De Moivre's Theorem MCQ Quiz in বাংলা - Objective Question with Answer for De Moivre's Theorem - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Mar 11, 2025

পাওয়া De Moivre's Theorem उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন De Moivre's Theorem MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest De Moivre's Theorem MCQ Objective Questions

Top De Moivre's Theorem MCQ Objective Questions

De Moivre's Theorem Question 1:

For z ∈ \(\mathbb{C}\), if (1 + z)n = 1 + nc1z + nc2z2 + .... ncnzn and \(\sum_{r=0}^{100} 100 \mathrm{c}_r(\sin \mathrm{r} x)=\left(2 \cos \frac{x}{2}\right)^{100} \sin k x\), then k =

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

Answer (Detailed Solution Below)

Option 3 : 50

De Moivre's Theorem Question 1 Detailed Solution

Concept:

Euler's theorem:

e = cosθ + isinθ 

De Moivre's Theorem:

(cosθ + isinθ)ncos(nθ) + isin(nθ) 

Calculation:

Given, \((1+z)^n=1+{}^nC_1z+{}^nC_2z^2+\cdots+{}^nC_nz^n=\sum_{r=0}^{n}{}^{n}C_rz^r\)

Putting z = eir, we get:

\((1+e^{ir})^n=1+{}^nC_1e^{ir}+{}^nC_2e^{2ir}+\cdots+{}^nC_ne^{nir}=\sum_{r=0}^{n}{}^{n}C_r(e^{ix})^r\)...(i)

Now \(\sum_{r=0}^{100}{}^{100}C_rz^r(\sin rx)\)

\(\rm Img[\sum_{r=0}^{100}{}^{100}C_rz^r(e^{ix})^r]\), where Img(z) represents the imaginary part of z

\(\rm Img[(1+e^{ix})^{100}]\)

\(\rm Img[(1+\cos x +i\sin x)^{100}]\)

\(\rm Img\left[\left(2\cos^2\left(\frac{x}{2} \right )+2i\sin\left(\frac{x}{2} \right )\cos\left(\frac{x}{2} \right )\right)^{100}\right]\)

\(\rm Img\left[\left(2cos\left(\frac{x}{2} \right )\right)^{100}\left(\cos\left(\frac{x}{2} \right )+i\sin\left(\frac{x}{2} \right)\right)^{100}\right]\)

\(\rm \left(2\cos\left(\frac{x}{2} \right )\right)^{100}Img\left[\cos\left(\frac{100x}{2} \right )+i\sin\left(\frac{100x}{2} \right)\right]\)

\(\rm \left(2\cos\left(\frac{x}{2} \right )\right)^{100}Img\left[\cos50x+i\sin50x\right]\)

\(\rm \left(2\cos\left(\frac{x}{2} \right )\right)^{100}\sin50x\) = \(\left(2 \cos \left(\frac{x}{2}\right)\right)^{100} \sin k x\) (given)

Comparing both sides we get, k = 50

∴ The value of k is 50.

De Moivre's Theorem Question 2:

If z = e and \(\dfrac{3 \cos 3 \theta + 2\cos2\theta + 5\cos5\theta}{3 \sin3\theta +2\sin2\theta + 5\sin5\theta}\) = \(\dfrac{{i\sum\limits_{r = 0}^{10} {a_rz^r} }}{{\sum\limits_{r = 0}^{10} {b_rz^r} }}\) 

then  \(\dfrac{\Bigg({\sum\limits_{r = 0}^{10} {a_r + } \sum\limits_{r = 0}^{10} {b_r} }\Bigg)}{{10}}\)=

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

Answer (Detailed Solution Below)

Option 2 : 1

De Moivre's Theorem Question 2 Detailed Solution

Concept:

  • If \(\rm \frac{a}{b}=\frac{c}{d}\) then by componendo and dividendo rule  \(\rm \frac{a+b}{a-b}=\frac{c+d}{c-d}\)

Calculation:

Given z = e and  \(\dfrac{3 \cos 3 θ + 2\cos2θ + 5\cos5θ}{3 \sin3θ +2\sin2θ + 5\sin5θ}\) = \(\dfrac{{i\sum\limits_{r = 0}^{10} {a_rz^r} }}{{\sum\limits_{r = 0}^{10} {b_rz^r} }}\)

⇒   \(\dfrac{{\sum\limits_{r = 0}^{10} {a_rz^r} }}{{\sum\limits_{r = 0}^{10} {b_rz^r} }}\) = \(\dfrac{3 \cos 3 θ + 2\cos2θ + 5\cos5θ}{i(3 \sin3θ +2\sin2θ + 5\sin5θ)}\)

We know that if \(\rm \frac{a}{b}=\frac{c}{d}\) then \(\rm \frac{a+b}{a-b}=\frac{c+d}{c-d}\)

⇒ \(\rm\frac{{\sum\limits_{r = 0}^{10} {a_rz^r} }+{\sum\limits_{r = 0}^{10} {b_rz^r} }}{{\sum\limits_{r = 0}^{10} {a_rz^r} }-{\sum\limits_{r = 0}^{10} {b_rz^r} }}\) = \(\rm\frac{(3 \cos 3 θ + 2\cos2θ + 5\cos5θ)+i(3 \sin3θ +2\sin2θ + 5\sin5θ)}{(3 \cos 3 θ + 2\cos2θ + 5\cos5θ)-i(3 \sin3θ +2\sin2θ + 5\sin5θ)}\)

⇒ \(\rm\frac{{\sum\limits_{r = 0}^{10} {a_rz^r} }+{\sum\limits_{r = 0}^{10} {b_rz^r} }}{{\sum\limits_{r = 0}^{10} {a_rz^r} }-{\sum\limits_{r = 0}^{10} {b_rz^r} }}\) = \(\rm\frac{3e^{iθ}+2e^{iθ}+5e^{iθ}}{3e^{-iθ}+2e^{-iθ}+5e^{-iθ}}\)  [e = cos θ + i sin θ ]

⇒ \(\rm\frac{{\sum\limits_{r = 0}^{10} {z^r(a_r + b_r)} }}{{\sum\limits_{r = 0}^{10} {z^r(a_r -b_r)} }}\) = \(\rm\frac{3e^{iθ}+2e^{iθ}+5e^{iθ}}{3e^{-iθ}+2e^{-iθ}+5e^{-iθ}}\) 

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 .

⇒ \(\frac{\Bigg({\sum\limits_{r = 0}^{10} {a_r + } \sum\limits_{r = 0}^{10} {b_r} }\Bigg)}{{10}}\)\(\frac{2+3+5}{10}\) = 1

Required value of the expression is 1

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