JEE MainMathematics

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Applying Rolle's Theorem to Symmetric Functions

Q: Why did we evaluate f'(x) at x = 1?

Because the function has symmetry about x=1, so 2-1 = 1, making g'(1) - g'(1) = 0 automatically vanish, providing a third zero of f'(x).

Let g: R → R be a non constant twice differentiable such that g'(1/2)=g'(3/2). If a real valued function f is defined as f(x)=1/2[g(x)+g(2-x)], then

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Af^(' ')(x)=0 for atleast two x in (0,2)Bf^(' ')(x)=0 for exactly one x in (0,1)Cf^(' ')(x)=0 for no x in (0,1)Df'(3/2)+f'(1/2)=1

Step-by-step solutionView

Correct answer

By differentiating $f (x)$ and using the given derivative condition along with Rolle's Theorem on two disjoint intervals inside $(0, 2)$ , we find that $f^{''} (x) = 0$ for at least two values of $x$ in $(0, 2)$ .

Option analysis

Why each option works or fails

A · f^(' ')(x)=0 for atleast two x in (0,2)

This is the correct statement. Differentiating $f (x)$ yields $f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]$ . Evaluating at $x = 1 / 2, 1, 3 / 2$ shows that $f^{'} (1 / 2) = f^{'} (1) = f^{'} (3 / 2) = 0$ . Applying Rolle's Theorem on $(1 / 2, 1)$ and $(1, 3 / 2)$ guarantees at least two roots for $f^{''} (x) = 0$ in $(0, 2)$ .

B · f^(' ')(x)=0 for exactly one x in (0,1)

Assuming that existence of at least one root by Rolle's Theorem guarantees uniqueness without checking higher-order constraints. Rolle's Theorem guarantees existence ('at least one'), not uniqueness ('exactly one'). Without information on higher derivatives, the number of roots in $(0, 1)$ could be more than one.

C · f^(' ')(x)=0 for no x in (0,1)

Overlooking that $f^{'} (1 / 2) = 0$ and $f^{'} (1) = 0$ , which immediately guarantees at least one root of $f^{''} (x) = 0$ in $(1 / 2, 1) \subset (0, 1)$ by Rolle's Theorem. Check the values of $f^{'} (x)$ at $x = 1 / 2$ and $x = 1$ ; since both equal zero, Rolle's Theorem directly implies $f^{''} (x) = 0$ must have at least one root in $(1 / 2, 1)$ .

D · f'(3/2)+f'(1/2)=1

Incorrectly applying the chain rule to $g (2 - x)$ , missing the negative sign and concluding the sum is non-zero. The derivative of $g (2 - x)$ with respect to $x$ is $- g^{'} (2 - x)$ . Evaluating gives $f^{'} (1 / 2) = 0$ and $f^{'} (3 / 2) = 0$ , so their sum is $0$ , not $1$ .

Reviewed route

Solution

StepWorking

01Given

$g : R \to R$ is twice differentiable with $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ , and $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ .

02Goal

Determine the roots and properties of $f^{'} (x)$ and $f^{''} (x)$ on the interval $(0, 2)$ .

03Approach

Differentiate $f (x)$ with respect to $x$ using the chain rule, evaluate $f^{'} (x)$ at $x = 1 / 2$ , $x = 1$ , and $x = 3 / 2$ , and then apply Rolle's Theorem to $f^{'} (x)$ on sub-intervals.

04Execute

Differentiating $f (x)$ gives: $f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]$ Evaluating at specific points: $f^{'} (\frac{1}{2}) = \frac{1}{2} [g^{'} (\frac{1}{2}) - g^{'} (\frac{3}{2})] = 0$ $f^{'} (1) = \frac{1}{2} [g^{'} (1) - g^{'} (1)] = 0$ $f^{'} (\frac{3}{2}) = \frac{1}{2} [g^{'} (\frac{3}{2}) - g^{'} (\frac{1}{2})] = 0$

05Execute

Since $f^{'} (x)$ is differentiable on $R$ : 1. On $[\frac{1}{2}, 1]$ , $f^{'} (\frac{1}{2}) = f^{'} (1) = 0$ , so by Rolle's Theorem, there exists $c_{1} \in (\frac{1}{2}, 1)$ such that $f^{''} (c_{1}) = 0$ . 2. On $[1, \frac{3}{2}]$ , $f^{'} (1) = f^{'} (\frac{3}{2}) = 0$ , so by Rolle's Theorem, there exists $c_{2} \in (1, \frac{3}{2})$ such that $f^{''} (c_{2}) = 0$ . Thus, $f^{''} (x) = 0$ for at least two values of $x \in (\frac{1}{2}, \frac{3}{2}) \subset (0, 2)$ .

✓Verify

Check options: Option A states $f^{''} (x) = 0$ for at least two $x \in (0, 2)$ , which matches our findings in $(\frac{1}{2}, 1)$ and $(1, \frac{3}{2})$ . Option D gives $f^{'} (3 / 2) + f^{'} (1 / 2) = 0 + 0 = 0 \neq 1$ .

