JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding Unknown Coefficients from a Shared Extreme Point

Q: Why are we allowed to just subtract the two equations to find the common root?

If a value x = x_0 satisfies both P(x) = 0 and Q(x) = 0, it must satisfy any linear combination P(x) - Q(x) = 0. Subtracting eliminates the x^2 term, isolating x_0 directly.

If the functions f(x) = (x^3)/3 + 2bx + (ax^2)/2 and g(x) = (x^3)/3 + ax + bx^2, a ≠ 2b have a common extreme point, then a + 2b + 7 is equal to:

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What feels right?

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A3/2B3C4D6

Step-by-step solutionView

Correct answer

Setting $f^{'} (x) = g^{'} (x) = 0$ yields $x = 1$ , which requires $a + 2 b = - 1$ , making $a + 2 b + 7 = 6$ .

Option analysis

Why each option works or fails

A · 3/2

Solving $x (a - 2 b) - (a - 2 b) = 0$ incorrectly as $x = - 1$ , leading to an erroneous calculation of $a + 2 b$ . Factor out $(a - 2 b)$ carefully: $(x - 1) (a - 2 b) = 0$ with $a \neq 2 b$ gives $x = 1$ , not $x = - 1$ .

B · 3

Assuming that finding $a + 2 b = - 1$ directly yields a final answer of $7 - 4 = 3$ by confusing the coefficients. Substitute the found value directly into the requested expression: $a + 2 b + 7 = - 1 + 7 = 6$ .

C · 4

Setting $f^{'} (x) - g^{'} (x) = 0$ while dropping the linear coefficient, yielding $a + 2 b = - 3$ instead of $- 1$ . Ensure the derivatives $f^{'} (x) = x^{2} + a x + 2 b$ and $g^{'} (x) = x^{2} + 2 b x + a$ are subtracted term-by-term without sign errors: $(a - 2 b) x - (a - 2 b) = 0$ .

D · 6

This is the correct option. Correct: equating the derivative conditions gives the shared critical point at $x = 1$ , which leads to $1 + a + 2 b = 0$ , giving $a + 2 b + 7 = 6$ .

Reviewed route

Solution

StepWorking

01Given

$f (x) = \frac{x^{3}}{3} + \frac{a x^{2}}{2} + 2 b x$ and $g (x) = \frac{x^{3}}{3} + b x^{2} + a x$ , with $a \neq 2 b$ , having a common extreme point.

02Goal

Find the value of $a + 2 b + 7$ .

03Approach

For a polynomial function, an extreme point occurs where its derivative is zero. Since $f (x)$ and $g (x)$ share a common extreme point, the quadratic equations $f^{'} (x) = 0$ and $g^{'} (x) = 0$ share a common root. Subtract the two equations to locate this common root and substitute back.

04Execute

Differentiate both functions: $f^{'} (x) = x^{2} + a x + 2 b = 0$ $g^{'} (x) = x^{2} + 2 b x + a = 0$ Subtract the two equations: $(x^{2} + a x + 2 b) - (x^{2} + 2 b x + a) = 0 ⟹ (a - 2 b) x - (a - 2 b) = 0$ $(a - 2 b) (x - 1) = 0$ Since $a \neq 2 b$ , we have $a - 2 b \neq 0$ , which uniquely gives the common root $x = 1$ .

05Execute

Substitute the common root $x = 1$ into $f^{'} (x) = 0$ : $1^{2} + a (1) + 2 b = 0 ⟹ 1 + a + 2 b = 0 ⟹ a + 2 b = - 1$ Now evaluate the requested expression: $a + 2 b + 7 = - 1 + 7 = 6$

✓Verify

Substituting $x = 1$ into $g^{'} (x) = 0$ gives $1 + 2 b + a = 0$ , which is consistent. Thus $a + 2 b + 7 = 6$ holds without contradiction.

