The minimum value of alpha for which the equa | vedclass

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(B) Let $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.Let $t = \sin x$. Since $x \in (0, \frac{\pi}{2})$,we have $t \in (0, 1)$.Then $f(t) = \frac{4}

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Let $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.Let $t = \sin x$. Since $x \in (0, \frac{\pi}{2})$,we have $t \in (0, 1)$.Then $f(t) = \frac{4}{t} + \frac{1}{1-t}$.To find the minimum value,we differentiate $f(t)$ with respect to $t$:$f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2}$.Setting $f'(t) = 0$,we get $\frac{1}{(1-t)^2} = \frac{4}{t^2}$,which implies $(1-t)^2 = \frac{t^2}{4}$.Taking the square root,$1-t = \frac{t}{2}$ (since $t \in (0, 1)$),so $1 = \frac{3t}{2}$,which gives $t = \frac{2}{3}$.Since $t = \frac{2}{3}$ is in $(0, 1)$,the minimum value is $f(\frac{2}{3}) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9$.Thus,the minimum value of $\alpha$ for which the equation has at least one solution is $9$.

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

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