JEE Advanced 2023 · Paper 2 · Mathematics Q7

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MathematicsMultiple correct+4 / -2Question 7 of 51

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

There exists a function

f \in S

such that

X_f = 0

For every function

f \in S

, we have

X_f \le 2

There exists a function

f \in S

such that

X_f = 2

There does NOT exist any function

f

S

such that

X_f = 1

Answer and worked solution

Official answer: A, B, C

Let $g(x)=f(x)-x$ . Then $g''=f''>0$ on $(-1,1)$ , so $g$ is strictly convex there. $X_f$ is the number of zeros of $g$ in $(-1,1)$ .

(B) True. A strictly convex function has at most two zeros: by Rolle's theorem, three zeros would force $g''$ to vanish somewhere.

(A) True. $f(x)=x^2+1$ : $x^2-x+1$ has no real root, so $X_f=0$ .

(C) True. $f(x)=2x^2$ : $2x^2=x$ at $x=0$ and $x=\frac12$ , so $X_f=2$ .

(D) False. $f(x)=x^2$ : $x^2=x$ at $0$ and $1$ , and only $0$ lies in $(-1,1)$ , so $X_f=1$ .

Topics:Rolle's and mean value theorems Monotonicity

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JEE Advanced 2023 · Paper 2, Mathematics question 7