Animated Solution for Mathematics - Differentiation: Let $p(x)$ be a polynomial of degree 4 having extremum at $x=1,2$ and ${\lim }_{x\to 0}(1+\frac{p(x)}{x^2})=2$. Then the value of $p(2)$ is

Imagine you are standing before a mathematical landscape, looking at a fourth-degree polynomial, $p(x)$. It is a smooth, flowing curve, but it hides secrets in its turning points—its extrema—at $x=1$ and $x=2$.

We are also given a cryptic clue: a limit condition as $x$ approaches zero. This is not just a problem; it is a detective story.

We start with the limit:

If we subtract 1 from both sides, we get:

Think about what this means for the structure of $p(x)$. A general fourth-degree polynomial looks like $p(x)=a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0$.

If $a_0$ (the constant) or $a_1$ (the linear coefficient) were non-zero, the term $\frac{p(x)}{x^2}$ would contain $\frac{a_0}{x^2}$ or $\frac{a_1}{x}$, both of which explode to infinity as $x\to 0$. But our limit is finite!

This forces $a_0=0$ and $a_1=0$. Furthermore, the limit of $\frac{a_2{x}^2}{x^2}$ is simply $a_2$. Since the limit is 1, we must have $a_2=1$.

Our polynomial has been stripped down to its essential core:

Now that we have the skeleton of our polynomial, we need to find the values of $a$ and $b$. We are told that $p(x)$ has extrema at $x=1$ and $x=2$.

In the language of calculus, this means the slope of the tangent line—the derivative $p^{\prime }(x)$—must be zero at these points. Let us differentiate $p(x)=x^2+a{x}^3+b{x}^4$ with respect to $x$:

This derivative is the key to unlocking the values of $a$ and $b$. We have two conditions: $p^{\prime }(1)=0$ and $p^{\prime }(2)=0$.

Substituting $x=1$ into our derivative, we get $2(1)+3a(1{)}^2+4b(1{)}^3=0$, which simplifies to:

Now, for $x=2$, we have $2(2)+3a(2{)}^2+4b(2{)}^3=0$, which becomes $4+12a+32b=0$. Dividing by 4, we get $1+3a+8b=0$, or:

We are left with a beautiful, simple system of linear equations: 1) $3a+4b=-2$ 2) $3a+8b=-1$

Subtracting the first from the second, the $3a$ terms vanish, leaving us with $4b=1$, so $b=\frac{1}{4}$.

Substituting this back into the first equation, $3a+4(\frac{1}{4})=-2$, we get $3a+1=-2$, which means $3a=-3$, so $a=-1$.

We have found our coefficients! The polynomial is:

If you look closely, this can be factored as:

Finally, we evaluate $p(2)$. Substituting $x=2$ into our factored form, we get:

The value is 0. It is a satisfying conclusion to our journey, showing how limits, derivatives, and algebra weave together to reveal the hidden nature of the function.