Animated Solution for Mathematics - Differentiation: For every twice differentiable function $f:R\to [-2,2]$ with $(f(0){)}^2+(f^{\prime }(0){)}^2=85$, which of the following statement(s) is (are) TRUE ?

Welcome, fellow explorer of the mathematical universe! Today, we are going to dissect a problem that at first glance might seem like a chaotic mess of constraints, but as we peel back the layers, it reveals a structure of pure, crystalline elegance.

We are dealing with a twice-differentiable function $f:\mathbb{R}\to [-2,2]$ with a peculiar condition at the origin:

The Constraint at the Origin

First, let's ground ourselves. We are given that $f(x)$ is trapped within the interval $[-2,2]$, which implies that for any $x$, $|f(x)|\le 2$. Consequently, at $x=0$, we must have $(f(0){)}^2\le 4$.

Now, look at our given constraint: $(f(0){)}^2+(f^{\prime }(0){)}^2=85$. If we isolate the derivative term, we get:

Since $(f(0){)}^2$ is at most 4, the derivative squared must be at least $85-4=81$. This implies $|f^{\prime }(0)|\ge 9$.

This is an incredibly steep tangent! This immediately tells us that $f'(0) e 0$. Because the function is twice differentiable, the derivative is continuous, meaning the slope will maintain its sign in a small neighborhood around zero.

A function with a non-zero derivative in an interval is strictly monotonic, and thus, one-one. Option A is confirmed true.

The MVT Insight

Now, let's tackle Option B. We need to find a point $x_0\in (-4,0)$ such that $|f^{\prime }(x_0)|\le 1$.

This is a classic setup for the Lagrange Mean Value Theorem (L.M.V.T.). Let's apply it on the interval $[-4,0]$. The theorem guarantees the existence of a point $x_0$ such that:

The magnitude of this slope is $\frac{|f(0)-f(-4)|}{4}$. By the triangle inequality, $|f(0)-f(-4)|\le |f(0)|+|f(-4)|\le 2+2=4$.

Thus, $|f^{\prime }(x_0)|\le \frac{4}{4}=1$. The logic is airtight. Option B is also true.

The Auxiliary Function Masterstroke

Finally, we arrive at the most thrilling part: Option D. We need to find $\alpha \in (-4,4)$ such that $f(\alpha )+f^{\prime \prime }(\alpha )=0$.

This looks like a differential equation, but it's actually a hint to construct an auxiliary function. Let's define $g(x)=(f(x){)}^2+(f^{\prime }(x){)}^2$. We know $g(0)=85$.

Using the same MVT logic we used for Option B, we can find points $x_0\in (-4,0)$ and $x_1\in (0,4)$ where $|f^{\prime }(x_0)|\le 1$ and $|f^{\prime }(x_1)|\le 1$. At these points:

So, $g(x)$ starts at a value $\le 5$ at $x_0$, climbs to 85 at $x=0$, and drops back to $\le 5$ at $x_1$. By the Extreme Value Theorem, $g(x)$ must attain a local maximum at some point $\alpha$ between $x_0$ and $x_1$.

At this local maximum, $g^{\prime }(\alpha )=0$. Let's differentiate $g(x)$:

Setting this to zero at $\alpha$, we get:

We already proved that $f^{\prime }(\alpha )$ cannot be zero because $g(\alpha )\ge 85$. Therefore, the only possibility is $f(\alpha )+f^{\prime \prime }(\alpha )=0$.

And there it is! The elegance of calculus in action. We have navigated the constraints, applied the fundamental theorems, and arrived at the truth. Option D is true.