Question -
Answer -
Given f(x) = x2┬атАУ x + 1
Now by differentiating the given equation with respect to x, weget
тЗТ┬а
тЗТ┬аfтАЩ(x)= 2x тАУ 1
Taking different region from (0, 1)
Let x тИИ (0, ┬╜)
тЗТ┬а2xтАУ 1 < 0
тЗТ┬аfтАЩ(x)< 0
Thus f(x) is decreasing in (0, ┬╜)
Let x тИИ (┬╜, 1)
тЗТ┬а2xтАУ 1 > 0
тЗТ┬аfтАЩ(x)> 0
Thus f(x) is increasing in (┬╜, 1)
Therefore, from above condition we find that
тЗТ┬аf(x) is decreasing in┬а(0, ┬╜) ┬аand increasing in┬а(┬╜, 1)
Hence, condition for f(x) neither increasing nor decreasing in(0, 1)