Animated Solution for Mathematics - Differentiation: Let $f$ and $g$ be twice differentiable functions on $\mathbb{R}$ such that $f^{\prime \prime }(x)=g^{\prime \prime }(x)+6x$, $f^{\prime }(1)=4g^{\prime }(1)-3=9$, $f(2)=3g(2)=12$. Then which of the following is NOT true?

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, might look like a chaotic mess of two unknown functions, $f(x)$ and $g(x)$.

But here is the secret: in the world of advanced calculus, complexity is often just a mask for a deeper, simpler truth. We are given $f^{\prime \prime }(x)=g^{\prime \prime }(x)+6x$, along with specific values for the derivatives and the functions themselves at certain points.

The key to this puzzle is to stop looking at $f(x)$ and $g(x)$ as two separate entities. Instead, let us define a new function, $h(x)=f(x)-g(x)$. By doing this, we collapse the problem into a single, manageable variable. This is the first step of a true mathematician—reducing the noise to find the signal.

Integrating the Truth

Now that we have our function $h(x)$, let us look at the differential equation. We have $f^{\prime \prime }(x)-g^{\prime \prime }(x)=6x$, which is simply $h^{\prime \prime }(x)=6x$. This is a beautiful, simple second-order differential equation.

To find $h(x)$, we must integrate twice. First, we integrate $h^{\prime \prime }(x)$ to find $h^{\prime }(x)$:

We need to find the constant $C$. We are given $f^{\prime }(1)=9$ and $4g^{\prime }(1)-3=9$. Solving the second equation gives $g^{\prime }(1)=3$.

Thus, $h^{\prime }(1)=f^{\prime }(1)-g^{\prime }(1)=9-3=6$. Substituting $x=1$ into our expression for $h^{\prime }(x)$, we get $3(1{)}^2+C=6$, which yields $C=3$. So, $h^{\prime }(x)=3x^2+3$.

Now, we integrate once more to find $h(x)$:

To find $D$, we use the values at $x=2$. We know $f(2)=12$ and $3g(2)=12$, so $g(2)=4$.

Thus, $h(2)=f(2)-g(2)=12-4=8$. Substituting $x=2$ into $h(x)$, we get $2^3+3(2)+D=8$, which simplifies to $8+6+D=8$, giving us $D=-6$. Our final, elegant function is $h(x)=x^3+3x-6$.

Testing the Options

With $h(x)=x^3+3x-6$ in our toolkit, we can now evaluate each option with absolute confidence.

Option 1 asks about $g(-2)-f(-2)$, which is simply $-h(-2)$. Calculating $h(-2)=(-2{)}^3+3(-2)-6=-8-6-6=-20$. Thus, $-h(-2)=20$. This is true.

Option 3 claims $|f^{\prime }(x)-g^{\prime }(x)|\le 6\Rightarrow -1\le x\le 1$. This is $|h^{\prime }(x)|\le 6$. Since $h^{\prime }(x)=3x^2+3$, which is always positive, we have $3x^2+3\le 6$, leading to $3x^2\le 3$, or $x^2\le 1$. This is true.

Option 4 suggests a root exists in $(1,1.5)$. We check $h(1)=1+3-6=-2$ and $h(1.5)=3.375+4.5-6=1.875$. Since the function changes sign, the Intermediate Value Theorem guarantees a root. This is also true.

Finally, Option 2 claims $|f(x)-g(x)|<8$ for $-1<x<2$. But we found $h(-1)=-1-3-6=-10$. The absolute value $|h(-1)|=10$, which is not less than $8$. Therefore, Option 2 is the false statement.