Mathematics Test

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Mathematics Test - 3

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Question 1
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Let f be a function satisfying f(xy) = f(x)/y for all positive real numbers. If f(500) = 3, then what is the value of f(600)?

A
A. 2.0

B
B. 3.6

C
C. 2.5

D
D. 4.0

Solution

Given, f(xy) =

Also, f(500) = 3

⇒ f(100 × 5) = = 3

⇒ f(100) = 15

∴ f(600) = f(100 × 6)

=

= 2.5
Question 2
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If z = , then

A
A. Re(z) = 0

B
B. Im(z) = 0, Re(z) <0

C
C. Re(z) = 0, Im(z) > 0

D
D. Re(z) > 0, Im(z) <0

Solution

Let

Then

∴ Im(z) = 0, Re(z)
Question 3
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The sum of the cubes of three consecutive natural numbers is divisible by

A
10

B
9

C
13

D
14

Solution

Let n, n + 1, n + 2 be three consecutive numbers.

Then, P(n) = n³ + (n + 1)³ + (n + 2)³

We prove by induction on n.

P(1) = 1 + 8 + 27 = 36, which is divisible by 9.

Let the result be true for k, i.e. P(k) = k³ + (k + 1)³ + (k + 2)³ is divisible by 9.

For k + 1, we have

P(k + 1) = (k + 1)³ + (k + 2)³ + (k + 3)³

= (k + 1)³ + (k + 2)³ + k³ + 9k² + 27k + 27

= k³ + (k +1)³ + (k + 2)³ + 9(k² + 3k + 3) = P(k) + 9(k² + 3k + 3), which is divisible by 9.

Hence, P(n) is divisible by 9 for all natural numbers.
Question 4
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What is the sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ……. up to n terms?

A
A. n(n + 1)(n + 2)/3

B
B. n(n + 1)(n + 2)(n + 3)/4

C
C. (n + 1)(n + 2)(n + 3)/6

D
D. None of these

Solution

nth term = n(n + 1)(n + 2) = n(n² + 3n + 2) = n³ + 3n² + 2n. Let the sum be S, then

S =

S = [{n(n + 1)/(2)}² + 3n(n + 1)(2n + 1)/6 + 2n(n + 1)/2]

S = n(n + 1)[n(n + 1)/4 + (2n + 1)/2 + (1)]

S = n(n + 1)[(n² + n + 4n + 2 + 4)/4]

S = n(n + 1)(n + 2)(n + 3)/4
Question 5
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Let f(x) = . If f(x) is continuous in , then is

A
A. -1/2

B
B. 1/2

C
C. 1

D
D. -1

Solution

Since f is continuous, so

= -1/2
Question 6
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The locus of the mid-points of the focal chords of the parabola is another parabola with which of the following equations?

A
y² = 2a(x - a)

B
y² = a(x - a)

C
y² = 4a(x - 1)

D
None of these

Solution

Let PQ be the focal chord of the parabola. Let the coordinates of P and Q be (at₁² , 2at₁ ) and (at₂² , 2at₂ ), respectively.

Then, t₁ t₂ = -1

Let (h, k) be coordinates of the mid-point of PQ. Then,
Question 7
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In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .

A
A. 2/3

B
B. 1/3

C
C. 3/4

D
D. 1/2

Solution

We know that, mean = np = 18 and variance = n p q = 12

Now,

∴ p = 1 - q = 1 - (⅔) = (⅓)
Question 8
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The particular solution of the differential equation log(dy/dx) = x, when x = 0, y = 1 is ...

A
A. y = e^x + 2

B
B. y = -e^x

C
C. y = -e^x + 2

D
D. y = e^x

Solution

Wo have, differential equations,

dy = e^x dx

Integrating on both sides, we get ∫dy = ∫e^x dx

⇒ y = e^x + C …(i)

On putting x = 0, y = 1 is Eq. (i), we get

1 = e⁰ + C ⇒ C = 0

Now, particular solution of the given differential is y = e^x
Question 9
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If the function

is continuous at x = 0. then k = ...

