Using the principal values of the inverse trigonometric functions, the sum of the maximum and the mi

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

This is a common point of confusion! In this problem, we are not dealing with energy conservation, but rather the properties of inverse trigonometric functions. We don't 'reject' a root; rather, we analyze the domain of the function. Since $\sec^{-1}x$ is only defined for $|x| \ge 1$, its range is restricted to $[0, \pi]$ excluding $\frac{\pi}{2}$. We must respect these boundaries when finding the maximum and minimum.

Q: Why can't we just use the range of $\sec^{-1}x$ and $\csc^{-1}x$ separately?

That is a classic trap! You cannot treat them as independent variables because they are linked by the identity $\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$. If you treat them as independent, you ignore the constraint that they must sum to $\frac{\pi}{2}$. Always look for these hidden constraints in JEE problems.

Q: How do I know if the maximum is at the endpoint or the vertex?

Great question. The expression $E$ is a quadratic in $a$, which represents a parabola opening upwards. The vertex of a parabola is its minimum point. The maximum value on a closed interval will always occur at the endpoint that is furthest from the vertex. By finding the vertex and checking the distance to the boundaries of the domain, you can easily determine where the maximum occurs.

Question

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16((\sec^{-1}x)^{2}+(\csc^{-1}x)^{2})$ is:

Options:

(A) $24\pi^{2}$

(B) $22\pi^{2}$

(C) $31\pi^{2}$

(D) $18\pi^{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, might seem like a chaotic mess of inverse trigonometric functions.

We are given the expression $E = 16((\sec^{-1}x)^{2}+(\csc^{-1}x)^{2})$ and asked to find the sum of its maximum and minimum values.

It looks intimidating, but remember: in the world of JEE Advanced, complexity is often just a mask for elegance.

### The Golden Key: Unlocking the Identity

The first step in any great journey is orientation. We see two different inverse trigonometric functions, $\sec^{-1}x$ and $\csc^{-1}x$.

If we try to treat them as independent variables, we will get lost in a labyrinth of derivatives and inequalities. Instead, we must look for the hidden connection.

Do you remember the identity $\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$? This is our golden key.

It tells us that these two functions are not independent; they are locked in a rigid relationship. By using this identity, we can replace $\csc^{-1}x$ with $\frac{\pi}{2} - \sec^{-1}x$.

Suddenly, the problem transforms from a multivariable nightmare into a single-variable algebraic exercise.

### The Transformation: From Trigonometry to Algebra

Let us introduce a dummy variable, $a = \sec^{-1}x$. We must be careful with the domain: the principal value branch of $\sec^{-1}x$ is $[0, \pi]$, excluding $\frac{\pi}{2}$.

So, our variable $a$ lives in the domain $a \in [0, \pi] \setminus \{\frac{\pi}{2}\}$.

Now, substitute this into our expression $E$:

$$E = 16\left(a^2 + \left(\frac{\pi}{2} - a\right)^2\right)$$

Expanding this, we get $E = 16(a^2 + \frac{\pi^2}{4} - \pi a + a^2)$, which simplifies to $E = 16(2a^2 - \pi a + \frac{\pi^2}{4})$.

Distributing the $16$, we arrive at the quadratic equation:

$$E = 32a^2 - 16\pi a + 4\pi^2$$

### The Geometry: Visualizing the Parabola

Now that we have a quadratic, we can use the power of geometry. We want to find the extrema of this parabola by completing the square.

Factoring out $32$ from the first two terms, we get $E = 32(a^2 - \frac{\pi}{2}a) + 4\pi^2$. Adding and subtracting $(\frac{\pi}{4})^2$, we reach the vertex form:

$$E = 32\left(a - \frac{\pi}{4}\right)^2 + 2\pi^2$$

This tells us everything we need to know. The parabola opens upwards, and its vertex is at $a = \frac{\pi}{4}$.

Since $\frac{\pi}{4}$ is within our domain, the **minimum value** of $E$ is simply the constant term:

$$E_{min} = 2\pi^2$$

### The Final Stretch: Finding the Maximum

For a parabola, the maximum on a closed interval occurs at the endpoint furthest from the vertex. Our domain is $[0, \pi] \setminus \{\frac{\pi}{2}\}$.

The vertex is at $a = \frac{\pi}{4}$. The distance to the left endpoint $a = 0$ is $\frac{\pi}{4}$, while the distance to the right endpoint $a = \pi$ is $\frac{3\pi}{4}$.

Since $\pi$ is further away, the maximum must occur at $a = \pi$. Substituting $a = \pi$ into our vertex form:

$$E_{max} = 32\left(\pi - \frac{\pi}{4}\right)^2 + 2\pi^2 = 32\left(\frac{3\pi}{4}\right)^2 + 2\pi^2$$

$$E_{max} = 32\left(\frac{9\pi^2}{16}\right) + 2\pi^2 = 18\pi^2 + 2\pi^2 = 20\pi^2$$

Finally, the sum of the maximum and minimum values is:

$$20\pi^2 + 2\pi^2 = \mathbf{22\pi^2}$$

And there you have it! A complex problem, solved with nothing but a fundamental identity and the beauty of quadratic geometry.

Animated Solution for Mathematics - Inverse Trigonometric Functions: Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16 ((sec^{- 1} x)^{2} + (csc^{- 1} x)^{2})$ is:

Select Answer:

Visualized Solution

Analyze the Expression

Inverse Trigonometric Identity

Variable Substitution

Substitute into Expression

Expand the Squares

Simplify the Quadratic

Complete the Square - Step 1

Complete the Square - Step 2

Vertex Form of the Quadratic

Calculate Minimum Value

Analyze Domain for Maximum

Calculate Maximum Value

Final Summation

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Golden Key

The Transformation

The Geometry

The Final Stretch

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