The minimum value of alpha for which the equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 ..........

Similar Questions

Let the radius and height of a right circular cylinder be related as $r^2 + h = 6$. If the volume of the cylinder is maximum,then the value of $\frac{r}{h}$ is:

Find all points of local maxima and local minima of the function $f$ given by $f(x) = x^3 - 3x + 3$.

If for a function $f(x)$,$f'(a) = 0$ and $f''(a) = 0$,then $x = a$ is:

Let $f(x) = \int\limits_0^x \frac{\cos t}{t} dt, x > 0$. Then $f(x)$ has:

The set of all values of $a$ for which the function $f(x) = (a^2 - 3a + 2) \left( \cos^2 \frac{x}{4} - \sin^2 \frac{x}{4} \right) + (a - 1)x + \sin 1$ does not possess critical points is

Vedclass Test Series

Exam Paper Generator

Online Exam Module