Question

Mark the correct alternative of the following. At $x=\dfrac{5\pi }{6};f\left( x \right)=2\sin 3x+3\cos 3x$ is?
(a) 0
(b) Maximum
(c) Minimum
(d) None of these

Answer 1

Hint: A curve is given in question. First differentiate it then equate the differentiation of the curve to zero. By that you get few roots for differentiation. By basic differentiation properties those are the points where the curve attains maximum or minimum. To find which is maximum and which is minimum we must again differentiate the differentiated curve. By getting the second differential equation. Find their value at each root. If the second differential >0, then it is a point where we get minima or else it becomes maxima.

Complete step-by-step solution -
Given curve in question to which we have find the conditions is:
$y=2\sin 3x+3\cos 3x$
Given point in question for which we need explanation is:
$x=\dfrac{5\pi }{6}$
By plucking this x value into the expression, we get:
$y=2\sin \dfrac{15\pi }{6}+3\cos \dfrac{15\pi }{6}$
By simplifying the above equation of sin and cos, we get,
$\begin{aligned} & y=2.1+3.0 \\\ & y=2 \\\ \end{aligned}$
Now we find the differentiation of the curve y by:
By differentiation on both sides of the curve equation, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( 2\sin 3x+3\cos 3x \right)$
By general knowledge of differentiation, we can say:
$d\left( \sin ax \right)=\left( a \right)\left( \cos ax \right)\left( dy \right);d\left( \cos bx \right)=\left( b \right)\left( -\sin bx \right)\left( dx \right)$
By using above equation, we get:
$\dfrac{dy}{dx}=6\cos x-9\sin 3x$
By equating it to zero, we get:
$6\cos 3x-9\sin 3x=0$
By adding the term $9\sin 3x$on both sides of equation, we get
$2\cos 3x=3\sin 3x$
By dividing with cos3x term on both sides of equation, we get
$2=3\tan 3x$
By dividing with 3 on both sides of equation, we get:
$\tan 3x=\dfrac{2}{3}$
By applying ${{\tan }^{-1}}$ on both sides of equation we get:
$3x={{\tan }^{-1}}\left( \dfrac{2}{3} \right)\Rightarrow x-\dfrac{1}{3}{{\tan }^{-1}}\dfrac{2}{3}$
So, by comparing this x given in question, we get
$\dfrac{5\pi }{6}\ne \dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{2}{3} \right)$
So here in this question for a given value the differentiation of function is not zero. So we don’t need the value of second differentiation because it is not satisfying the first condition so we can conclude here itself.
So, the given x is neither maximum nor minimum and also the curve does not get value 0 at the given point.
Therefore, none of the options are true. So, we get none of these as answers.
Option (d) is the correct answer.

Note: While substituting value of $x=\dfrac{5\pi }{6}$, be careful to convert the value back to range $0 < x < \dfrac{\pi }{2}$ as the value of sine and cosine are found easily in this range. In this Problem there is no need to find second derivative as our critical point is $\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{2}{3} \right)$ where we get minima or maxima. so at the given point we will not get any minima and maxima.