Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the least value of $a$ for which $$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a$$ has atleast one solution in the interval $(0,\frac{\pi}{2})$?

I first calculate $f'(x)$ and put it equal to $0$ to find out the critical points. This gives $$\sin(x)=\frac{2}{3}$$ as $\cos(x)$ is not $0$ in $(0,\frac{\pi}{2})$. I calculate $f''(x)$ and at $\sin(x)=\frac{2}{3}$, I get a minima. Now to have at least one solution, putting $\sin(x)=\frac{2}{3}$ in the main equation, I get $f=9-a$, which should be greater than or equal to $0$. I then get the 'maximum' value of $a$ as $9$. Where did I go wrong? [Note the function is $f(x)=LHS-RHS$ of the main equation.]

share|improve this question
1  
You're almost there. The minimum of the function in the interval is $9$, which is $a$. You are done. –  Ron Gordon Feb 19 '13 at 13:15
    
The question needs the minimum value of $a$. Do you think it can be wrong( it could have asked instead for maximum value)? –  Ashish Gaurav Feb 19 '13 at 13:18
    
No, I think it is OK. –  Ron Gordon Feb 19 '13 at 13:19
1  
You can save yourself some writing and maybe some confusion by noting that over $(0,\frac \pi 2), \sin x$ varies over $(0,1)$ so define $y=\sin x$ and work with $y$. –  Ross Millikan Feb 19 '13 at 14:10

1 Answer 1

up vote 2 down vote accepted

One possible approach: Find a common denominator, then :

$$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a\iff \frac{4(1- \sin x) + \sin x}{\sin x - \sin^x} = a$$ $$ \iff 4-3\sin x = a(\sin x - \sin^2 x)\tag{$\sin x \neq 0$}$$

Now write the equation as a quadratic equation in $\sin x$:

$$a\sin^2 x - (3 + a)\sin x + 4 = 0 $$

You can solve for when the equation has a real solution (by determining when the discriminant is greater than or equal to 0). $$b^2 - 4ac \geq 0 \iff (3+a)^2 - 16 a \geq 0 \iff a^2 -10a + 9 \geq 0 \iff (a - 1)(a-9) \geq 0$$

Then determine which values of $a$ satisfy the inequality and give in the desired interval.

share|improve this answer
    
Doing your way, I get the quadratic as $a{sin(x)}^2-(a+3)sin(x)+4=0$, and since $sin(x)$ is real, so $D>=0$ This gives $a>=9$ or $a<=1$, where $a>0$. Any explanation why we reject the latter(I mean it could also be a correct range for $a$, if this were not the question)? –  Ashish Gaurav Feb 19 '13 at 13:22
    
@AshishGaurav Note the left hand side of the equation is at least $4$. –  David Mitra Feb 19 '13 at 13:31
    
Thanks, this worked--! –  Ashish Gaurav Feb 19 '13 at 13:32
    
Great! You're welcome! –  amWhy Feb 19 '13 at 13:33
    
You don't need to solve the equation (and your factorization is incorrect). You need to determine when the quadratic has a real solution. So, determine when the discriminant is nonnegative. –  David Mitra Feb 19 '13 at 13:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.