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The following equation has been provided and the question asks for number of distinct roots between $(1,2)$ of the following equation.

$3x^2-12x+11+\frac{1}{5}\left ( x^3-6x^2+11x-6 \right)$

I tried to solve it using the derivative but couldn't proceed further. I took derivative because there is a relationship between number of roots of a function and it's derivative. Please help me with this. Thanks in advance.

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  • $\begingroup$ How does the derivate look? $\endgroup$ – razvanelda Jan 3 '18 at 11:54
  • $\begingroup$ $6x-12+\frac{1}{5}\left (3x^2-12x+11 \right)$ $\endgroup$ – RAHUl JHa Jan 3 '18 at 11:56
  • $\begingroup$ Can we infer something from that derivative? $\endgroup$ – RAHUl JHa Jan 3 '18 at 11:57
  • $\begingroup$ Tidy it up using completing the square and look to see what values it takes between 1 and 2. $\endgroup$ – Paul Jan 3 '18 at 11:58
  • $\begingroup$ Yes, you can see the monotonicity of the function. If the derivate is positive the function is increasing, if it is negative decreasing. Try to found how it is in the interval (1, 2). If you can't see this, derivate once more and find the monocity of the first derivate. From the second derivate you can also see how is the function, convex or concave, which gives you the number of solutions. Try the second derivate. $\endgroup$ – razvanelda Jan 3 '18 at 12:01
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Let $$f(x)=3x^2-12x+11+\frac{1}{5}\left ( x^3-6x^2+11x-6 \right).$$ Thus, $$f(x)=\frac{1}{5}(x^3+9x^2-49x+49).$$

It's obvious that $f$ has a negative root.

Also, since $f(1)>0$ and $f(2)<0$, there is a root in $(1,2)$,

but it's an unique root there because there is the last root for $x>2$.

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Put $P(x)=x^3-6x^2+11x-6$ It is easy to see that $P(x)=(x-1)(x-2)(x-3)$. Now let $Q(x)$ your polynomial, we have $Q(x)=P^{\prime}(x)+\frac{P(x)}{5}$. Let $f(x)=P(x)\exp(x/5)$. We get that $f^{\prime}(x)=Q(x)\exp(x/5)$. Now $f(1)=f(2)=f(3)=0$. By the mean value theorem, the derivative of $f$ has a zero say $c_2\in ]1,2[$, and another one $c_3$ in $]2,3[$. In addition, we have $f(x)\to 0$ if $x\to -\infty$. Hence $f$ has a local extremum on $]-\infty,1[$, say $c_1$, and the derivative of $f$ is zero at $c_1$. The $c_k$ are zeros of $Q$, they are distincts, and as $Q$ is of degree $3$, we have all zeros of $Q$, and there is only one in $]1,2[$.

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Hint:

On taking derivative you know that the function is monotonic in (1,2), So 1 or 0 roots.

Now Integrate the equation and apply Lagrange's Mean Value theorem.

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