The minimum value of a for which the equation 4/sinx + 1/(1-sinx) = a has at least one solution in (

Q: Why can't we just use AM-GM inequality here?

AM-GM is a powerful tool, but it requires all terms to be positive. While our terms are positive, the variables are linked. Using AM-GM directly on $\frac{4}{t} + \frac{1}{1-t}$ would give a constant, but it might not be the minimum if the variables $t$ and $1-t$ cannot be equal. Calculus is the safer, more robust path.

Q: Why is the domain $t \in (0, 1)$?

Because $x$ is in $(0, \frac{\pi}{2})$, $\sin x$ ranges from $0$ to $1$. Since the interval is open, $t$ cannot be $0$ or $1$, as that would make the denominators zero, which is undefined.

Q: What happens if $a < 9$?

If $a < 9$, the horizontal line $y = a$ lies below the minimum point of the function. Since the function never reaches a value lower than $9$, the line will never intersect the curve, meaning there are no solutions for $x$.

Question

The minimum value of $a$ for which the equation $\frac{4}{\sin x} + \frac{1}{1-\sin x} = a$ has at least one solution in $(0, \frac{\pi}{2})$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the equation:

$$\frac{4}{\sin x} + \frac{1}{1-\sin x} = a$$

We must find the minimum value of $a$ for which this equation holds at least one solution in the interval $x \in (0, \frac{\pi}{2})$. This is equivalent to finding the minimum value of the function $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$ within the specified domain.

### The Transformation

Trigonometry can sometimes obscure the underlying algebraic beauty. Let us simplify our perspective by substituting $t = \sin x$.

Since $x \in (0, \frac{\pi}{2})$, our new variable $t$ is strictly confined to the interval $(0, 1)$. This is a crucial observation, as ignoring this boundary would lead us into mathematical territory that does not exist for our problem.

Our function now becomes:

$$f(t) = \frac{4}{t} + \frac{1}{1-t}$$

As $t$ approaches $0$ or $1$, the denominators shrink, causing the function to soar toward infinity. Consequently, there must be a minimum value located somewhere within the interval $(0, 1)$.

### The Calculus of Discovery

To find this minimum, we turn to the derivative. We seek the critical point where the slope of the tangent is zero.

Differentiating $f(t)$ with respect to $t$:

$$f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2}$$

Setting $f'(t) = 0$ leads us to the following equation:

$$\frac{1}{(1-t)^2} = \frac{4}{t^2}$$

Taking the square root of both sides, we obtain:

$$\frac{1}{1-t} = \frac{2}{t}$$

We ignore the negative root because $t$ and $1-t$ are both positive in our domain. Solving this yields $t = 2 - 2t$, which simplifies to $3t = 2$, or $t = \frac{2}{3}$. This critical point sits perfectly within our valid interval $(0, 1)$.

### Final Calculation

Now that we have the location of the minimum, we evaluate the function at $t = \frac{2}{3}$:

$$f\left(\frac{2}{3}\right) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9$$

The minimum value of the function is $9$. Geometrically, this means the lowest point on our curve is at height $9$.

If our horizontal line $y = a$ is any lower than $9$, it will never touch the curve. Therefore, the **minimum value of $a$** for which a solution exists is exactly **$9$**.

Animated Solution for Mathematics - Differentiation: The minimum value of $a$ for which the equation $\frac{4}{s i n x} + \frac{1}{1 - s i n x} = a$ has at least one solution in $(0, \frac{π}{2})$ is

Enter Numerical Value:

Visualized Solution

Defining the Function $f (x)$

Substitution $t = sin x$

The Transformed Function $f (t)$

Differentiating $f (t)$

Calculating $f^{'} (t)$

Setting $f^{'} (t) = 0$

Solving for $t$

Finding the Critical Point

Verifying the Minimum

Evaluating the Minimum Value

Final Calculation

Visualizing the Equation $f (t) = a$