Animated Solution for Mathematics - Differentiation: If m is the minimum value of k for which the function $f(x)=x\sqrt{kx-x^2}$ is increasing in the interval $[0,3]$ and M is the maximum value of f in $[0,3]$ when k = m, then the ordered pair (m, M) is equal to :

Unlocking the Parametric Function

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are dissecting the behavior of a function that hides its secrets behind a parameter $k$.

The function $f(x)=x\sqrt{kx-x^2}$ might look intimidating at first glance, but it is a beautiful example of how parameters dictate the landscape of a curve. Our journey involves two distinct phases: first, finding the critical parameter $m$ that forces this function to grow, and second, finding the peak of that growth, $M$.

The Domain Constraint

Before we even touch the calculus, we must respect the boundaries of our mathematical universe. The function $f(x)=x\sqrt{kx-x^2}$ involves a square root, and in the real number system, we cannot take the square root of a negative number.

Thus, we require $kx-x^2\ge 0$. Factoring this, we get $x(k-x)\ge 0$.

Since we are interested in the interval $x\in [0,3]$, we know $x$ is non-negative. For the product to be non-negative, the second factor, $k-x$, must also be non-negative. This implies $x\le k$.

If our function must be defined for the entire interval $[0,3]$, then $k$ must be at least $3$. This is our first anchor point: $k\ge 3$. Without this, our function would simply vanish into the complex plane before we reached our destination.

The Calculus Engine

Now, let us find the slope of our function. We want to know where it is increasing, which means we need the derivative $f^{\prime }(x)$.

Since $f(x)$ is the product of $x$ and $\sqrt{kx-x^2}$, we apply the Product Rule:

Let $u=x$ and $v=\sqrt{kx-x^2}$. The derivative of $u$ is $1$. For $v$, we use the Chain Rule: the derivative of $\sqrt{g(x)}$ is $\frac{g^{\prime }(x)}{2\sqrt{g(x)}}$.

Here, $g(x)=kx-x^2$, so $g^{\prime }(x)=k-2x$. Putting it all together, we get:

The Inequality Dance

To simplify this, we find a common denominator, which is $2\sqrt{kx-x^2}$. The expression becomes:

Expanding the numerator, we have $2kx-2x^2+kx-2x^2$, which simplifies beautifully to $3kx-4x^2$. Thus, our derivative is:

For the function to be increasing, we need $f^{\prime }(x)\ge 0$. Since the denominator is always positive in our interval, we only need the numerator to be non-negative: $3kx-4x^2\ge 0$.

Dividing by $x$ (which is positive in our interval), we get $3k-4x\ge 0$, or $k\ge \frac{4x}{3}$. This condition must hold for all $x\in [0,3]$.

The maximum value of $\frac{4x}{3}$ occurs at $x=3$, giving us $k\ge \frac{4(3)}{3}=4$. Therefore, the minimum value $m$ is $4$.

The Final Ascent

With $m=4$, our function becomes $f(x)=x\sqrt{4x-x^2}$. Because we have ensured the function is increasing on $[0,3]$, the maximum value $M$ must occur at the rightmost endpoint, $x=3$.

Plugging this in:

We have arrived! The ordered pair $(m,M)$ is $(4,3\sqrt{3})$. This problem is a testament to the power of systematic analysis. By respecting the domain, carefully applying the rules of calculus, and solving the resulting inequality, we have tamed the function.