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Let f(x)=a x^5+b x^4+c x^3+d x^2+e x, where a, b, c, d, e \in R \text { and } f(x)=0 has a positive root. Then, 

Option: 1

f^{\prime}(x)=0  has a root \alpha_1 such that 0<\alpha_1<\alpha_0

 


Option: 2

f^{\prime}(x)=0 has at least one real root 


Option: 3

f^{\prime}(x)=0 has at least two real roots

 


Option: 4

All of the above


Answers (1)

best_answer

It is given that \alpha is a positive root of f(x) and by inspection, we have f(0)=0. Therefore, x=0 \text { and } x=\alpha are the roots of f(x)=0.

By Rolle's theorem, f(x)=0 has root \alpha _{1} between 0 and \alpha \text { i.e., } 0<\alpha_1<\alpha \text {. }

So option (a) is correct.

Clearly, f(x)=0 is a fourth degree equation in x and imaginary roots always occur in pairs. Since, x=\alpha_1 is a root of f^{\prime}(x)=0

Therefore, f^{\prime \prime}(x)=0 will have another real root, \alpha _{2} (say)

Now, \alpha _{1}\: \text{and}\: \alpha _{2} are real roots of f^{\prime}(x)=0.

Therefore, by Rolle's theorem f^{\prime \prime}(x)=0 will have a real root between \alpha _{1}\: \text{and}\: \alpha _{2}

Thus, option (b) is correct.

We have see that x=0 \: \text{and}\: x=\alpha are two real roots of f(x)=0. As f(x)=0 is a fifth degree equation, it will have at least three real roots. Consequently, by Rolle's theorem f\prime(x)=0 will have at least two real roots. 

Thus, option (c) is also correct.  

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shivangi.shekhar

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