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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Mathematics
Chapter
Differentiation

JEE Advanced 2013 · Paper 1 · Mathematics · Differentiation

Let f (x) = x π x , x > 0 . Then for all natural numbers n , f ^ ′...

Let f \left(\text{x}\right) = \text{x} \text\sin \pi \text{x}, x>0\text{x} > 0. Then for all natural numbers n,f′(x)\text{n} \text{,} {f }^{′} \left(\text{x}\right) vanishes atLet f (x) = x π x , x > 0 . Then for all natural numbers n , f ^ ′ (x) vanishes at

Options

  1. AA unique point in the interval (n,n+12)\left(\text{n} \text{,} \text{n} + \dfrac{1}{2}\right)
  2. BA unique point in the interval (n+12,n+1)\left(\text{n} + \dfrac{1}{2} \text{,} \text{n} + 1\right)
  3. CA unique point in the interval (n, n + 1)
  4. DTwo points in the interval (n, n + 1)

Correct answer

B. A unique point in the interval (n+12,n+1)\left(\text{n} + \dfrac{1}{2} \text{,} \text{n} + 1\right)

Step-by-step solution

f(x)=x sinπx,x>0\text{f} \left(\text{x}\right) = \text{x sin} \pi \text{x} , \text{x} > 0

As f′(x)f ' \left(x\right) vanishes,
{\text{f}}^{′} \left(\text{x}\right) = \text\sin \pi \text{x} + \pi x cos \pi \text{x} = 0
\text\tan \pi \text{x} = - \pi \text{x}

We will draw the graph of tan \pix & - \pi x.

And their intersection points will be the points, where f′(x)f ' \left(x\right) vanishes.
Let f (x) = x π x , x > 0 . Then for all natural numbers n , f ^ ′ (x) vanishes at

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