Let f(x) = 3√(x-2) + √(4-x) be a real valued function. If α and β are respectively the minimum and t

Q: Why can't we just use calculus to find the maximum and minimum by setting the derivative to zero?

You absolutely can! However, differentiating this function involves the chain rule and results in a more complex expression that you then have to set to zero and solve. While valid, it is computationally heavier and prone to algebraic errors. The trigonometric substitution method is an elegant, 'JEE-style' shortcut that transforms the problem into a standard form, saving you time and mental energy.

Q: Why do we check the boundaries for the minimum value instead of just the critical points?

Great question! When we transform the function into f(theta) = a*sin(theta) + b*cos(theta), we are working within a restricted domain for theta (from 0 to pi/2). The function a*sin(theta) + b*cos(theta) is monotonic or has its peak within the interval, but the minimum often occurs at the boundaries of the domain when the function is restricted to a specific interval. Always check your endpoints!

Q: How do I know when to use trigonometric substitution in these types of problems?

Look for the structure! Whenever you see square roots containing linear expressions like sqrt(x-a) + sqrt(b-x), check if the sum of the terms inside the roots is a constant. If (x-a) + (b-x) = constant, you can almost always use a substitution like x-a = constant * sin^2(theta) to simplify the expression into a trigonometric function.

Question

Let $f(x) = 3\sqrt{x-2} + \sqrt{4-x}$ be a real valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$, then $\alpha^2 + 2\beta^2$ is equal to

Options:

(A) 42

(B) 38

(C) 24

(D) 44

Sigma Percentile · Accepted Answer

### The Dance of the Square Roots

Welcome, future engineers! Today, we are going to unravel a problem that looks like a tangled mess of radicals but hides a beautiful, rhythmic structure. We are dealing with the function $f(x) = 3\sqrt{x-2} + \sqrt{4-x}$.

At first glance, it might seem intimidating, but in the world of JEE Advanced, intimidation is just a mask for elegance waiting to be discovered.

### Phase 1: Defining the Arena

Before we perform any magic, we must define where our function lives. For $f(x)$ to output real numbers, the expressions inside our square roots must be non-negative.

This gives us two simple inequalities:
1. $x - 2 \ge 0 \implies x \ge 2$
2. $4 - x \ge 0 \implies x \le 4$

Combining these, we find our domain: $x \in [2, 4]$. Our function is trapped in this narrow corridor on the $x$-axis. Knowing the boundaries is the first step to mastering any function.

### Phase 2: The Trigonometric Insight

Now, look closely at the terms inside the roots: $(x-2)$ and $(4-x)$. If we add them together, the $x$ terms vanish, leaving us with $2$.

This is not a coincidence! It is a massive hint. In trigonometry, we know that $\sin^2	heta + \cos^2	heta = 1$.

If we multiply by $2$, we get $2\sin^2	heta + 2\cos^2	heta = 2$. This suggests that we can map our $x$ values to an angle $	heta$ such that:
$$x-2 = 2\sin^2	heta$$
$$4-x = 2\cos^2	heta$$

Since $x$ ranges from $2$ to $4$, our angle $	heta$ will range from $0$ to $\frac{\pi}{2}$.

### Phase 3: The Transformation

Let's substitute these into our function. The expression $f(x)$ transforms into:
$$f(	heta) = 3\sqrt{2\sin^2	heta} + \sqrt{2\cos^2	heta}$$

Simplifying this, we get:
$$f(	heta) = 3\sqrt{2}\sin	heta + \sqrt{2}\cos	heta$$

Suddenly, the scary radicals are gone, replaced by a clean, standard trigonometric expression of the form $a\sin	heta + b\cos	heta$. We know that the maximum value of such a function is $\sqrt{a^2 + b^2}$.

### Phase 4: Finding the Extremes

For our function, $a = 3\sqrt{2}$ and $b = \sqrt{2}$. The maximum value, $\beta$, is:
$$\beta = \sqrt{(3\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{18 + 2} = \sqrt{20}$$

What about the minimum, $\alpha$? Since we are restricted to $	heta \in [0, \frac{\pi}{2}]$, we must check the boundaries.

At $	heta = 0$:
$$f(0) = 3\sqrt{2}(0) + \sqrt{2}(1) = \sqrt{2}$$

At $	heta = \frac{\pi}{2}$:
$$f\left(\frac{\pi}{2}ight) = 3\sqrt{2}(1) + \sqrt{2}(0) = 3\sqrt{2}$$

Comparing $\sqrt{2}$ and $3\sqrt{2}$, the minimum $\alpha$ is clearly $\sqrt{2}$.

### The Grand Finale

We have our values: $\alpha = \sqrt{2}$ and $\beta = \sqrt{20}$. The problem asks for $\alpha^2 + 2\beta^2$.

Substituting our values, we get:
$$\alpha^2 + 2\beta^2 = (\sqrt{2})^2 + 2(\sqrt{20})^2 = 2 + 2(20) = 2 + 40 = \mathbf{42}$$

And there it is! A complex function reduced to a simple, elegant integer. Remember, in physics and math, when things look complicated, look for the hidden symmetry. Keep practicing, and keep falling in love with the process!

Animated Solution for Mathematics - Functions: Let $f (x) = 3 x - 2 + 4 - x$ be a real valued function. If $α$ and $β$ are respectively the minimum and the maximum values of $f$ , then $α^{2} + 2 β^{2}$ is equal to

Select Answer:

Visualized Solution

Function and Objective

Finding the Domain

Domain Interval

Trigonometric Substitution

Applying the Substitution

Transforming the Function

Maximum Value Formula

Calculating $β$ (Maximum)

Finding the Minimum Value

Evaluating Boundaries

Identifying $α$ (Minimum)

Setting Up the Final Expression

Final Calculation

The Sigma Insight: Domain and Range of a Function

The Dance of the Square Roots

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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