Animated Solution for Mathematics - Differentiation: Let $a$ be an integer such that all the real roots of the polynomial $2x^5+5x^4+10x^3+10x^2+10x+10$ lie in the interval $(a,a+1)$ Then, $|a|$ is equal to

Welcome, brave explorer of the mathematical landscape. Today, we stand before a formidable-looking beast: a fifth-degree polynomial, $f(x)=2x^5+5x^4+10x^3+10x^2+10x+10$.

At first glance, the sheer power of $x^5$ might make you want to reach for a calculator or despair at the thought of finding its roots. But hold your ground! In the world of JEE Advanced, we do not fear polynomials; we analyze them.

Our goal is to find an integer $a$ such that the real root lies in $(a,a+1)$. Let us embark on this journey together.

Our Compass

To understand the soul of this function, we must ask: does it oscillate wildly, or does it climb steadily? The tool for this is the derivative, $f^{\prime }(x)$.

By applying the power rule, we find:

Notice the beautiful symmetry in the coefficients? They are all multiples of $10$. This is not a coincidence; it is a hint from the problem setter. Let us factor out the $10$:

Now, look closely at the expression inside the parentheses. It is a symmetric polynomial. If you have been practicing your algebra, you might recognize this as a perfect square.

Let us verify:

Yes! It is exactly the same. So, we have:

This is a revelation! Since any real number squared is non-negative, $f^{\prime }(x)$ is always $\ge 0$.

Furthermore, the quadratic $x^2+x+1$ has a discriminant $\Delta =1^2-4(1)(1)=-3$. Since the discriminant is negative and the leading coefficient is positive, the quadratic $x^2+x+1$ is always positive for all real $x$.

Therefore, $f^{\prime }(x)$ is strictly positive for all real $x$. This means our function $f(x)$ is strictly increasing. It never turns back; it is a monotonic function.

Because $f(x)$ is a polynomial of odd degree, it must have at least one real root. And because it is strictly increasing, it can have at most one real root. Thus, it has exactly one real root.

Now, we need to find where this root lies. We use the Intermediate Value Theorem. We need to find two values, $x_1$ and $x_2$, such that $f(x_1)<0$ and $f(x_2)>0$.

Let us test some integer values. Let us try $x=-1$:

Since $3>0$, the root must be to the left of $-1$. Let us try $x=-2$:

Since $-34<0$, the root must be to the right of $-2$. Therefore, the root lies in the interval $(-2,-1)$.

The problem states the root lies in $(a,a+1)$. Comparing $(-2,-1)$ with $(a,a+1)$, we see that $a=-2$.

The question asks for $|a|$, which is $|-2|=2$.

We have successfully navigated the problem. Remember, in JEE Advanced, it is not about knowing the most complex formula, but about understanding the fundamental properties of functions. Keep practicing, keep exploring, and you will master these concepts.