The number of solutions of the equation frac{ | vedclass

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

Question

(D) Let $f(x) = \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}}$.Using Leibniz's rule,$\frac{d}{dx} f(x) = \frac{1}{\sqrt{1 - (\sin x)^2}} \cdot \fra

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $f(x) = \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}}$.Using Leibniz's rule,$\frac{d}{dx} f(x) = \frac{1}{\sqrt{1 - (\sin x)^2}} \cdot \frac{d}{dx}(\sin x) - \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot \frac{d}{dx}(\cos x)$.Since $\sqrt{1 - \sin^2 x} = |\cos x|$ and $\sqrt{1 - \cos^2 x} = |\sin x|$,the expression becomes $\frac{\cos x}{|\cos x|} - \frac{-\sin x}{|\sin x|} = 	ext{sgn}(\cos x) + 	ext{sgn}(\sin x)$.In the interval $[0, \pi]$,$\sin x$ is always positive,so $	ext{sgn}(\sin x) = 1$.Thus,the equation is $	ext{sgn}(\cos x) + 1 = 2\sqrt{2}$.Since $	ext{sgn}(\cos x)$ can only be $1, -1,$ or $0$,the sum $	ext{sgn}(\cos x) + 1$ can only be $2, 0,$ or $1$.Since $2\sqrt{2} \approx 2.828$,there is no value of $x$ such that $	ext{sgn}(\cos x) + 1 = 2\sqrt{2}$.Therefore,the number of solutions is $0$.

The number of solutions of the equation $\frac{d}{dx} \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}} = 2\sqrt{2}$ in the interval $[0, \pi]$ is:

Similar Questions

Let $f$ be a differentiable function in $\left(0, \frac{\pi}{2}\right)$. If $\int\limits_{\cos x}^{1} t^{2} f(t) d t = \sin^{3} x + \cos x - 1$,then $\frac{1}{\sqrt{3}} f^{\prime}\left(\frac{1}{\sqrt{3}}\right)$ is equal to

$\mathop {Limit}\limits_{x \to {x_1}} \,\,\frac{x}{{x - {x_1}}}\,\,\int\limits_{{x_1}}^x {f(t)} \, dt$ is equal to :

$\int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$ is equal to -

$\int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x \, dx =$

If $\int_0^{t^2} xf(x)dx = \frac{2}{5}t^5, t > 0,$ then $f\left( \frac{4}{25} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module