JEE Advanced 2023 — Paper 2 — MATH-II solution

JEE Advanced2023Paper 2MATH-II

Let $S$ be the set of all twice differentiable functions $f$ from $R$ to $R$ such that $\frac{d ^{2} f}{d x ^{2}} (x) > 0$ for all $x \in (- 1, 1)$ . For $f \in S$ , let $X_{f}$ be the number of points $x \in (- 1, 1)$ for which $f (x) = x$ . Then which of the following statements is(are) true?

A
There exists a function $f \in S$ such that $X_{f} = 0$
B
For every function $f \in S$ , we have $X_{f} \leq 2$
C
There exists a function $f \in S$ such that $X_{f} = 2$
D
There does NOT exist any function $f$ in $S$ such that $X_{f} = 1$

Solution

Define $φ (x) = f (x) - x$ . Then $φ^{''} (x) = f^{''} (x) > 0$ on $(- 1, 1)$ , so $φ^{'}$ is strictly increasing and has at most one zero. Therefore $φ$ is convex on $(- 1, 1)$ and can vanish at most twice. This gives $X_{f} \leq 2$ , so option (B) is true.

For (A): take $f (x) = x^{2} + 1$ . Then $f^{''} (x) = 2 > 0$ and $f (x) - x = x^{2} - x + 1 > 0$ for all real $x$ (discriminant negative), so $X_{f} = 0$ .

For (C): take $f (x) = 2 x^{2} - \frac{1}{2}$ . Then $f^{''} (x) = 4 > 0$ and $f (x) - x = 2 x^{2} - x - 1/2 = 0$ gives $x = \frac{1 \pm 5}{4}$ , both lying in $(- 1, 1)$ . So $X_{f} = 2$ .

For (D): take $f (x) = x^{2}$ . Then $f^{''} (x) = 2 > 0$ and $f (x) = x$ has roots $x = 0$ and $x = 1$ , only $x = 0$ lies in $(- 1, 1)$ , giving $X_{f} = 1$ . Hence (D) is false.

$∴$ Correct options are (A), (B) and (C).

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