Animated Solution for Mathematics - Matrices and Determinants: Let M and m respectively be the maximum and the minimum values of $f(x)=|\begin{array}{ccc}1+si{n}^{2}x& co{s}^{2}x& 4sin4x\\ si{n}^{2}x& 1+co{s}^{2}x& 4sin4x\\ si{n}^{2}x& co{s}^{2}x& 1+4sin4x\end{array}|$ Then $M^4-m^4$ is equal to:

Facing the Determinant

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare. You see a $3\times 3$ determinant filled with trigonometric functions like ${\sin }^{2}x$, ${\cos }^{2}x$, and $4\sin 4x$.

It is designed to make you panic. It is designed to make you think, "Oh no, I have to expand this massive expression."

But here is the secret of the JEE Advanced: the complexity is often just a mask. Our job is to peel back that mask and find the elegant simplicity underneath.

Row Operations

When you encounter a determinant like this, your first instinct should not be to expand. Your instinct should be to observe. Look at the columns; do you see the repetition?

The terms ${\sin }^{2}x$, ${\cos }^{2}x$, and $4\sin 4x$ are dancing across the rows. This is not a coincidence; it is an invitation. We are going to use the most powerful tool in our linear algebra arsenal: row operations.

By performing $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we are essentially performing a surgical strike on the complexity of the matrix.

Let us look at the second row. We take $R_2$ and subtract $R_1$. The first element becomes ${\sin }^{2}x-(1+{\sin }^{2}x)=-1$. The second element becomes $(1+{\cos }^{2}x)-{\cos }^{2}x=1$. The third element becomes $4\sin 4x-4\sin 4x=0$.

Look at that! We have created a zero. Now, let us do the same for the third row. $R_3-R_1$ gives us $-1$ in the first column, $0$ in the second, and $1$ in the third.

Our determinant has transformed from a terrifying beast into a sleek, manageable form:

Now that we have these zeros, expanding the determinant is no longer a chore; it is a pleasure. Expanding along the first row, we get:

This simplifies beautifully to $f(x)=1+{\sin }^{2}x+{\cos }^{2}x+4\sin 4x$. And here, the magic happens.

We invoke the most fundamental identity in trigonometry: ${\sin }^{2}x+{\cos }^{2}x=1$. The entire expression collapses into $f(x)=1+1+4\sin 4x$, which is simply $f(x)=2+4\sin 4x$.

Do you see it? The entire complex determinant was just a fancy way of writing a simple, shifted sine wave.

Finding the Extremes

Now we are in the home stretch. We have $f(x)=2+4\sin 4x$. We need the maximum value $M$ and the minimum value $m$.

We know that the range of $\sin \theta$ is $[-1,1]$. Therefore, for the maximum value $M$, we set $\sin 4x=1$, giving us $M=2+4(1)=6$.

For the minimum value $m$, we set $\sin 4x=-1$, giving us $m=2+4(-1)=-2$.

The question asks for $M^4-m^4$. This is where we must be careful.

Subtracting these, we get $1296-16=1280$.

There you have it. We took a problem that looked like it required hours of calculation and solved it with a few clever observations and basic identities. This is the essence of JEE Advanced—it is not about brute force; it is about seeing the underlying structure.