Applications Of Derivatives Question 50
Question: A function g(x) is defined as $ g(x)=\frac{1}{4}f(2x^{2}-1)+\frac{1}{2}f(1-x^{2}) $ and $ f(x) $ is an increasing function. Then g(x) is increasing in the interval
Options:
A) $ (-1,1) $
B) $ ( -\sqrt{\frac{2}{3},}0 )\cup ( \sqrt{\frac{2}{3}},\infty ) $
C) $ ( -\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}} ) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ g’(x)=xf’(2x^{2}-1)-xf’(1-x^{2}) $
$ =x(f’(2x^{2}-1)-f’(1-x^{2})) $
$ g’(x)>0 $ If $ x>0,,2x^{2}-1>1-x^{2} $ (as f- is an increasing function) or $ 3x^{2}>2 $ or $ x\in ( -\infty ,-\sqrt{\frac{2}{3}} )\cup ( \sqrt{\frac{2}{3}},\infty ) $ or $ x\in ( \sqrt{\frac{2}{3}},\infty ) $ If $ x<0,2x^{2}-1<1-x^{2} $ or $ 3x^{2}<2 $ or $ x\in ( -\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}} )orx\in ( -\sqrt{\frac{2}{3}},0 ) $
BETA
