TIFR CSE 2012 | Part A | Question: 12

Question

For the polynomial $p(x)= 8x^{10}-7x^{3}+x-1$ consider the following statements (which may be true or false)

1. It has a root between $[0, 1].$
2. It has a root between $[0, -1].$
3. It has no roots outside $(-1, 1).$

Which of the above statements are true?

• A. Only (i).
• B. Only (i) and (ii).
• C. Only (i) and (iii).
• D. Only (ii) and (iii).
• E. All of (i), (ii) and (iii).

Answer 1

At $f(0)$ it is negative, at $f(1)$ it is positive, and at $f(-1)$ it is positive, which means there will be roots between $(0,1)$ and $(-1,0)$. Any values below $-1$ and above $1$ will always yield positive values for $f(x)$, which means no roots available.

Correct Answer: $E$

Comments

roots of the function means the value of x for which f(x)=0??

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Yes

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Great answer.
If $x\gt 1$, then $8x^{10} \gt 7x^3$, so $p(x) \gt 0$.
If $x\lt-1$, then $\underbrace{8x^{10}}_{\gt0}+\underbrace{x}_{\lt 0} \gt0$ and $\underbrace{-7x^3}_{\gt0}-1\gt 0$, so again $p(x)\gt0$

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your approach is not 100% correct , we must find p`(x)  and show that p`(x) (means slope) always positive for x >1 and always -ve for x< -1

Answer 2

Use Intermediate Value Theorem (Bolzano's Theorem).  https://en.wikipedia.org/wiki/Intermediate_value_theorem

It states that if:

1.)  F(x) is continuous in [a,b]  

2.) F(a)*F(b) < 0 then,

there exists X such that  F(X) = 0

so in this question just put the values and check if product at two boundary values is negative or not.
In option III it is not negative at for those ranges value is always positive and thus the function never cuts the axis and hence no root.

ANSWER : Option E   (All true)