ISI maths

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ISI maths

chirag
f (x) = (x − a) + (x − b) + (x − c) , . a < b < c
The number of real roots of f ( ) x = 0 is
(A) 3; (B) 2; (C) 1; (D) 0.
Can someone tell me how to solve these types of questions?
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Re: ISI maths

Amit Goyal
Administrator
You can rewrite the function as f(x) = 3x - a -b - c. It is a straight line with slope 3. It will intersect the horizontal axis exactly once at x = (a+b+c)/3, so it has 1 root.
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Re: ISI maths

chirag
In reply to this post by chirag
Sorry sir i uploaded the wrong question. There was a typing error in the question.
Question is this--->
F(x)=(x-a)^3+(x-b)^3+(x-c)^3 where a<b<c
The number of real roots of F(x)=0 are
A)3 ,B)2 ,C)1 ,D)0
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Re: ISI maths

Abhitesh
F'(x) > 0 => Function is always increasing, so only one root.
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