Mathematics Test

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Mathematics Test - 7

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Weekly Quiz Competition

Question 1
2.5 / -0.83

The area of the figure bounded by y = e^x , y = e^−x and the straight line x = 1 is

A
e + 2/e

B
e –1/e

C
e + 1/e - 2

D
none of these

Solution
Question 2
2.5 / -0.83

The area bounded by the parabolas y = (x+1)² and y = (x −1)² and the line y = (1/4) is equal to

A
4/3

B
1/3

C
1/6

D
4

Solution

Required area :
Question 3
2.5 / -0.83

The solution of the differential equation is :

Solution

dy/dx + 3y = 2e³ x
p = 3, q = 2e³ x
∫p.dx = 3x
Integrating factor (I.F) = e³ x
y(I.F) = ∫Q(I.F) dx
ye³ x = ∫2e³ x e³ x dx
ye³ x = ∫2e⁶ x dx
ye³ x = 2 ∫e⁶ x dx
ye³ x = 2/6[e⁶ x] + c
ye³ x = ⅓[e⁶ x] + c
Dividing by e³ x, we get
y = ⅓[e³ x] + ce^-(3x)
Question 4
2.5 / -0.83

The integrating factor of differential equation is :

A
log (1/x)

B
log x2

C
log x

D
-log x

Solution

xlog x dy/dx + y = 2logx
⇒dy/dx + y/(xlogx) = 2/x...(1)
Put P = 1/x logx
⇒∫PdP = ∫1/x logx dx
= log(logx)
∴I.F.= e^∫PdP = e^log (logx)
= logx
Question 5
2.5 / -0.83

The solution of the differential equation x dy = (2y + 2x⁴ + x² ) dx is:

A
y = x4+ x2log |x2+ cx2

B
y = x3+ 2x + cx2

C
y = x3+ x log |x| + cx

D
y = x4+ x2log |x| + cx2

Solution

xdy = (2y + 2x⁴ + x² )dx
→dy/dx −(2x)y = 2x³ + x
This differential is of the form y ′+P(x)y=Q(x) which is the general first order linear differential equation, where P(x) and Q(x) are continuous function defined on an interval.
The general solution for this is y ∙I.F = ∫I.F ×Q(x)dx
Where I.F = e ∫P(x)dx is the integrating factor of the differential equation.
I.F = e ∫P(x)dx
= e^∫−2 /xdx
= e(−2 ∙lnx)
= e^ln (x⁻² )
= x⁻²
Thus y(1/x² ) = ∫1/x² (2x³ + x)dx
=∫(2x + 1/x)dx
= x² + lnx + C
⟹y = x⁴ + x² lnx + c
Question 6
2.5 / -0.83

The angle between the vectors is: is :

A

B

C

D
0

Solution

a = 6i - 3j + 2k b = 2i + j - 2k
a.b = 12 - 3 - 4 = 5
|a| = [(6)² + (-3)² + (2)² ]^1/2
|a| = [36 + 9 + 4]^½
|a| = (49)^½
|a| = 7
|b| = [(2)² + (1)² + (-2)² ]^½
|b| = [4 + 1 + 4]^½
|b| = 3
Cos θ= (a.b)/|a||b|
= 5/(7)(3)
= 5/21
θ= cos^-1 (5/21)
Question 7
2.5 / -0.83

If are two vectors, such that , then = ……

A
3

B
√7

C
√5

D
√3

Solution

|a - b|² = |a|² + |b|² - 2|a||b|
|a - b|² = (3)² + (2)² - 2(5)
|a - b|² = 9 + 4 - 10
|a - b|² = 3
|a - b| = (3)^½.
Question 8
2.5 / -0.83

The projection of the vector on the vector is:

A
10/ √6

B
10/ √17

C
2

D
None of the above

Solution

Projection = (A.B)/|B|
= [(i + 2j + k) . (2i + 3j + 2k)]/[(2)² + (3)² + (2)² ]^½
= (2 + 6 + 2)/[4 + 9 + 4]^½
= 10/(17)^1/2
Question 9
2.5 / -0.83

An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is:

A
1/2

B
3/10

C
1/10

D
3/5

Solution
Question 10
2.5 / -0.83

A bag ‘A ’contains 2 white and 3 red balls and a bag ‘B ’contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. The probability that it was drawn from bag B is:

A
27/52

B
3/5

C
25/52

D
5/9

Solution

If we assume that it is equally likely that the ball is drawn from either bag, then we proceed as follows. Let R be the event that a red ball is drawn, and let B be the event that bag B is chosen for the drawing. Then
P(B) = P(not B) = 1/2
P(R|B) = 5/9 = P(RB) / P(B) = 2 P(RB),
P(R|not B) = 3/5 = P(R not B) / P(not B) = 2 P(R not B)
P(R) = P(RB) + P(R not B) = (1/2)( 5/9 + 3/5 ) = 26/45
P(B|R) = P(RB) / P(R) = (5/9)(1/2) / (26/45) = 25/52