JEE Advanced 2023 — Paper 2 — Question 7

Step 1: Let $g(x)=f(x)-x$. Then $g''(x)=f''(x)>0$ on $(-1,1)$, so $g$ is strictly convex on $(-1,1)$. Step 2: A strictly convex function can have at most two zeros in an interval.

Hence $X_f \le 2$ for all $f \in S$. Statement B is true. Step 3: To show existence of $f$ with $X_f=0$, choose $g(x)=x^2+ \frac{1}{2}$.

Then $g''(x)=2>0$ and $g(x)>0$ for all $x$, so $f(x)=g(x)+x = x^2+x+\frac{1}{2}$ has no fixed points. Statement A is true. Step 4: To show existence of $f$ with $X_f=2$, choose $g(x)=x^2- \frac{1}{4}$.

Then $g''(x)=2>0$ and $g(x)=0$ at $x=\pm \frac{1}{2}$, both in $(-1,1)$.

So $f(x)=g(x)+x = x^2+x-\frac{1}{4}$ has exactly two fixed points. Statement C is true. Step 5: To show existence of $f$ with $X_f=1$, choose $g(x)=x^2$.

Then $g''(x)=2>0$ and $g(x)=0$ only at $x=0$. So $f(x)=g(x)+x = x^2+x$ has exactly one fixed point.

Thus statement D is false. Step 6: Therefore, statements A, B, C are true; D is false.