A
A. ± 1/8

B
B. ± 4

C
C. ± 2

D
D. ± 8

Solution

We have function,

16, x = 0 is continuous at x = 0
Question 10
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The maximum value of z = 6x + 8y subject to x - y ≥0, x + 3y ≤12, x ≥0, y ≥0 is.....

A
72

B
42

C
96

D
24

Solution

We have, z = 6x + 8y subject to constraints x - y ≥0, x + 3y ≤12, x ≥0, y ≥0.

On taking given constraints as equations, we get the following graph.

Intersecting point of the line x-y = 0 and x + 3y = 12 is B(3, 3),

Here, OABO is the required feasible region

Whoso comer points are O(0, 0), A(12, 0) and B(0, 4) Now,

∴Maximum value of Z is 72.
Question 11
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If p and q are true and r and s are false statements, then which of the following is true?

A
A. (q ∧ r ) v (~ p ∧ s)

B
B. (∼p → q)↔(r ∧ s)

C
C. (p → q ) v ( r ↔ s)

D
D. (p ∧ ∼ r) ∧ (∼ q v s)

Solution

We have statements p, q → T and r, s → F

Option (a) (q ∧ r) v (∼p ∧ s) = (T ∧ F) v (F ∧ F) = F v F = F

Option (b) (-P → q) ↔ (r ∧ s) = (p v q) → (r v s) (∵ p → q = -p v q)

Option (d) is similar to option (a).
Question 12
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If f( x) = (x), where (x) is the greatest integer not greater than x, then f’(1) = ...

A
A. 1

B
B. 2

C
C. 0

D
D. -1

Solution

We have, f(x) = (x)
Question 13
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If A = |x| x ∈N, x is a prime number less than |12| and B = |x | x ∈N, x is a factor of 10). then A ∩B = ...

A
(2)

B
(2, 5)

C
(2, 5, 10)

D
(1, 2, 5, 10)

Solution

We have, A = {x |x ∈N, x| is a prime number less than 12} = (2, 3, 5, 7, 11) and B = (x|x ∈N, x is a factor of 10) = (1, 2, 5, 10)

∴A ∩B = (2, 5)
Question 14
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If A, B, C and D are (3, 7, 4), (5, -2, 3), (-4, 5, 6) and (1, 2, 3) respectively, then the volume of the parallelepiped with AB, AC and AD as the coterminous edges, is ....... cubic units.

A
91

B
94

C
92

D
93

Solution

We have

∴The volume of the parallelepiped with AB, AC, and AD on the coterminous edges

= |2 (2 + 10) + 9(7 + 4) -1(35 - 4)|

= |2 (12) + 9(11) - 1(31)|

= |(24 + 99 - 31)|

= |92| = 92 cubic units
Question 15
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If then p = ...

A
A. 6

B
B. 1

C
C. 3

D
D. 12

Solution

We have

Now,

put cosx + sinx = t

(-sinx + cosx)dx = dt

∴ p = 6
Question 16
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Equations of planes parallel to the plane x - 2y + 2z + 4 = 0 which are at a distance of one unit from the point (1, 2, 3) are ............

A
A. x + 2y + 2z = -6, x + 2y + 2z = 5

B
B. x - 2y - 6 = 0, x - 2y + z = 6

C
C. x + 2y + 2z = 6, x + 2y + 2z = 0

D
D. x - 2y + 2z = 0, x - 2y + 2z - 6 = 0

Solution

Given equation of plane x - 2y + 2z + 4 = 0 …(i)

Equation of plane parallel to plane (i) is x - 2y + 2z + k = 0 …(ii)

Since, distance of plane (i) from point (1, 2, 3) is 1 unit

∴

⇒

⇒

⇒ |3+ k| = 3 ⇒ 3 + k = ±3

⇒ k = 0, -6

∴ Required equation of plane are x - 2y+ 2z + 0 = 0 and x - 2y + 2z - 6 = 0

⇒ x - 2y + 2z = 0 and x - 2y + 2z - 6 = 0