Mathematics Test

Result

Mathematics Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If a,b,c are vectors such that [abc]=4 , then =[a*b b*c c*a]=

A
16

B
64

C
4

D
8

Solution

[a*b b*c c*a] = (a*b) - [(b*c)*(c*a)]

=(a*b) . ([bca]c - [bcc]a)= (a*b) . ([bca]c - 0)

=[bca][abc]=[abc][abc]=4.4 = 16
Question 2
1 / -0

If the n^th roots of unity are z₁ and z₂ then arg ((z₁)/z₂) is a multiple of:

A
nπ

B
3π/n

C
2π/n

D
None of the above

Solution
Question 3
1 / -0

A
(1, 2)

B
(0, 0)

C
(1, 1)

D
(2, 1)

Solution
Question 4
1 / -0

Find the multiplication of the two complex numbers (4 + 2i) and (1 + 8i).

A
12 + 34i

B
-12 + 34i

C
34 + 12i

D
-34 + 12i

Solution
Question 5
1 / -0

The value of (sin245^°−sin215^°)

A
√3/2

B
1/2

C
1

D
√3/4

Solution

We know that,

[sin²A−sin²B=sin(A+B).sin(A−B)

= sin(45°+15°).sin(45°−15°)

= sin60º.sin30° = $\frac{\sqrt{3}}{2} . \frac{1}{2} = \frac{\sqrt{3}}{4}$
Question 6
1 / -0

Find the value of 50 terms of the given sequence:
$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + . . . . . .$ 50 terms.

A
51 - 2⁴⁹

B
49 + 2^-50

C
50 - 2^-50

D
50 + 2⁵⁰

Solution

In the given sequence, the numerator is in the difference of 2,i.e, 1,3,7,15...up to 50 terms. Likewise in denominator, the increase is in multiples of 2,i.e, 2,4,8,16...up to 50 terms.

The given sequence can be written as,
Question 7
1 / -0

Find how many four letter word can be formed out of the word “PINEAPPLES”?

A
840

B
5040

C
720

D
5120

Solution

We know that the number of permutations of n dissimilar things, taking r at a time is
Question 8
1 / -0

Evaluate $\int e^{(2+2 \ln x)} d x$

A

$\frac{e^{z} x}{3}+c$

B

$\frac{e^{z} x^{3}}{2}+c$

C

$\frac{e^{2} x}{2}+c$

D

$\frac{e^{I n 2 x} x^{3}}{3}+c$

Solution
Question 9
1 / -0

Find the mode for the following distribution:

Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Frequency 5 9 8 12 28 20 12 11

A
46.67

B
56.67

C
36.67

D
26.67

Solution
Question 10
1 / -0

If f: R → R is given by f(x) = 2x + 5, then the function is said to be

A
many-to-one

B
One-to-one

C
Onto

D
Inverse

Solution

The given function is f(x) = 2x + 5,

Let us assume, f(a) = f(b)

f(a) = 2a + 5 and f(b) = 2b + 5

then f(a) = f(b) ⇒ 2a + 5 = 2b + 5

⇒ 2a = 2b,

Therefore, a = b.

So, this function is one-to-one.