,3, 4,5, 6,7, 8,9

∴ Probability of one selected coupon to have number ≤ 9 = 9/15 = 3/5

Hence, the required probability

(Using multiplicative principle for independent events)
Question 10
4 / -1

If O is the origin, OP and OQ are distinct tangents to the circle x²+ y²+ 2gx + 2fy + c = 0, then the circumcentre of the triangle OPQ is

A
(−g,−f)

B
(g, f)

C
(−f, −g)

D
None of these

Solution

Tangents drawn from the point O, meet the cirle at P & Q and C be the centre of given circle. Then points O, P, C and Q are concyclic. That means any circle passing through O, P and Q also passes through C and OC is the diameter for this circle. Hence, mid-point of OC is the circumcentre of triangle OPQ.