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If $f(x_{i}).f(x_{i+1})< 0$ then

  1. There must be a root of $f(x)$ between $x_i$ and $x_{i+1}$
  2. There need not be a root of $f(x)$ between $x_{i}$ and $x_{i+1}$
  3. There fourth derivative of $f(x)$ with respect to $x$ vanishes at $x_{i}$
  4. The fourth derivative of $f(x)$ with respect to $x$ vanishes at $x_{i+1}$
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2 Answers

Best answer
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17 votes
As $f(x_{i}).f(x_{i+1}) <0$

Means one of them is positive and one of them in negative . as their multiplication is negative.

So, when you draw the graph for $f(x)$ where $x_{i}\leq x \leq x_{i+1} $. Definitely $F(x)$ will cut the $X$- axis.

So, there will definitely a root of $F(x)$ between $x_{i}$ and $x_{i+1}.$

Correct Answer: A.
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here ,option A is true because ,

f(xi) * f(xi+1) < 0 ; which means one of them should be negative i.e both must have opposite sign,  

and if there exist the opposite that means they will cut the x-axis and when they cut the x axis then there exist roots we all know that 

Answer:

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