1.If y=tan−1(1+a2x2−1ax)y = \tan^{-1} \left( \frac{\sqrt{1+a^2x^2-1}}{ax} \right)y=tan−1(ax1+a2x2−1), then (1+a2x2)y′′+2a2xy′(1 + a^2x^2)y'' + 2a^2xy'(1+a2x2)y′′+2a2xy′ is equal toa.−2a2-2a^2−2a2b.1(2x+1)2a2\frac{1}{(2x+1)^2} a^2(2x+1)21a2c.2a22a^22a2d.0Login to continueOnly logged in users canattempt or see the solution.
