Question

The least value of \(a\), for which the function \[ \frac{4}{\sin x} + \frac{1}{1-\sin x} = a \] has at least one solution in the interval \(\left(0,\frac{\pi}{2}\right)\), is:

• A. 9
• B. 4
• C. 5
• D. 1

Hint:

For least value problems, convert to a single variable and minimize using derivatives.

Answer 1

The Correct Option is A

To find the least value of \( a \) such that the function \(\frac{4}{\sin x} + \frac{1}{1-\sin x} = a\) has at least one solution in the interval \( \left(0,\frac{\pi}{2}\right) \), we need to explore the behavior of the function.

1. First, let's analyze each term separately for \( x \in \left(0,\frac{\pi}{2}\right) \).
   • \(\sin x\) varies from 0 to 1 as \( x \) moves from \( 0 \) to \(\frac{\pi}{2}\).
   • The first term, \(\frac{4}{\sin x}\), approaches infinity as \( \sin x \to 0^+\) and approaches 4 as \(\sin x \to 1^-\).
   • The second term, \(\frac{1}{1-\sin x}\), also approaches infinity as \(\sin x \to 1^- \) and approaches 1 as \( \sin x \to 0^+\).
2. We can summarize \( f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x} \) as having:
   • \(\lim_{x \to 0^+} f(x) = \infty\)
   • \(\lim_{x \to \frac{\pi}{2}^-} f(x) = \infty\)
3. Check the critical point by defining \( y = \sin x \), where \( y \in (0, 1) \).
   • \(f(y) = \frac{4}{y} + \frac{1}{1-y}\).
   • To find the minimum value, take the derivative \(f'(y)\) and set it to zero:
     • \(f'(y) = -\frac{4}{y^2} + \frac{1}{(1-y)^2} = 0\).
     • Solving for \( y \), we find: \(\frac{4}{y^2} = \frac{1}{(1-y)^2}\)
     • This implies: \(4(1-y)^2 = y^2\).
     • Expanding and solving: \(4 - 8y + 4y^2 = y^2\)
     • Simplies to: \( 3y^2 - 8y + 4 = 0 \).
     • Solve the quadratic equation using the quadratic formula: \(y = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6}\).
     • The roots are \( y = \frac{8 + 4}{6} = 2 \) and \( y = \frac{8 - 4}{6} = \frac{2}{3}\). Only \( y = \frac{2}{3} \) is valid since \( y \leq 1 \).
4. Evaluate \( f(y) \) at \( y = \frac{2}{3} \):
   • \( f\left(\frac{2}{3}\right) = \frac{4}{\frac{2}{3}} + \frac{1}{1 - \frac{2}{3}} = 6 + 3 = 9 \).
5. Thus, the least value of \( a \) for which the function has a solution is \(9\).

Therefore, the correct answer is 9.