Hints that build this answer step by step

What is the correct expression for the first derivative $f^{'} (x)$ ?
$f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]$
Evaluating $f^{'} (x)$ at $x = \frac{1}{2}$ , $x = 1$ , and $x = \frac{3}{2}$ gives which values?
$f^{'} (\frac{1}{2}) = 0$ , $f^{'} (1) = 0$ , and $f^{'} (\frac{3}{2}) = 0$
Applying Rolle's Theorem to $f^{'} (x)$ on the intervals $[\frac{1}{2}, 1]$ and $[1, \frac{3}{2}]$ , what can we conclude about $f^{''} (x) = 0$ ?
$f^{''} (x) = 0$ for at least one $x \in (\frac{1}{2}, 1)$ and at least one $x \in (1, \frac{3}{2})$ , so at least two roots in $(0, 2)$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why did we evaluate

f^{'} (x)

x = 1

Because the function has symmetry about $x = 1$ , so $2 - 1 = 1$ , making $g^{'} (1) - g^{'} (1) = 0$ automatically vanish, providing a third zero of $f^{'} (x)$ .

Answer

By differentiating $f (x)$ and using the given derivative condition along with Rolle's Theorem on two disjoint intervals inside $(0, 2)$ , we find that $f^{''} (x) = 0$ for at least two values of $x$ in $(0, 2)$ .

Why each option works or fails

A · correct: f^(' ')(x)=0 for atleast two x in (0,2) - This is the correct statement. Differentiating $f (x)$ yields $f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]$ . Evaluating at $x = 1 / 2, 1, 3 / 2$ shows that $f^{'} (1 / 2) = f^{'} (1) = f^{'} (3 / 2) = 0$ . Applying Rolle's Theorem on $(1 / 2, 1)$ and $(1, 3 / 2)$ guarantees at least two roots for $f^{''} (x) = 0$ in $(0, 2)$ .
B: f^(' ')(x)=0 for exactly one x in (0,1) - Assuming that existence of at least one root by Rolle's Theorem guarantees uniqueness without checking higher-order constraints. Rolle's Theorem guarantees existence ('at least one'), not uniqueness ('exactly one'). Without information on higher derivatives, the number of roots in $(0, 1)$ could be more than one.
C: f^(' ')(x)=0 for no x in (0,1) - Overlooking that $f^{'} (1 / 2) = 0$ and $f^{'} (1) = 0$ , which immediately guarantees at least one root of $f^{''} (x) = 0$ in $(1 / 2, 1) \subset (0, 1)$ by Rolle's Theorem. Check the values of $f^{'} (x)$ at $x = 1 / 2$ and $x = 1$ ; since both equal zero, Rolle's Theorem directly implies $f^{''} (x) = 0$ must have at least one root in $(1 / 2, 1)$ .
D: f'(3/2)+f'(1/2)=1 - Incorrectly applying the chain rule to $g (2 - x)$ , missing the negative sign and concluding the sum is non-zero. The derivative of $g (2 - x)$ with respect to $x$ is $- g^{'} (2 - x)$ . Evaluating gives $f^{'} (1 / 2) = 0$ and $f^{'} (3 / 2) = 0$ , so their sum is $0$ , not $1$ .

Step-by-step solution

Given: $g : R \to R$ is twice differentiable with $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ , and $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ .
Goal: Determine the roots and properties of $f^{'} (x)$ and $f^{''} (x)$ on the interval $(0, 2)$ .
Approach: Differentiate $f (x)$ with respect to $x$ using the chain rule, evaluate $f^{'} (x)$ at $x = 1 / 2$ , $x = 1$ , and $x = 3 / 2$ , and then apply Rolle's Theorem to $f^{'} (x)$ on sub-intervals.
Execute: Differentiating $f (x)$ gives: $f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]$ Evaluating at specific points: $f^{'} (\frac{1}{2}) = \frac{1}{2} [g^{'} (\frac{1}{2}) - g^{'} (\frac{3}{2})] = 0$ $f^{'} (1) = \frac{1}{2} [g^{'} (1) - g^{'} (1)] = 0$ $f^{'} (\frac{3}{2}) = \frac{1}{2} [g^{'} (\frac{3}{2}) - g^{'} (\frac{1}{2})] = 0$
Execute: Since $f^{'} (x)$ is differentiable on $R$ : 1. On $[\frac{1}{2}, 1]$ , $f^{'} (\frac{1}{2}) = f^{'} (1) = 0$ , so by Rolle's Theorem, there exists $c_{1} \in (\frac{1}{2}, 1)$ such that $f^{''} (c_{1}) = 0$ . 2. On $[1, \frac{3}{2}]$ , $f^{'} (1) = f^{'} (\frac{3}{2}) = 0$ , so by Rolle's Theorem, there exists $c_{2} \in (1, \frac{3}{2})$ such that $f^{''} (c_{2}) = 0$ . Thus, $f^{''} (x) = 0$ for at least two values of $x \in (\frac{1}{2}, \frac{3}{2}) \subset (0, 2)$ .
Verify: Check options: Option A states $f^{''} (x) = 0$ for at least two $x \in (0, 2)$ , which matches our findings in $(\frac{1}{2}, 1)$ and $(1, \frac{3}{2})$ . Option D gives $f^{'} (3 / 2) + f^{'} (1 / 2) = 0 + 0 = 0 \neq 1$ .

Applying Rolle's Theorem to Symmetric Functions

What feels right?

By differentiating $f (x)$ and using the given derivative condition along with Rolle's Theorem on two disjoint intervals inside $(0, 2)$ , we find that $f^{''} (x) = 0$ for at least two values of $x$ in $(0, 2)$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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