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 May 19, 2025
Latest Power Series and Taylor Series MCQ Objective Questions
Power Series and Taylor Series Question 1:
\(\rm \frac{3}{1!}+\frac{5}{3!}+\frac{7}{5!}+\frac{9}{7!}+...\) बराबर है
Answer (Detailed Solution Below)
Power Series and Taylor Series Question 1 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
Power Series and Taylor Series Question 2:
\(\rm \frac{3}{1!}+\frac{5}{3!}+\frac{7}{5!}+\frac{9}{7!}+...\) बराबर है
Answer (Detailed Solution Below)
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
Power Series and Taylor Series Question 3:
\(\rm d\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!} \ldots\right) \)/dx किसका समतुल्य है?