Mathematics Test 109

Result

Mathematics Test 109

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
4 / -1

The negation of p → (~ p ∨ q) is

A
p ∨ (p ∨ ~ q)

B
p → ~ (p ∨ q)

C
p → q

D
p ∧ ~ q

Solution

Consider ~ [p → (~ p v q)] ≡ p ∧ ~ (~ p ∨ q)]

≡ p ∧ (p ∧ ~ q) ≡ p ∧ p ∧ ~ q ≡ p ∧ ~ q
Question 2
4 / -1

If the algebraic sum of deviations of 20 observation from 30 is 20, then the mean of observation is :

A
30

B
30.1

C
29

D
31

Solution
Question 3
4 / -1

The system of equations

kx + (k + 1)y + (k – 1)z = 0

(k + 1)x+ ky + (k + 2)z = 0

(k – 1)x + (k + 2)y + kz = 0

has a non trivial solution for

A
Exactly three real values of k

B
Exactly two real values of k

C
Exactly one real value of k

D
Infinite number of values of k

Solution
Question 4
4 / -1

The natural numbers are grouped as follows {1}, {2, 3, 4}, {5, 6, 7, 8, 9}, .... , then the first element of the n^th group is-

A
n² – 1

B
n² + 1

C
(n – 1)² – 1

D
(n – 1)² + 1
Question 5
4 / -1

If α, β, γ are the real roots of the equation x³ – 3px² + 3qx – 1 = 0, then the centroid of the triangle whose vertices are

A
p, –q

B
(–p, q)

C
(p, q)

D
((p/2, q/2))
Question 6
4 / -1

The sum of the binomial coefficients in the expansion of lies between 200 and 400 and the term independent of x equals 448. The value of a is :-

A
1

B
2

C
1/2

D
3

Solution
Question 7
4 / -1

Let z₁ & z₂ be non zero complex number satisfying the equation z₁² – 2z₁z₂ + 2z₂² = 0, then geometrical nature of the triangle whose vertices are the origin & the points representing z₁ & z₂ is :-

A
An isosceles right angle triangle

B
An equilateral triangle

C
An isosceles which is not right angled

D
None of these

Solution
Question 8
4 / -1

Letters of the word GIGANTIC are arranged to form all possible words, then the probability that a word formed starts either with a G or a vowel is

A
1/8

B
3/8

C
5/8

D
7/8

Solution
Question 9
4 / -1

Let f(x) = –x² + x + P, where P is a real number. If g(x) = [f(x)] and g(x) is discontinuous at x = 1/2, then P can't be (where [·] is greatest integer function)

A
1/2

B
3/4

C
7/4

D
- 1/4

Solution
Question 10
4 / -1

For f : R → R, f(x) = x⁴ – 8x³ + 22x² – 24x, the sum of all local extreme values of f(x) is equal to

A
–9

B
–8

C
–17

D
–26

Solution

f(x) = x⁴ – 8x³ + 22x² – 24x

f'(x) = 4x³ – 24x² + 44x – 24

f'(x) = 4[x³ – 6x² + 11x – 6]

f'(x) = 4(x – 1) (x – 2) (x – 3)

L_max ⇒ x = 2

L_min ⇒ x = 1, 3

∴ f(1) + f(2) + f(3) = –26