Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
$f(x)$ is defined for $x \geq 0$ and has a continuous derivative. It satisfies $f(0) = 1, f'(0) = 0$ and $(1 + f(x))f''(x) = 1 + x$. The values $f(1)$ can't take is/are:
Step-by-Step Solution
Key Concept: Using positivity of $f$ and its derivatives, combined with integration of inequalities to bound $f(1)$.
Since $1 + x$ is never zero and $f(0) = 1$, we have $f(x) > 0$ for all $x$, making $f$ strictly increasing. This gives $1 + f(x) \geq 2$ for all $x$. From the differential equation $F''(x) = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x)$, dividing by $F(x) = f(x)g(x)$ yields $\frac{F''}{F} = \frac{f''}{f} + \frac{g''}{g}$. We have $f''(x) \leq \frac{1+x}{2}$, so integrating twice gives $f(x) \leq f(0) + \frac{x^2}{4} + \frac{x^4}{12}$, which means $f(1) \leq 1 + \frac{1}{4} + \frac{1}{12} = \frac{4}{3}$.
Correct Answer: 1,2,3,4