Notifications
Clear all
0
18/01/2022 1:30 pm
Topic starter
If 0 < x < 1 and y = \(\frac{1}{2}\)x2 + \(\frac{2}{3}\)x3 + \(\frac{3}{4}\)x4 + ....., then the value of e1+y at x = \(\frac{1}{2}\) is
(1) \(\frac{1}{2}e^2\)
(2) 2e
(3) \(\frac{1}{2}\)
(4) 2e2
1 Answer
0
18/01/2022 1:41 pm
Correct answer: (1) \(\frac{1}{2}e^2\)
Explanation:
y = \(\Big(1 - \frac{1}{2}\Big)x^2\) + \(\Big(1 - \frac{1}{3}\Big)x^3\) + .....
= (x2 + x3 + x4 + ......) - \(\Big(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4}+....\Big)\)
= \(\frac{x^2}{1-x}\) + x - \(\Big(x - \frac{x^2}{2} + \frac{x^3}{3}+....\Big)\)
= \(\frac{x}{1-x}+\ell n(1 - x)\)
x = \(\frac{1}{2}\) ⇒ y = 1 - ℓn2
e1+y = e1+1-ℓn2
= e2-ℓn2
= \(\frac{e^2}{2}\)