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If the interval in which the function f(x) = xx2+1\frac{x}{x^2+1} is strictly increasing is (-a, a), then a is equal to

Solution

Correct Option: 1

To find the value of aa such that the function f(x)=xx2+1f(x) = \frac{x}{x^2+1} is strictly increasing in the interval (a,a)(-a, a), we need to find where f(x)>0f'(x) > 0.

Using the quotient rule with u=xu = x and v=x2+1v = x^2 + 1:

f(x)=1(x2+1)x(2x)(x2+1)2f'(x) = \frac{1 \cdot (x^2+1) - x \cdot (2x)}{(x^2+1)^2}

f(x)=x2+12x2(x2+1)2f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2}

f(x)=1x2(x2+1)2f'(x) = \frac{1 - x^2}{(x^2+1)^2}


For the function to be strictly increasing:

1x2(x2+1)2>0\frac{1 - x^2}{(x^2+1)^2} > 0

The denominator (x2+1)2(x^2+1)^2 is always positive.

Therefore, we need:

1x2>01 - x^2 > 0

1>x21 > x^2

x2<1x^2 < 1


The inequality x2<1x^2 < 1 gives:

1<x<1-1 < x < 1

The interval is (1,1)(-1, 1).


Comparing with the given interval (a,a)(-a, a):

a=1a = 1

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