JEE Advanced 2018 · Paper 1, Mathematics question 4

Since $|f(0)|\le2$, we get $f'(0)^2\ge85-4=81$, so $|f'(0)|\ge9$.

• (A) $f'$ is continuous and $f'(0)\ne0$, so $f'$ keeps one sign on some interval around 0. There $f$ is strictly monotonic, hence one-one. True.
• (B) By the MVT on $[-4,0]$ there is $x_0$ with
  $$|f'(x_0)|=\frac{|f(0)-f(-4)|}4\le\frac44=1.$$
  True.
• (C) Take $f(x)=2\sin\left(\dfrac{\sqrt{85}}2x\right)$. It satisfies the conditions, but $\lim_{x\to\infty}f(x)$ does not exist. False.
• (D) Let $g=f^2+f'^2$, with $g(0)=85$. By (B) there is $x_1\in(-4,0)$ with $g(x_1)\le4+1=5$. Similarly there is $x_2\in(0,4)$ with $g(x_2)\le5$. So $g$ has a maximum at some interior $\alpha\in(x_1,x_2)$, with $g(\alpha)\ge85$, and
  $$g'(\alpha)=2f'(\alpha)\left(f(\alpha)+f''(\alpha)\right)=0.$$
  Since $f(\alpha)^2\le4$, $f'(\alpha)^2\ge81$, so $f'(\alpha)\ne0$. Hence $f(\alpha)+f''(\alpha)=0$. True.

(A), (B), (D)