Hi Ankit, the solution is correct. When a>0, $$ax^2$$ >=0 and when b<0, -b|x| >=0. Hence, f(x) is always greater than or equal to zero. Hence, the minimum value is 0 when x=0.

can we use derivatives?

Is the question x tending to 0.

Hi Aninda,
You are correct at x = 0, f(x) will be zero, the question is asking whether zero will be the minimum value of the expression or the maximum value of the expression.

Okay, I got it now ...

Hi Shreya, we have expanded the solution. Please check it out. We are just considering the case where a>0 and b<0 and in that case, f(x) is attaining its minimum value.

Hi Sriram,

the question checks for the value of x>0 and x<0.

hope this helps.

Hi gaurav, it just asks about the nature of graph function will make on either side of 0.

Hope this helps.

Hi Dibyajyoti,

In this case, we have assumed x<0. Also, b<0, therefore bx will be positive.

Hope this helps.

Hi Tanveer,

For a>0, b<0 and x<0,

we know that x is negative, so after removing modulus, we can write f(x) as $$ax^2$$+bx. In this, we know that both x and b are negative so f(x)>0.

Hope this helps.

Hi,

It will not be maximum at x=0, it will increase with the increase in the value of x in both the directions.

Hope this helps.

Hi Akshay Acharya,
In option C, it is mentioned that f(x) is minimised whenever a > 0 and b > 0
In option D, it is mentioned that f(x) is minimised whenever a > 0 and b < 0

If a > 0 and b < 0, we can conclude that f(x) is minimised at x = 0
If a > 0 and b > 0,
Case 1: x > 0
f(x) = $$ax^2-bx$$
a > 0 and b > 0; therefore, we cannot conclude if f(x) is greater than or less than zero.
Therefore, we cannot say that f(x) is minimised at x = 0 if a > 0 and b > 0
Hope this helps!

Hi Arpit Jain,
b > 0 or b < 0 is used to determine whether f(x) is greater than or less than zero.
Case 1: x > 0, a > 0, b > 0
f(x) = $$ax^2-bx$$
a > 0 and b > 0; therefore, we cannot conclude if f(x) is greater than or less than zero. It depends on the values of $$ax^2$$ and $$bx$$
Case 2: x > 0, a > 0, b < 0
f(x) = $$ax^2-bx$$
If b < 0, then -bx is positive. This implies f(x) > 0
Hope this helps!

Hi Aditya Pandey,
We have to check all the given options. We cannot get to the answer by directly looking at the options. Upon solving such questions, you speed in improve. Don't worry about that, keep practicing.
I hope this helps!!

i'm saying that sir started with the correct option which turned out to hold,are options the only way or is there a diff approach.

Hi Aditya,
You must use options to solve these types of problems and check all the options. In this question, there are two cases of a > 0, b > 0 and a > 0, b < 0. Sir chose one case and obtained the correct answer where the condition given in the option holds, and he did not check for the other case.
Hope this helps!

Got it now.

Hi PRITHA SINGH,
We are glad you figured it out on your own. Feel free to post your further doubts.