Power Series and Taylor Series MCQ Quiz in हिन्दी - Objective Question with Answer for Power Series and Taylor Series - मुफ्त [PDF] डाउनलोड करें

Last updated on Jun 12, 2025

पाईये Power Series and Taylor Series उत्तर और विस्तृत समाधान के साथ MCQ प्रश्न। इन्हें मुफ्त में डाउनलोड करें Power Series and Taylor Series MCQ क्विज़ Pdf और अपनी आगामी परीक्षाओं जैसे बैंकिंग, SSC, रेलवे, UPSC, State PSC की तैयारी करें।

Latest Power Series and Taylor Series MCQ Objective Questions

Power Series and Taylor Series Question 1:

\(\rm d\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!} \ldots\right) \)/dx किसका समतुल्य है?

  1. sin(x)
  2. ex
  3. cos(x)
  4. tan(x)

Answer (Detailed Solution Below)

Option 3 : cos(x)

Power Series and Taylor Series Question 1 Detailed Solution

Power Series and Taylor Series Question 2:

\(\rm \frac{3}{1!}+\frac{5}{3!}+\frac{7}{5!}+\frac{9}{7!}+...\) बराबर है

  1. e
  2. \(\rm \frac{3e-e^{-1}}{2}\)
  3. \(\rm \frac{3e+e^{-1}}{2}\)
  4. \(\rm \frac{3e}{2}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{3e-e^{-1}}{2}\)

Power Series and Taylor Series Question 2 Detailed Solution

व्याख्या:

ex = \(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\)

x = 1 और x = -1 रखने पर

e = \(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\)

और

e-1 = \(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\)

इसलिए,

3e - e-1 = (\(3+\frac{3}{1!}+\frac{3}{2!}+\frac{3}{3!}+...\)) - (\(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\))

= 2e + 2

Top Power Series and Taylor Series MCQ Objective Questions

\(\rm d\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!} \ldots\right) \)/dx किसका समतुल्य है?

  1. sin(x)
  2. ex
  3. cos(x)
  4. tan(x)

Answer (Detailed Solution Below)

Option 3 : cos(x)

Power Series and Taylor Series Question 3 Detailed Solution

Download Solution PDF

Power Series and Taylor Series Question 4:

\(\rm \frac{3}{1!}+\frac{5}{3!}+\frac{7}{5!}+\frac{9}{7!}+...\) बराबर है

  1. e
  2. \(\rm \frac{3e-e^{-1}}{2}\)
  3. \(\rm \frac{3e+e^{-1}}{2}\)
  4. \(\rm \frac{3e}{2}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{3e-e^{-1}}{2}\)

Power Series and Taylor Series Question 4 Detailed Solution

व्याख्या:

ex = \(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\)

x = 1 और x = -1 रखने पर

e = \(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\)

और

e-1 = \(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\)

इसलिए,

3e - e-1 = (\(3+\frac{3}{1!}+\frac{3}{2!}+\frac{3}{3!}+...\)) - (\(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\))

= 2e + 2

Power Series and Taylor Series Question 5:

\(\rm d\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!} \ldots\right) \)/dx किसका समतुल्य है?

  1. sin(x)
  2. ex
  3. cos(x)
  4. tan(x)

Answer (Detailed Solution Below)

Option 3 : cos(x)

Power Series and Taylor Series Question 5 Detailed Solution

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