Practicing Success
Let $y=log(x+x\sqrt{x^2+1})$, and $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}=c$. Then identify the correct statements about the values of a, b and c: (A) a = 1 + x2 (B) b = 0 (C) c = 0 (D) b = x (E) c = 2 Choose the correct answer from the options given below: |
(A), (B), (E) only (A), (B), (C) only (A), (C), (D) only (A), (B), (D) only |
(A), (C), (D) only |
Differentiable w.r.t.x. $\frac{dy}{dx}=\frac{d[log(x+\sqrt{x^2+1})]}{d(x+\sqrt{x^2+1})}×\frac{d[x+\sqrt{x^2+1}]}{dx}$ $\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+1}}×[1+\frac{1}{2\sqrt{x^2+1}}×2x]$ $\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+1}}×\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}=\frac{dy}{dx}\frac{1}{\sqrt{x^2+1}}$ $\sqrt{x^2+1}\frac{dy}{dx}=1$ ....(ii) Differentiable eq.(ii) w.r.t.x $\sqrt{x^2+1}.\frac{d^2y}{dx^2}+\frac{dy}{dx}×\frac{d}{dx}\sqrt{x^2+1}=0$ $\frac{d}{dx}(u.v)=u.\frac{dv}{dx}+v.\frac{du}{dx}$ $\sqrt{x^2+1}.\frac{d^2y}{dx^2}+\frac{dy}{dx}×\frac{1}{2+\sqrt{x^2+1}}×2x=0$ $\sqrt{x^2+1}.\frac{d^2y}{dx^2}+\frac{x.\frac{dy}{dx}}{\sqrt{x^2+1}}=0$ $\frac{(x^2+1)\frac{d^2y}{dx^2}+x.\frac{dy}{dx}}{\sqrt{x^2+1}}=0$ $(x^2+1)\frac{d^2y}{dx^2}+x.\frac{dy}{dx}=0$ So value for a = x2 + 1, b = x, c = 0 |