Question

f(x) = $\lim_{n \rightarrow \infty}$ (2x + 4x³ +……..+ 2ⁿx ^2n–1) (0 < x < 1) then $\int_{}^{}{f(x)dx}$ is equal to

• A. log$\left( \frac{1}{\sqrt{1 - x^{2}}} \right)$ + c
• B. log $\sqrt{1 - 2x^{2}}$ + c
• C. log$\left( \frac{1}{\sqrt{1 - 2x^{2}}} \right)$ + c
• D. None of these

Answer 1

Answer: (C)

log$\left( \frac{1}{\sqrt{1 - 2x^{2}}} \right)$ + c

Answer 2

Solve series

s = 2x + 4x³ +…+ 2ⁿx^2n–1 ® Total term (n) ® G.P.

s = 2(x) $\frac{\lbrack(2x^{2})^{n} - 1\rbrack}{(2x^{2} - 1)}$ ® first term = 2x

® C.R. = 2x²

s = (2x) $\frac{(2x)\lbrack 2^{n}x^{2n} - 1\rbrack}{(2x^{2} - 1)}$, ${}_{n \rightarrow \infty}^{0 < x < 1}$ x²ⁿ ® 0

f(x) = s = $\frac{2x}{1 - 2x^{2}}$

so $\int_{}^{}\frac{2x}{1 - 2x^{2}}$dx put 1 – 2x² = t

– 4x dx = dt

2x dx = – 1/2 dt

= – $\frac{1}{2}\int_{}^{}\frac{dt}{t}$= – $\frac{1}{2}$ log (t) + c

= – $\frac{1}{2}$ log (1 – 2x²) + c

= log $\left( \frac{1}{\sqrt{1 - 2x^{2}}} \right)$ + c