Animated Solution for Mathematics - Functions: Let $f:\mathbb{R}\to \mathbb{R}$ be a function defined by $f(x)={\log }_{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is

Welcome, future engineer. Today, we are not just solving a problem; we are dissecting a mathematical machine.

When you look at $f(x)={\log }_{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}$, do not see a terrifying wall of symbols. See a two-stage rocket.

The inner stage is a trigonometric engine, and the outer stage is a logarithmic controller. To understand the flight path—the range of the function—we must understand how the inner stage feeds the outer one.

Taming the Trigonometric Beast

Let us isolate the inner part. We define $k=\sqrt{2}(\sin x-\cos x)$.

This is a classic structure. Whenever you see $a\sin x+b\cos x$, your intuition should immediately scream, "Harmonic Addition Theorem!"

We want to collapse this into a single sine wave. We calculate the amplitude:

Thus, our expression becomes $k=2\sin (x-\frac{\pi }{4})$.

Now, pause and breathe. We know the sine function, regardless of the phase shift, is trapped between $-1$ and $1$.

Therefore, our variable $k$ is strictly bounded: $-2\le k\le 2$. This is the "fuel" for our logarithmic stage.

If we do not respect these bounds, the entire rocket fails.

The Logarithmic Bridge

Now, we look at the outer function: $f(x)={\log }_{\sqrt{m}}(k+m-2)$.

The problem gives us a gift: the range of $f(x)$ is $[0,2]$. This means the output of our log function is trapped between $0$ and $2$.

We write this as an inequality:

To solve this, we must strip away the logarithm. We convert it to exponential form.

Assuming our base $\sqrt{m}>1$, the inequality signs remain stable. We raise the base to the power of the bounds:

This simplifies beautifully to $1\le k+m-2\le m$.

The Final Convergence

We are almost there. We need to isolate $k$ to see what range the logarithmic function demands of it.

Subtracting $(m-2)$ from all parts, we get:

This simplifies to $3-m\le k\le 2$.

Here is the moment of truth. We have two descriptions of the range of $k$.

From our trigonometric analysis, we know $k\in [-2,2]$. From our logarithmic analysis, we know $k\in [3-m,2]$.

For the function to exist as defined, these two ranges must be identical. The upper bounds already match ($2=2$).

Therefore, the lower bounds must also match: $3-m=-2$.

Solving this linear equation, we find $m=5$.

This problem is a masterclass in constraints. It teaches you that in complex systems, the output of one stage is the input of the next.

By respecting the bounds of the trigonometric function and aligning them with the logarithmic requirements, we did not just find an answer; we solved a puzzle of logical consistency.

Keep this mindset, and no JEE problem will ever be too daunting.