A function y = f(x) satisfies the condition f | vedclass

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

Question

(A) The given differential equation is $f'(x) \sin x + f(x) \cos x = 1$. This can be written as $\frac{d}{dx} (f(x) \sin x) = 1$. Integrating both sid

Vedclass · Accepted Answer

$\frac{\pi}{2} < I < \frac{\pi^2}{4}$

Vedclass · Answer

(A) The given differential equation is $f'(x) \sin x + f(x) \cos x = 1$. This can be written as $\frac{d}{dx} (f(x) \sin x) = 1$. Integrating both sides with respect to $x$,we get $f(x) \sin x = x + C$. Thus,$f(x) = \frac{x + C}{\sin x}$. Since $f(x)$ is bounded as $x ightarrow 0$,we have $\lim_{x ightarrow 0} \frac{x + C}{\sin x}$ must be finite. Using $\lim_{x ightarrow 0} \frac{x}{\sin x} = 1$,the limit is finite only if $C = 0$. Therefore,$f(x) = \frac{x}{\sin x}$. We need to evaluate $I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \, dx$. For $x \in (0, \frac{\pi}{2}]$,we know that $\sin x This implies $\frac{1}{	an x} Thus,$f(x) > 1$ for $x \in (0, \frac{\pi}{2}]$. Also,using the graph or Taylor series,$f(x) = \frac{x}{\sin x}$ is an increasing function on $(0, \frac{\pi}{2}]$ with $f(0^+) = 1$ and $f(\frac{\pi}{2}) = \frac{\pi}{2}$. Since $1 This gives $\frac{\pi}{2} < I < \frac{\pi^2}{4}$.

$A$ function $y = f(x)$ satisfies the condition $f'(x) \sin x + f(x) \cos x = 1$,where $f(x)$ is bounded as $x \rightarrow 0$. If $I = \int_{0}^{\frac{\pi}{2}} f(x) \, dx$,then:

Similar Questions

The solution of the differential equation $\frac{dy}{dx} = \frac{\sin y + e^x}{\ln y - x \cos y}$ is:

The solution of $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$ is

The solution of $\frac{d^2y}{dx^2} = \cos x - \sin x$ is

The solution of the differential equation $\frac{d^2 y}{d x^2}+y=0$ is

The curve amongst the family of curves represented by the differential equation,$(x^2 - y^2)dx + 2xy\, dy = 0$ which passes through $(1, 1)$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module