
Find the value of $x$ for which the function $y=a\left( 1-\cos x \right)$ attains maximum value.
(a) $\pi $
(b) $\dfrac{\pi }{2}$
(c) $-\dfrac{\pi }{2}$
(d) $-\dfrac{\pi }{6}$
Answer
611.4k+ views
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 second differential > 0, then it is a point where maximum or else it becomes minimum.
Complete step-by-step answer:
Given equation of curve to which maximum point to be found:
$y=a\left( 1-\cos x \right)$
Firstly, we have to differentiate with respect to $x$ both sides:
$\dfrac{dy}{dx}=\dfrac{d}{dx}a\left( 1-\cos x \right)$
By basic knowledge of differentiation, we know that this is true:
$\dfrac{d}{dx}kx=k$ (k = constant)
$\dfrac{d}{dx}\cos x=-\sin x$
By substituting these into the equation above, we get:
$\dfrac{dy}{dx}=a\sin x$
For getting maximum we must equate it to zero now.
By equating the differentiation to zero, we get that:
$a\sin x=0$
By solving the above equation, we get $x$ values to be:
$x=0,\pi $
These 2 can’t be the maximum point or minimum point of the curve.
To find which of these is maximum we must differentiate again:
$\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( a\sin x \right)$
By basic knowledge of differentiation, we know that this:
$\begin{align}
& \dfrac{d}{dx}\left( \sin x \right)=\cos x \\
& \Rightarrow y''\left( x \right)=a\cos x \\
\end{align}$
By substituting $0,\pi $ into this equation we get:
$\begin{align}
& x=0\Rightarrow y''\left( 0 \right)=a \\
& x=\pi \Rightarrow y''\left( \pi \right)=-a \\
\end{align}$
As the second differentiation is negative at $x=\pi $ it is the part of maximum for the curve given in the question.
Therefore $x=\pi $ is the maximum point for the curve.
Option (a) is the correct answer.
Note: Be careful while calculating the second differentiating equation as it defines the point being maximum or minimum.
Complete step-by-step answer:
Given equation of curve to which maximum point to be found:
$y=a\left( 1-\cos x \right)$
Firstly, we have to differentiate with respect to $x$ both sides:
$\dfrac{dy}{dx}=\dfrac{d}{dx}a\left( 1-\cos x \right)$
By basic knowledge of differentiation, we know that this is true:
$\dfrac{d}{dx}kx=k$ (k = constant)
$\dfrac{d}{dx}\cos x=-\sin x$
By substituting these into the equation above, we get:
$\dfrac{dy}{dx}=a\sin x$
For getting maximum we must equate it to zero now.
By equating the differentiation to zero, we get that:
$a\sin x=0$
By solving the above equation, we get $x$ values to be:
$x=0,\pi $
These 2 can’t be the maximum point or minimum point of the curve.
To find which of these is maximum we must differentiate again:
$\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( a\sin x \right)$
By basic knowledge of differentiation, we know that this:
$\begin{align}
& \dfrac{d}{dx}\left( \sin x \right)=\cos x \\
& \Rightarrow y''\left( x \right)=a\cos x \\
\end{align}$
By substituting $0,\pi $ into this equation we get:
$\begin{align}
& x=0\Rightarrow y''\left( 0 \right)=a \\
& x=\pi \Rightarrow y''\left( \pi \right)=-a \\
\end{align}$
As the second differentiation is negative at $x=\pi $ it is the part of maximum for the curve given in the question.
Therefore $x=\pi $ is the maximum point for the curve.
Option (a) is the correct answer.
Note: Be careful while calculating the second differentiating equation as it defines the point being maximum or minimum.
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