Question

Let f be an odd function defined on the set of real numbers such that for x ≥ 0, f(x) = 3 sin x + 4 cos x. Then f(x) at $x=-\frac{11x}{6}$ is equal to:
(a) $\frac{3}{2}+2\sqrt{3}$
(b) $-\frac{3}{2}+2\sqrt{3}$
(c) $\frac{3}{2}-2\sqrt{3}$
(d) $-\frac{3}{2}-2\sqrt{3}$

Answer 1

Answer:

The required value is

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

option  (c) is correct

Step-by-step explanation:

Concept used:

A function f(x) is said to be odd if  f(-x)=-f(x)

Given:

$f(x)= 3\:sinx+4\:cosx$

To find:

$f(\frac{-11\pi}{6})$

since f(x) is odd,

$f(\frac{-11\pi}{6})=-f(\frac{11\pi}{6})$

$=-[3\:sin(\frac{11\pi}{6})+4\:cos(\frac{11\pi}{6})]$

$=-[3\:sin(2\pi-\frac{\pi}{6})+4\:cos(2\pi-\frac{\pi}{6})]$

$=-[-3\:sin\frac{\pi}{6}+4\:cos\frac{\pi}{6}]$

$=-[-3(\frac{1}{2})+4(\frac{\sqrt{3}}{2})]$

$=-[\frac{-3}{2}+2\sqrt{3}]$

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

Answer 2

Answer:

3/2 - 2√3

Step-by-step explanation:

If a function f(x) is odd-function, f(-x) = - f(x)

f(-x) = - f(x)     ⇒ - f(-x) = f(x)

  In the question, for x = -11π*/6

f(x) = - f(-x)

f(-11π/6) = - [3sin{-(-11π/6) + 4cos(-(-11π/6))]

    = - [3sin(11π/6) + 4cos(11π/6)]

    = - 3sin(2π - π/6) - 4cos(2π -π/6)

    = - 3sin(-π/6) - 4cos(-π/6)

    = - 3[-sin(π/6)] - 4cos(π/6)

    = - 3[-(1/2)] - 4(√3/2)

    = 3/2 - 2√3