Mathematics Test

Result

Mathematics Test - 12

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Weekly Quiz Competition

Question 1
2.5 / -0.83

A
0

B
1

C
does not exist

D
none of these

Solution
Question 2
2.5 / -0.83

Let f(x) = x –[x], then f ‘(x) = 1 for

A
all x ∈I

B
all x ∈(R −I)

C
all x ∈R

D
all x ∈Rx- {0]

Solution

f(x) = x -[x] is derivable at all x ∈ R –I , and f ‘(x) = 1 for all x ∈ R –I .
Question 3
2.5 / -0.83

The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 5x² + 22x + 35^. Find the marginal revenue, when x = 7, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant

A
Rs 7

B
Rs 127

C
Rs 92

D
Rs 48

Solution
Question 4
2.5 / -0.83

The volume of cube is increasing at the constant rate of 3 cm³ /s. Find the rate of change of edge of the cube when its edge is 5 cm.

A
25 cm3/sec

B
25 cm/s

C
1/25 cm/s

D
1/25 cm3/s

Solution

Let V be the instantaneous volume of the cube.
dV/dt = 3 cm³ /s
Let x be the side of the cube.
V = x³
dV/dt = 3x² * (dx/dt)
3 = 3*(5² )*(dx/dt)
So, dx/dt = 1/25 cm/s
Question 5
2.5 / -0.83

The total cost associated with the production of x units of a product is given by C(x) = 5x² + 14x + 6. Find marginal cost when 5 units are produced

A
Rs.64

B
Rs.70

C
Rs.50

D
Rs.(10x + 14)

Solution

C(x) = 5x² + 14x + 6
Marginal Cost M(x) = C ’(x)
M(x) = 10x +14
So, M(5) = 50 + 14
= 64 Rs
Question 6
2.5 / -0.83

The function is monotonically decreasing in

A
(2, ∞)

B
(0, 1)

C
(0, 1) and (2, ∞)

D
(–∞, ∞)

Solution

f(x)= {−(x −1)/x² x <1, x(x −1)/x² x ≥1}
f '(x)= {(x −2)/x³ x <1, −(x −2)/x³ x ≥1}
x <1, if f ′(x)<0 (for f(x) to be monotonically decreasing
⇒(x −2)/x³ <0
⇒x ∈(0,2)
But x <1 ⇒x ∈(0,1)
For x ≥1, if f ′(x)<0
⇒−(x −2)]/x³ <0
⇒(x −2)/x³ >0
⇒x ∈(−∞,0)∪(2,∞)
But, x ≥1 ⇒x ∈(2,∞)
Hence, x ∈(0,1)∪(2,∞)
Question 7
2.5 / -0.83

The true set of real values of x for which the function, f(x) = x l n x – x + 1 is positive is

A
(1, ∞)

B
(1/e, ∞)

C
[e, ∞)

D
(0, 1) and (1, ∞)

Solution

f(x)=xlnx −x+1
f(1)=0
On differentiating w.r.t x, we get
f ′(x)=x*1/x+lnx −1
=lnx
Therefore,f ′(x)>0 for allx ∈(1,∞)
f ′(x)<0 for allx ∈(0,1)
lim x →0 f(x)=1
f(x)>0
for allx ∈(0,1)∪(1,∞)
Question 8
2.5 / -0.83

The value of is:

A
10

B
17/2

C
7/2

D
5

Solution

∫(-3 to 3) (x+1)dx

= ∫(-3 to -1) (x+1)dx + ∫(-1 to 3) (x+1) dx

= [x² + x](-3 to -1) + [x² + x](-1 to 3)

= [½- 1 - (9/2 - 3)] + [9/2 + 3 - (½- 1)]

= -[-4 + 2] + [4 + 4]

= -[-2] + [8]

= 10
Question 9
2.5 / -0.83

A

B

C

D

Solution

Option d is correct, because it is the property of definite integral
∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a –x) dx
Question 10
2.5 / -0.83

The numbers 2, 4, 6, 8 and 10 are written separately on five slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Then, the sample space for the experiment is:

A

B

C

D

Solution

Elements {2,4,6,8,10}
A person draws two slips from the box, one after the other, without replacement:
Sample Space ={(2,4) (2,6) (2,8) (2,10) (4,2) (4,6) (4,8) (4,10) (6,2) (6,4) (6,8) (6,10) (8,2) (8,4) (8,6) (8,10) (10,2) (10,4) (10,6) (10,8)}