Notifications
Clear all
0
27/12/2021 12:00 pm
Topic starter
If f(x) = cos[2tan-1(sin(cot-1\(\sqrt{\frac{1-x}{x}}\)))]
(a) f'(x)(x-1)2 - 2(f(x))2 = 0
(b) f'(x)(x-1)2 + 2(f(x))2 = 0
(c) f'(x)(x+1)2 + 2(f(x))2 = 0
(d) f'(x)(x+1)2 - 2(f(x))2 = 0
1 Answer
0
27/12/2021 12:10 pm
Correct answer: (c) f'(x)(x+1)2 + 2(f(x))2 = 0
Explanation:
f(x) = \(cos\Big[2tan^{-1}\Big( sin \Big( cot^{-1}\sqrt{\frac{1-x}{x}}\Big) \Big) \Big]\)
= cos (2tan-1√x)
= \(\Big(\frac{2}{1 + x}\Big)\) - 1
∴ f'(x) = \(\frac{-2}{(1 -x)^2}\)
⟹ f'(x)(x+1)2 = -2
This post was modified 4 years ago 6 times by Reyana09