Hints that build this answer step by step

What condition must a common extreme point $x_{0}$ satisfy for both $f (x)$ and $g (x)$ ?
$f^{'} (x_{0}) = 0$ and $g^{'} (x_{0}) = 0$
Computing $f^{'} (x) = x^{2} + a x + 2 b$ and $g^{'} (x) = x^{2} + 2 b x + a$ , what is the value of the common critical point $x_{0}$ given $a \neq 2 b$ ?
$x_{0} = 1$
Using $x_{0} = 1$ in $f^{'} (x_{0}) = 0$ , what is the value of $a + 2 b + 7$ ?
$6$

✓ 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: 9 September 2026

Quick checks

Students also ask

Why are we allowed to just subtract the two equations to find the common root?

If a value $x = x_{0}$ satisfies both $P (x) = 0$ and $Q (x) = 0$ , it must satisfy any linear combination $P (x) - Q (x) = 0$ . Subtracting eliminates the $x^{2}$ term, isolating $x_{0}$ directly.

Answer

Setting $f^{'} (x) = g^{'} (x) = 0$ yields $x = 1$ , which requires $a + 2 b = - 1$ , making $a + 2 b + 7 = 6$ .

Why each option works or fails

A: 3/2 - Solving $x (a - 2 b) - (a - 2 b) = 0$ incorrectly as $x = - 1$ , leading to an erroneous calculation of $a + 2 b$ . Factor out $(a - 2 b)$ carefully: $(x - 1) (a - 2 b) = 0$ with $a \neq 2 b$ gives $x = 1$ , not $x = - 1$ .
B: 3 - Assuming that finding $a + 2 b = - 1$ directly yields a final answer of $7 - 4 = 3$ by confusing the coefficients. Substitute the found value directly into the requested expression: $a + 2 b + 7 = - 1 + 7 = 6$ .
C: 4 - Setting $f^{'} (x) - g^{'} (x) = 0$ while dropping the linear coefficient, yielding $a + 2 b = - 3$ instead of $- 1$ . Ensure the derivatives $f^{'} (x) = x^{2} + a x + 2 b$ and $g^{'} (x) = x^{2} + 2 b x + a$ are subtracted term-by-term without sign errors: $(a - 2 b) x - (a - 2 b) = 0$ .
D · correct: 6 - This is the correct option. Correct: equating the derivative conditions gives the shared critical point at $x = 1$ , which leads to $1 + a + 2 b = 0$ , giving $a + 2 b + 7 = 6$ .

Step-by-step solution

Given: $f (x) = \frac{x^{3}}{3} + \frac{a x^{2}}{2} + 2 b x$ and $g (x) = \frac{x^{3}}{3} + b x^{2} + a x$ , with $a \neq 2 b$ , having a common extreme point.
Goal: Find the value of $a + 2 b + 7$ .
Approach: For a polynomial function, an extreme point occurs where its derivative is zero. Since $f (x)$ and $g (x)$ share a common extreme point, the quadratic equations $f^{'} (x) = 0$ and $g^{'} (x) = 0$ share a common root. Subtract the two equations to locate this common root and substitute back.
Execute: Differentiate both functions: $f^{'} (x) = x^{2} + a x + 2 b = 0$ $g^{'} (x) = x^{2} + 2 b x + a = 0$ Subtract the two equations: $(x^{2} + a x + 2 b) - (x^{2} + 2 b x + a) = 0 ⟹ (a - 2 b) x - (a - 2 b) = 0$ $(a - 2 b) (x - 1) = 0$ Since $a \neq 2 b$ , we have $a - 2 b \neq 0$ , which uniquely gives the common root $f^{'} (x) = x^{2} + a x + 2 b = 0$ .
Execute: Substitute the common root $x = 1$ into $f^{'} (x) = 0$ : $g^{'} (x) = x^{2} + 2 b x + a = 0$ Now evaluate the requested expression: $a + 2 b + 7 = - 1 + 7 = 6$
Verify: Substituting $x = 1$ into $g^{'} (x) = 0$ gives $1 + 2 b + a = 0$ , which is consistent. Thus $a + 2 b + 7 = 6$ holds without contradiction.

Finding Unknown Coefficients from a Shared Extreme Point

What feels right?

Setting $f^{'} (x) = g^{'} (x) = 0$ yields $x = 1$ , which requires $a + 2 b = - 1$ , making $a + 2 b + 7 = 6$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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Finding Unknown Coefficients from a Shared Extreme Point

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