JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Application of Derivatives

(A) Since $f$ is twice differentiable and maps to $[-2,2]$, it cannot be constant everywhere (as $(f(0){)}^2+(f^{\prime }(0){)}^2=85$ requires $|f(0)|>2$ or $|f^{\prime }(0)|>0$, contradiction if constant). There must exist intervals where $f$ is one-one. TRUE.

(B) By MVT on $[-4,0]$: $f^{\prime }(x_0)=\frac{f(-4)-f(0)}{-4}$. Since $-2\le f(-4),f(0)\le 2$, we have $-4\le f(-4)-f(0)\le 4$, so $|f^{\prime }(x_0)|\le 1$. TRUE.

(C) A function like $f(x)=\sin (\sqrt{85}x)$ satisfies the initial condition but ${\lim }_{x\to \infty }f(x)$ does not exist. FALSE.

(D) Let $g(x)=(f(x){)}^2+(f^{\prime }(x){)}^2$. Then $g(0)=85$. For $x\in (-4,0)$ and $(0,4)$, by (B) and boundedness, $g(x)\le 5$. So $g$ has a maximum at some $\alpha$. Then $g^{\prime }(\alpha )=2f^{\prime }(\alpha )(f(\alpha )+f^{\prime \prime }(\alpha ))=0$. If $f^{\prime }(\alpha )=0$, then $(f(\alpha ){)}^2=85$, impossible since $|f(\alpha )|\le 2$. So $f(\alpha )+f^{\prime \prime }(\alpha )=0$ and $f^{\prime }(\alpha )\ne 0$. TRUE.