Let f be an odd function defined on the set o | vedclass

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

Question

(C) Given that $f$ is an odd function,we have $f(-x) = -f(x)$ for all $x$ in the domain.For $x \geq 0$,$f(x) = 3 \sin x + 4 \cos x$.We need to find $f

Vedclass · Accepted Answer

$\frac{3}{2} - 2\sqrt{3}$

Vedclass · Answer

(C) Given that $f$ is an odd function,we have $f(-x) = -f(x)$ for all $x$ in the domain.For $x \geq 0$,$f(x) = 3 \sin x + 4 \cos x$.We need to find $f\left(-\frac{11\pi}{6}\right)$.Since $f$ is an odd function,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right)$.First,calculate $f\left(\frac{11\pi}{6}\right)$:$f\left(\frac{11\pi}{6}\right) = 3 \sin\left(\frac{11\pi}{6}\right) + 4 \cos\left(\frac{11\pi}{6}\right)$$f\left(\frac{11\pi}{6}\right) = 3 \sin\left(2\pi - \frac{\pi}{6}\right) + 4 \cos\left(2\pi - \frac{\pi}{6}\right)$$f\left(\frac{11\pi}{6}\right) = 3(-\sin\frac{\pi}{6}) + 4(\cos\frac{\pi}{6})$$f\left(\frac{11\pi}{6}\right) = 3(-\frac{1}{2}) + 4(\frac{\sqrt{3}}{2}) = -\frac{3}{2} + 2\sqrt{3}$.Therefore,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right) = -(-\frac{3}{2} + 2\sqrt{3}) = \frac{3}{2} - 2\sqrt{3}$.

Let $f$ be an odd function defined on the set of real numbers such that for $x \geq 0$,$f(x) = 3 \sin x + 4 \cos x$. Then $f(x)$ at $x = -\frac{11\pi}{6}$ is equal to:

Similar Questions

The real valued function $f(x) = \frac{|x-a|}{x-a}$ is

If $[x]^2-5[x]+6=0$,where $[x]$ denotes the greatest integer function,then

The graphs of the polynomial $x^{2}-1$ and $\cos x$ intersect

Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x$,$[x]^{2}+2[x+2]-7=0$ has

Draw the graph of the function $f: R \rightarrow R$ defined by $f(x) = x^{3}, x \in R$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module