Explanation

(b)
Given equation is $y = ax\cos\left( \frac{1}{x} + b\ \right)$ …..(i)
On differentiating Eq. (i), we get
$y_{1} = a\left\lbrack \cos\left( \frac{1}{x} + b\ \right)–x\sin{\left( \frac{1}{x} + b\ \right)\left( \ \frac{- 1}{x^{2}} \right)} \right\rbrack$
$\Rightarrow y_{1} = a\left\lbrack \cos\left( \frac{1}{x} + b\ \right) + \frac{1}{x}\sin\left( \frac{1}{x} + b\ \right) \right\rbrack$ …..(ii)
Again, on differentiating Eq. (ii), we get
$y_{2} = a\left\lbrack - \sin{\left( \frac{1}{x} + b\ \right)\left( \ - \frac{1}{x^{2}} \right) + \frac{1}{x}{\ cos}{\left( \frac{1}{x} + b\ \right)\left( \ - \frac{1}{x^{2}} \right) - \frac{1}{x^{2}}}}\sin\left( \frac{1}{x} + b\ \right) \right\rbrack$
$\Rightarrow y_{2} = \frac{- a}{x^{3}}\cos{\left( \frac{1}{x} + b\ \right) = \frac{- ax}{x^{4}}}\cos{\left( \frac{1}{x} + b\ \right) = \frac{- y}{x^{4}}}$
$\Rightarrow x^{4}y_{2} + y = 0$