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 $$x={\displaystyle \frac{-11\pi }{6}}$$ is equal to
A. $${\displaystyle \frac{-3}{2}}-2\sqrt{3}$$
B. $${\displaystyle \frac{3}{2}}-2\sqrt{3}$$
C. $${\displaystyle \frac{3}{2}}+2\sqrt{3}$$
D. $${\displaystyle \frac{-3}{2}}+2\sqrt{3}$$

Answer 1

Hint: Here we use the concept of odd functions which is $$f(-x)=-f(x)$$ and solve for the value at given x. We break the angle in such a way that it is added or subtracted from $$2\pi$$.

Complete step-by-step answer:
We have $$f(x)=3\sin x+4\cos x$$
Also, we know any function is an odd function if it satisfies $$f(-x)=-f(x)$$.
Now we have to find the value of the function at point $$x={\displaystyle \frac{-11\pi }{6}}$$
We have to find the value of $$f({\displaystyle \frac{-11\pi }{6}})$$
Since, f is an odd function, therefore, we can use the concept $$f(-x)=-f(x)$$. Substitute $$x={\displaystyle \frac{-11\pi }{6}}$$.
$$\Rightarrow f({\displaystyle \frac{-11\pi }{6}})=-f({\displaystyle \frac{11\pi }{6}})$$
Now we can break the angle inside the function as
$$\Rightarrow {\displaystyle \frac{11\pi }{6}}={\displaystyle \frac{12\pi -\pi }{6}}$$
Separating the fraction into two parts
$$\Rightarrow {\displaystyle \frac{11\pi }{6}}={\displaystyle \frac{12\pi }{6}}-{\displaystyle \frac{\pi }{6}}$$
Cancel out common factors from numerator and denominator
$$\Rightarrow {\displaystyle \frac{11\pi }{6}}=2\pi -{\displaystyle \frac{\pi }{6}}$$
Therefore, we can write
$$f({\displaystyle \frac{-11\pi }{6}})=-f(2\pi -{\displaystyle \frac{\pi }{6}})$$
Now we know $$f(x)=3\sin x+4\cos x$$
Put $$x=2\pi -{\displaystyle \frac{\pi }{6}}$$
$$\Rightarrow f({\displaystyle \frac{11\pi }{6}})=3\sin \left(2\pi -{\displaystyle \frac{\pi }{6}}\right)+4\cos \left(2\pi -{\displaystyle \frac{\pi }{6}}\right)$$ … (1)
Now we will use the quadrant graph to convert the angles.

Here we denote $$2\pi ={360}^{\circ },\pi ={180}^{\circ }$$
Now we calculate the values of both the functions on RHS of the equation using the quadrant diagram.
For $$\sin \left(2\pi -{\displaystyle \frac{\pi }{6}}\right)$$, we are subtracting from $$2\pi$$which goes into the fourth quadrant where sin function is negative.
So, the value of $$\sin \left(2\pi -{\displaystyle \frac{\pi }{6}}\right)=-\sin ({\displaystyle \frac{\pi }{6}})=-{\displaystyle \frac{1}{2}}$$ … (2)
For $$\cos \left(2\pi -{\displaystyle \frac{\pi }{6}}\right)$$, we are subtracting from $$2\pi$$which goes into the fourth quadrant where the cos function is positive.
So, the value of $$\cos \left(2\pi -{\displaystyle \frac{\pi }{6}}\right)=\cos ({\displaystyle \frac{\pi }{6}})={\displaystyle \frac{\sqrt{3}}{2}}$$ … (3)
Substitute the values from equation (2) and equation (3) in equation (1)
$$\Rightarrow f(x)=3({\displaystyle \frac{-1}{2}})+4({\displaystyle \frac{\sqrt{3}}{2}})$$
Multiply the terms in the bracket.
$$\begin{gathered} \Rightarrow f({\displaystyle \frac{11\pi }{6}})={\displaystyle \frac{-3}{2}}+{\displaystyle \frac{4\sqrt{3}}{2}} \\ \Rightarrow f({\displaystyle \frac{11\pi }{6}})={\displaystyle \frac{-3}{2}}+2\sqrt{3} \end{gathered}$$
So now $$f(-{\displaystyle \frac{11\pi }{6}})=-f({\displaystyle \frac{11\pi }{6}})$$
Therefore,
$$\begin{gathered} \Rightarrow f(-{\displaystyle \frac{11\pi }{6}})=-[{\displaystyle \frac{-3}{2}}+2\sqrt{3}] \\ \Rightarrow f(-{\displaystyle \frac{11\pi }{6}})={\displaystyle \frac{3}{2}}-2\sqrt{3} \end{gathered}$$
So, option B is correct.

Note: Students are likely to make mistakes while calculating the values from the quadrant diagram, keep in mind that we always move anti-clockwise as we add the angles, so when we subtract the angle we move backwards or clockwise to see which quadrant our function lies in.