Question

Let \(f\) be a twice differentiable function on \(\mathbb{R}\) such that \(f(0)=4\) and the function, \(g(x)=\frac{1}{2}f^{2}(x)+3x^{3}-5\) is monotonically increasing in \(\mathbb{R}\). Then \(f(1)\) cannot be :

• A. \(2\sqrt{3}\)
• B. \(3\sqrt{2}\)
• C. 4
• D. \(2\sqrt{2}\)

Hint:

A monotonically increasing function satisfies \(g(1)\ge g(0)\); this gives a lower bound on \(f(1)^2\).

Answer 1

The Correct Option is D

\(g\) is monotonically increasing on \(\mathbb{R}\), so \(g(1)\ge g(0)\).
\(g(0)=\frac12\cdot16+0-5=3\).
\(g(1)=\frac12f(1)^2+3-5=\frac12f(1)^2-2\).
So \(\frac12f(1)^2-2\ge3\), giving \(f(1)^2\ge10\).
Squares of the options: \((2\sqrt3)^2=12\), \((3\sqrt2)^2=18\), \(4^2=16\), \((2\sqrt2)^2=8\).
Only \(8\) is less than \(10\), so \(f(1)\) cannot be \(2\sqrt2\).
Answer: (4) \(2\sqrt{2}\)