JEE Advanced 2023 Paper 2 Q07 Mathematics Differentiation & Applications Maxima & Minima Hard

JEE Advanced 2023 Paper 2 · Q07 · Maxima & Minima

Let $S$ be the set of all twice differentiable functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that $\dfrac{d^2 f}{dx^2}(x) > 0$ 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?

  1. A. There exists a function $f \in S$ such that $X_f = 0$
  2. B. For every function $f \in S$, we have $X_f \le 2$
  3. C. There exists a function $f \in S$ such that $X_f = 2$
  4. D. There does NOT exist any function $f$ in $S$ such that $X_f = 1$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

$f$ is strictly convex on $(-1,1)$. Define $g(x) = f(x) - x$, then $g''(x) = f''(x) > 0$, so $g$ is convex. A strictly convex function intersects any line at most twice. Hence $X_f \le 2$ — (B) TRUE. Constructions: (A) take $f(x) = x^2 + 2$ (no fixed point in $(-1,1)$) — TRUE. (C) take $f(x) = x^2$, then $f(x) = x \Rightarrow x^2 - x = 0 \Rightarrow x = 0, 1$, but $1 \notin (-1,1)$. Try $f(x) = 4x^2 - 1$: $f(x)=x \Rightarrow 4x^2 - x - 1 = 0$, roots $(1\pm\sqrt{17})/8$, both real and in $(-1,1)$ — so $X_f = 2$ exists — (C) TRUE. (D) Counterexample: $f(x) = (x-0.5)^2$, $f(x)=x \Rightarrow (x-0.5)^2=x \Rightarrow x^2 - 2x + 0.25 = 0$, discriminant $> 0$, two roots — actually for $X_f=1$ tangent case exists, e.g. $f(x) = (x+0.5)^2 - 0.25$ tangent to $y=x$ at one point. So $X_f = 1$ IS achievable — (D) FALSE.

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