Question
Suppose that$$\frac{4}{\sin x}+\frac{1}{1-\sin x}=a$$has at least one solution on the interval $(0, \pi / 2)$. Then $a$ has minimum value of $x=$(a) $\sin ^{-1} 2 / 3$(b) $\sin ^{-1} 1 / 4$(c) $\cos ^{-1} 4 / 5$(d) 1
Step 1
We want to find the minimum value of $y$. Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 76 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The equation $\sin ^{4} x+\cos ^{4} x=a$ has a solution for (A) all of values of $a$ (B) $a=1$ (C) $a=\frac{1}{2}$ (D) $\frac{1}{2}<a<1$
The number of solutions of $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$ is (A) 1 (B) 0 (C) 2 (D) 4
Statement 1 Minimum value of $\sin ^{2} x+\cos ^{4} x$ is $\frac{3}{4}$. and Statement 2 Minimum value of $\left(a x^{2}+b x+c\right)$ where $a>0$ is $\frac{-\left(b^{2}-4 a c\right)}{4 a}$.
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD