If alpha in (2, 3),then the number of solutio | vedclass

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

Question

(B) Given the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$. Evaluating the integral: $[\sin(x + \alpha^2)]_{0}^{\alpha} = \sin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$. Evaluating the integral: $[\sin(x + \alpha^2)]_{0}^{\alpha} = \sin \alpha$. This simplifies to $\sin(\alpha + \alpha^2) - \sin(\alpha^2) = \sin \alpha$. Using the formula $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$,we get: $2 \cos(\frac{\alpha^2 + \alpha + \alpha^2}{2}) \sin(\frac{\alpha + \alpha^2 - \alpha^2}{2}) = \sin \alpha$. $2 \cos(\alpha^2 + \frac{\alpha}{2}) \sin(\frac{\alpha}{2}) = 2 \sin(\frac{\alpha}{2}) \cos(\frac{\alpha}{2})$. Since $\alpha \in (2, 3)$,$\sin(\frac{\alpha}{2}) 
eq 0$,so we can divide by $2 \sin(\frac{\alpha}{2})$: $\cos(\alpha^2 + \frac{\alpha}{2}) = \cos(\frac{\alpha}{2})$. This implies $\alpha^2 + \frac{\alpha}{2} = 2n\pi \pm \frac{\alpha}{2}$. Case $1$: $\alpha^2 + \frac{\alpha}{2} = 2n\pi + \frac{\alpha}{2} \Rightarrow \alpha^2 = 2n\pi \Rightarrow \alpha = \sqrt{2n\pi}$. For $\alpha \in (2, 3)$,$4 Case $2$: $\alpha^2 + \frac{\alpha}{2} = 2n\pi - \frac{\alpha}{2} \Rightarrow \alpha^2 + \alpha - 2n\pi = 0$. Using the quadratic formula: $\alpha = \frac{-1 \pm \sqrt{1 + 8n\pi}}{2}$. For $\alpha \in (2, 3)$,we check $n=1$: $\alpha = \frac{-1 + \sqrt{1 + 8\pi}}{2} \approx \frac{-1 + 5.12}{2} \approx 2.06$. Both values lie in $(2, 3)$. Thus,there are $2$ solutions.

If $\alpha \in (2, 3)$,then the number of solutions of the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$ is:

Similar Questions

Let $f(x) = \int\limits_0^x {(t^2 + 2t + 2)dt}$ where $x$ is the set of real numbers satisfying the inequation $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$. If the range of $f(x)$ is $[a, b]$,then $(a + b)$ is:

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} [x], & x \leq 2 \\ 0, & x>2 \end{cases}$,where $[x]$ is the greatest integer less than or equal to $x$. If $I=\int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$,then the value of $(4I-1)$ is

If $f(x)$ and $g(x)$ are inverse functions of each other such that $f(1) = 3$ and $f(3) = 1$,then $\int_{1}^{3} \left( g(x) + \frac{x}{f'(g(x))} \right) dx$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module