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

What is one of the maximum values of $f(x) = x + \sin(2x)$ in the interval $[0, 2\pi]$?

If $A=\{x : 9x \geq x^2+20\}$ and $f: A \rightarrow R$ is defined by $f(x)=2x^3-15x^2+36x-48$,then the maximum value of $f(x)$ is

Given $f(x) = -\frac{x^3}{3} + x^2 \sin(1.5a) - x \sin(a) \sin(2a) - 5 \sin^{-1}(a^2 - 8a + 17)$,then:

The function $f(x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at $x = $ ........

If $A = \{x \in R : \frac{\pi}{4} \leq x \leq \frac{\pi}{3}\}$ and $f(x) = \sin x - x$,then $f(A)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module