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Let$f\left( x \right) = x\sin \pi x,{\text{ }}x > 0$ . Then for all natural numbers $n,{\text{ }}f'\left( x \right)$ vanishes at:
(A). A unique point in the interval $\left( {n,n + \dfrac{1}{2}} \right)$
(B). A unique point in the interval $\left( {n + \dfrac{1}{2},n + 1} \right)$
(C). A unique point in the interval$\left( {n,n + 1} \right)$
(D). Two points in the interval $\left( {n,n + 1} \right)$

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Hint- To solve this question first we have to find the differentiation of the given function, then equate this to zero. and we get the required answer.

Complete step-by-step solution -

Now given that,
$f\left( x \right) = x\sin \pi x,{\text{ }}x > 0$
And we have to find $f'\left( x \right)$
Now to differentiate it we will apply product rule.
According to product rule,
$\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Now,
 $
  \dfrac{{df\left( x \right)}}{{dx}} = f'\left( x \right) = \sin \left( {\pi x} \right) \cdot \dfrac{{dx}}{{dx}} + x\dfrac{{d\sin \left( {\pi x} \right)}}{{dx}} \\
  {\text{or }}f'\left( x \right) = \sin \left( {\pi x} \right) + x\dfrac{{d\sin \left( {\pi x} \right)}}{{d\left( {\pi x} \right)}} \times \dfrac{{d\left( {\pi x} \right)}}{{dx}}{\text{ }}\left( {{\text{By chain rule}}} \right){\text{ }} \\
  {\text{or }}f'\left( x \right) = \sin \left( {\pi x} \right) + x\cos \left( {\pi x} \right) \cdot \pi \\
  {\text{or }}f'\left( x \right) = \sin \left( {\pi x} \right) + \cos \pi x\left( {\pi x} \right) \\
 $
Now we have to find $f'\left( x \right)$ vanishes at, for that we will put the value of $f'\left( x \right) = 0$
$
  f'\left( x \right) = \sin \left( {\pi x} \right) + \cos \pi x\left( {\pi x} \right) = 0 \\
  {\text{or - }}\left[ {\cos \left( {\pi x} \right)} \right]\pi x = \sin \left( {\pi x} \right) \\
  {\text{or }}\dfrac{{\sin \left( {\pi x} \right)}}{{\cos \left( {\pi x} \right)}} = - \pi x \\
  {\text{or }}\tan \left( {\pi x} \right) = - \pi x \\
 $
Now let’s draw the graph of $\tan \left( {\pi x} \right) = - \pi x$


Now from the graph it is pretty much clear that $ - \pi x$line is cutting $\tan \pi x$ for every $n + \dfrac{1}{2}$ but $n$ belongs to natural number, so it would be cutting a unique point between $\left( {n + \dfrac{1}{2},n + 1} \right)$ because as we see that it first cuts between $\dfrac{1}{2}$ to $1$ then it again cuts somewhere at $\dfrac{3}{2}$ and so on. Therefore there is a gap of $1$
$
   \Rightarrow \pi x \in \left( {\dfrac{{2n + 1}}{2}\pi ,\left( {n + 1} \right)\pi } \right) \\
   \Rightarrow x \in \left( {n + \dfrac{1}{2},n + 1} \right){\text{ and also }}x \in \left( {n,n + 1} \right). \\
 $
Therefore, the correct option will be (B) and (C).
Note- These types of questions can be solved by using the concept of maxima and minima. In this question we simply find the value of $f'\left( x \right)$ and then we observe it in the graph and we get our answer.