Mathematics Test 117

Result

Mathematics Test 117

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Weekly Quiz Competition

Question 1
4 / -1

A

B

C

D

Solution

Question 2

4 / -1

Let p and q be two statements then (~ p ∨ q) ∧ (~ p ∧ ~ q) is a

tautology

contradiction

neither tautology nor contradiction

None of these

Solution

p	q	~ p ∨ q	~ p ∧ ~ q	(~ p ∨ q ) ^ (~ p ∧ ~ q)
T	T	T	F	F
T	F	F	F	F
F	T	T	F	F
F	F	T	T	T

Question 3
4 / -1

Consider the line x-3/3 = y-2/1 = z-1/0 then which of the following statement is incorrect ?

A
the line is parallel to xy plane

B
the line is perpendicular to z-axis

C
the line passes through point (6,3, 1)

D
the line intersects the plane x– 3y + z = 3

Solution

It's D.R is {3, 1, 0}

i.e. option (1) & (2) are correct

Any point onto line is 3 + 3λ, 2 + λ, 1

hence for λ = 1, option (3) is correct

Line & plane are parallel to each other hence option (4) is incorrect.
Question 4
4 / -1

The proposition

A
Tautology

B
neither tautology nor contradiction

C
contradiction

D
None of these

Solution

P ~P P ⇒ ~ P ~ P ⇒ P (P ⇒ ~ P) ^ (~ P ⇒ P)

T F F T F

F T T F F
Question 5
4 / -1

Consider the set S = {1, 2, 3, 4}. Total number of subsets of S × S containing the elements (1, 2), (2, 3) and (3, 4) is :

A
7168

B
8192

C
6144

D
6384

Solution

Set S × S will have 4 × 4 = 16 elements. Number of elements except (1, 2), (2, 3) and (3, 4) in S × S = 16 – 3 = 13
Thus number of subsets of S × S containing (1, 2), (2, 3) and (3, 4) = 2¹³ = 8192.
Question 6
4 / -1

The following statement is equivalent to :

A

B

C

D

Solution
Question 7
4 / -1

Suppose f(x) is thrice differentiable polynomial function such that f(1) = 1, f(2) = 8, f (3) = 27 and f(4) = 64 then

A
f "' (x) = 6 ∀ x ∈ R

B
there always exists at least one x  ∈ (1, 4) such that f "' (x) = 6

C
there always exists at least one x  ∈ (2, 3) such that f ' (x) = 19 = f " (x) = f(x)

D
there always exists at least one x  ∈ (1, 2) such that f ' (x) = 19 = f " (x) = f(x)

Solution

Let g(x) = f(x) – x³

⇒ g(x) has at least 4 real roots which are x = 1, 2, 3, 4

⇒ g'(x) = 0 has atleast 3 real roots in (1, 4)

⇒ g''(x) = 0 has atleast 2 real roots in (1, 4)

⇒ g'''(x) = f '''(x) – 6 = 0 has atleast 1 real root in (1, 4)
Question 8
4 / -1

Solution of the differential equation ydx – xdy + 3x²y² ex³ dx = 0 is (where C is an arbitrary constant) :

A
x/y - ex³/ 3 = C

B
x/y + ex³= C

C
xy + ex³ = C

D
xy + ex³ / 3 + C

Solution
Question 9
4 / -1

A

B

C

D

Solution

We know that

Putting
α = θ, 2θ, 2² θ ..........., 2¹⁰ θ in (i),
we get
tan θ = (cot θ – 2 cot (2θ))
2 (tan(2θ)) = 2 (cot(2θ) – 2 cot (2² θ))
2² (tan(2²θ)) = 2² (cot(2²θ) – 2 cot (2³ θ))
-------------------------------------------------------
-------------------------------------------------------
2¹⁰ (tan(2¹⁰θ)) = 2¹⁰ (cot(2¹⁰θ) – 2 cot (2¹¹ θ))
adding
tanθ + 2 tan(2θ) + 2² tan(2²θ) + ........+
2¹⁰ tan(2¹⁰θ) = cotθ – 2¹¹ cot(2¹¹θ)
Question 10
4 / -1

Locus of a point which moves in such a way that ratio of its distance from x axis to y axis is 3 : 2, is S = 0 then which of the following point does not lie on S = 0

A
(4, –6)

B
(–3, –2)

C
(–2/3, 1)

D
(6, 9)

Solution

Locus will be : 3|x| = 2|y|