CAT 2004 Question 19

Question 19

Let $$f(x) = ax^2 - b|x|$$ , where a and b are constants. Then at x = 0, f(x) is

[CAT 2004]

Solution

$$f(x) = ax^2 - b|x|$$. When $$x=0, f(x) = 0$$
When $$a > 0$$ and $$b < 0$$,
For x > 0, $$f(x) = ax^2 - bx$$, will be greater than 0 as $$ax^2 > 0$$ and $$bx<0$$ as $$b$$ is negative and $$x$$ is positive.
For x < 0, $$f(x) = ax^2 + bx$$ will again be greater than 0 as $$ax^2 >0$$ and $$bx>0$$ as both $$b$$ and $$x$$ are negative. 

Therefore, the function $$f(x)$$ is positive when $$x<0$$ and when $$x>0$$ but becomes 0 when $$x=0$$.

Therefore, for $$a > 0$$ and $$b < 0$$, f(x) will attain its minimum value at $$x = 0$$.


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