The least value of a for which the equation f | vedclass

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(D) Let $f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}$. Let $t = \sin x$. Since $x \in (0, \pi/2)$,$t \in (0, 1)$.We define $g(t) = \frac{4}{t} + \f

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let $f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}$. Let $t = \sin x$. Since $x \in (0, \pi/2)$,$t \in (0, 1)$.We define $g(t) = \frac{4}{t} + \frac{1}{1 - t}$ for $t \in (0, 1)$.To find the minimum value,we differentiate $g(t)$ with respect to $t$:$g'(t) = -\frac{4}{t^2} + \frac{1}{(1 - t)^2}$.Setting $g'(t) = 0$ gives $\frac{1}{(1 - t)^2} = \frac{4}{t^2}$,which implies $(1 - t)^2 = \frac{t^2}{4}$.Taking the square root,$1 - t = \frac{t}{2}$ or $1 - t = -\frac{t}{2}$.For $t \in (0, 1)$,$1 - t = \frac{t}{2} \implies 1 = \frac{3t}{2} \implies t = \frac{2}{3}$.The value of the function at $t = \frac{2}{3}$ is $g(2/3) = \frac{4}{2/3} + \frac{1}{1 - 2/3} = 6 + 3 = 9$.Since $g(t) 	o \infty$ as $t 	o 0^+$ and $t 	o 1^-$,the range of $g(t)$ is $[9, \infty)$.Thus,the least value of $a$ for which the equation has at least one solution is $9$.

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

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