Mathematics Test 164

Result

Mathematics Test 164

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Weekly Quiz Competition

Question 1
4 / -1

The sum equals

A
– 2⁹⁸

B
2⁹⁸

C
– 2⁴⁹

D
2⁴⁹

Solution

Consider the expansion of (1 + x)⁹⁹ and put x = i and equate the real part to get option C
Question 2
4 / -1

Consider the polynomial equation x⁴ – 2x³ + 3x² – 4x + 1 = 0. Which one of the following statements describes correctly the solution set of this equation ?

A
four non real complex zeroes.

B
four positive zeroes

C
two positive and two negative zeroes

D
two real and two non real complex zeroes.

Solution
Question 3
4 / -1

If the sum and lie in [1, 15] then a + b + c equals

A
6

B
8

C
10

D
11

Solution
Question 4
4 / -1

The polynomial function f (x) satisfies the equation f (x) – f (x – 2) = (2x – 1)² for all x. If p and q are the coefficient of x² and x respectively in f (x), then p + q is equal to

A
0

B
5/6

C
4/3

D
1

Solution

∵ f'"(x) = f'"(x – 2)

⇒ f is a cubic polynomial function

∴ Let f(x) = ax³ + px² + qx + r

now use f(x) – f(x – 2) = 4x² – 4x + 1

compare the coefficients to get

a = 2/3, p = 1, q = – 1/6,

hence p + q = 5/6
Question 5
4 / -1

A real number x is chosen at random such that 0 < x < 100. The probability that where a and b are relatively primes and [x] denotes the greatest integer then (b – a) equals

A
0

B
1

C
2

D
4

Solution
Question 6
4 / -1

Let R be a relation defined in the set of real numbers by a R b ⇔ 1 + ab > 0. Then R is :

A
Equivalence relation

B
Transitive

C
Symmetric

D
Anti-symmetric

Solution

If (a, b) ∈ R

then (b, a) ∈ R

Hence it is a symmetric relation
Question 7
4 / -1

In an experiment with 15 observations on x, the following results were available

Σx² = 2830, Σx = 170

One observation that was 20, we found to be wrong and was replaced by the correct value 30. Then, the corrected variance is :

A
78.00

B
188.66

C
177.33

D
8.33

Solution

Given: N = 15

Σx² = 2830, Σx = 170

One observation 20 was replaced by 30, then

Σx² = 2830 – 400 + 900 = 3330

Σx = 170 – 20 + 30 = 180
Question 8
4 / -1

If p and q are two statements, then

A
tautology

B
contradiction

C
neither tautology nor contradiction

D
either tautology or contradiction

Solution
Question 9
4 / -1

If f be a continuous function on [0, 1], differentiable in (0, 1) such that f (1) = 0, then their exists some c (0, 1) such that

A
c f′ (c) – f (c) = 0

B
f′ (c) + c f (c) = 0

C
f′ (c) – f (c) = 0

D
c f′ (c) + f (c) = 0

Solution
Question 10
4 / -1

If |z| 4, then the maximum value of |iz + 3 – 4i| is equal to

A
2

B
4

C
3

D
9

Solution

|iz + 3 – 4i|

= |i||z – 3i – 4|

= |z – (4 + 3i)|

maximum value of |iz + 3 – 4i|

= PQ = OP + OQ

= 5 + 4 = 9