De Moivre's Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for De Moivre's Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 14, 2025

നേടുക De Moivre's Theorem ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക De Moivre's Theorem MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest De Moivre's Theorem MCQ Objective Questions

Top De Moivre's Theorem MCQ Objective Questions

De Moivre's Theorem Question 1:

If  are respectively the least positive and greatest negative integer values of  such that  then 

  1. 4
  2. 0
  3. 6
  4. 2

Answer (Detailed Solution Below)

Option 1 : 4

De Moivre's Theorem Question 1 Detailed Solution

Calculation

Using De Moivre's Theorem:

Comparing with -i, we have:

This implies:

⇒ 

⇒ 

⇒ 

⇒ 

For the least positive integer value of k, let n = 0:

Let n = 1:

Let n = 2:

So, m = 3.

For the greatest negative integer value of k, we can analyze the values of k for negative n.

For n = -1:

So, n = -1.

Hence option 1 is correct

De Moivre's Theorem Question 2:

For z ∈ , if (1 + z)n = 1 + nc1z + nc2z2 + .... ncnzn and , then k =

  1. 25
  2. 100
  3. 50
  4. 75

Answer (Detailed Solution Below)

Option 3 : 50

De Moivre's Theorem Question 2 Detailed Solution

Concept:

Euler's theorem:

e = cosθ + isinθ 

De Moivre's Theorem:

(cosθ + isinθ)ncos(nθ) + isin(nθ) 

Calculation:

Given, 

Putting z = eir, we get:

...(i)

Now 

, where Img(z) represents the imaginary part of z

 =  (given)

Comparing both sides we get, k = 50

∴ The value of k is 50.

De Moivre's Theorem Question 3:

If z = e and  =  

then  =

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 1

De Moivre's Theorem Question 3 Detailed Solution

Concept:

  • If  then by componendo and dividendo rule  

Calculation:

Given z = e and   = 

⇒    = 

We know that if  then 

⇒  = 

⇒  =   [e = cos θ + i sin θ ]

⇒  =  

We know zr = eirθ and in the R.H.S the value of r are 2, 3 and 5

∴ a0 + b0 = a1 + b1 = a4 + b4 = a6 + b6 = a7 + b7 = a8 + b8 = a9 + b9 = a10 + b10 = 0

∴ a2 + b2 = 2 and a3 + b3 = 3 and a5 + b5 = 5 .

⇒  = 1

Required value of the expression is 1

De Moivre's Theorem Question 4:

Comprehension:

Direction: If n ϵ Z (the set of integers), then

(cosθ + i sinθ)n = cos nθ + i sin nθ and

(cosθ - i sinθ)n = cos nθ - i sin nθ

If n is a positive integer, then (1 + i)n + (1 - i)n is equal to

  1. (√2)n - 2 cos 
  2. (√2)n - 2 sin 
  3. (√2)n + 2 cos 
  4. (√2)n + 2 sin 

Answer (Detailed Solution Below)

Option 3 : (√2)n + 2 cos 

De Moivre's Theorem Question 4 Detailed Solution

De Moivre's Theorem Question 5:

Comprehension:

Direction: If n ϵ Z (the set of integers), then

(cosθ + i sinθ)n = cos nθ + i sin nθ and

(cosθ - i sinθ)n = cos nθ - i sin nθ

If x +  = 2 cos θ, then xn is equal to

  1. 2 cos nθ 
  2. 2 sinnθ 
  3. cos 
  4. sin 

Answer (Detailed Solution Below)

Option 1 : 2 cos nθ 

De Moivre's Theorem Question 5 Detailed Solution

De Moivre's Theorem Question 6:

Comprehension:

Direction: If n ϵ Z (the set of integers), then

(cosθ + i sinθ)n = cos nθ + i sin nθ and

(cosθ - i sinθ)n = cos nθ - i sin nθ

If  = u + iv, where u and v all real numbers, then u is

  1. n cos ϕ 
  2. cos nϕ

Answer (Detailed Solution Below)

Option 2 : cos nϕ

De Moivre's Theorem Question 6 Detailed Solution

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