If the equation frac{x^2 + 5}{2} = x - 2cos(a | vedclass

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

Question

(B) Given the equation: $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.Rearranging the terms: $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.Multiply by $-1/2$: $-\

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(B) Given the equation: $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.Rearranging the terms: $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.Multiply by $-1/2$: $-\cos(ax + b) = \frac{x^2 + 5}{4} - \frac{x}{2} = \frac{x^2 - 2x + 5}{4}$.This simplifies to: $-\cos(ax + b) = \frac{(x - 1)^2}{4} + 1$.We know that the range of $-\cos(ax + b)$ is $[-1, 1]$.Also,for the right side,since $(x - 1)^2 \ge 0$,we have $\frac{(x - 1)^2}{4} + 1 \ge 1$.For the equation to have a solution,both sides must be equal to $1$.Thus,$\frac{(x - 1)^2}{4} + 1 = 1 \Rightarrow x = 1$.And $-\cos(a(1) + b) = 1 \Rightarrow \cos(a + b) = -1$.Therefore,$a + b = (2k + 1)\pi$ for $k \in I$. For $k=0$,$a + b = \pi$.

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

Similar Questions

If $2 + i\sqrt{3}$ is a root of the equation $x^2 + px + q = 0$,where $p$ and $q$ are real,then $(p, q) = $

The equation $e^x - x - 1 = 0$ has

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,then the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ is

The value of $p$ for which one root of the equation ${x^2} - 30x + p = 0$ is the square of the other,are

$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module