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Limits, Continuity and Differentiability: Mathematics | JEE Main

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

If a value satisfies both and , it must satisfy any linear combination . Subtracting eliminates the term, isolating directly.

If the functions

\mathrm{f}(\mathrm{x}) = \frac{\mathrm{x}^3}{3} + 2\mathrm{bx} + \frac{\mathrm{ax}^2}{2}

and

\mathrm{g}(\mathrm{x}) = \frac{\mathrm{x}^3}{3} + \mathrm{ax} + \mathrm{bx}^2

\mathrm{a} \neq 2\mathrm{b}

have a common extreme point, then

\mathrm{a} + 2\mathrm{b} + 7

is equal to:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{3}{2}

3

4

6

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

\frac{3}{2}

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

B ·

3

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

C ·

4

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

D ·

6

None. This is the correct option. Correct: equating the derivative conditions gives the shared critical point at $x = 1$ , which leads to $1 + a + 2b = 0$ , giving $a + 2b + 7 = 6$ .

Reviewed route

Solution

StepWorking

01given

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

02goal

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

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: $f'(x) = x^2 + ax + 2b = 0$ $g'(x) = x^2 + 2bx + a = 0$ Subtract the two equations: $(x^2 + ax + 2b) - (x^2 + 2bx + a) = 0 \implies (a - 2b)x - (a - 2b) = 0$ $(a - 2b)(x - 1) = 0$ Since $a \neq 2b$ , we have $a - 2b \neq 0$ , which uniquely gives the common root $x = 1$ .

05execute

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

✓verify

Substituting $x = 1$ into $g'(x) = 0$ gives $1 + 2b + a = 0$ , which is consistent. Thus $a + 2b + 7 = 6$ 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 $f'(x) = x^2 + ax + 2b$ and $g'(x) = x^2 + 2bx + a$ , what is the value of the common critical point $x_0$ given $a \neq 2b$ ?
$x_0 = 1$
Using $x_0 = 1$ in $f'(x_0) = 0$ , what is the value of $a + 2b + 7$ ?
$6$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $a + 2b = -1$ , making $a + 2b + 7 = 6$ .

Why each option works or fails

A: $\frac{3}{2}$ - Solving $x(a - 2b) - (a - 2b) = 0$ incorrectly as $x = -1$ , leading to an erroneous calculation of $a + 2b$ . Factor out $(a - 2b)$ carefully: $(x - 1)(a - 2b) = 0$ with $a \neq 2b$ gives $x = 1$ , not $x = -1$ .
B: $3$ - Assuming that finding $a + 2b = -1$ directly yields a final answer of $7 - 4 = 3$ by confusing the coefficients. Substitute the found value directly into the requested expression: $a + 2b + 7 = -1 + 7 = 6$ .
C: $4$ - Setting $f'(x) - g'(x) = 0$ while dropping the linear coefficient, yielding $a + 2b = -3$ instead of $-1$ . Ensure the derivatives $f'(x) = x^2 + ax + 2b$ and $g'(x) = x^2 + 2bx + a$ are subtracted term-by-term without sign errors: $(a - 2b)x - (a - 2b) = 0$ .
D · correct: $6$ - None. This is the correct option. Correct: equating the derivative conditions gives the shared critical point at $x = 1$ , which leads to $1 + a + 2b = 0$ , giving $a + 2b + 7 = 6$ .

Step-by-step solution

given: $f(x) = \frac{x^3}{3} + \frac{a x^2}{2} + 2bx$ and $g(x) = \frac{x^3}{3} + b x^2 + ax$ , with $a \neq 2b$ , having a common extreme point.
goal: Find the value of $a + 2b + 7$ .
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: $f'(x) = x^2 + ax + 2b = 0$ $g'(x) = x^2 + 2bx + a = 0$ Subtract the two equations: $(x^2 + ax + 2b) - (x^2 + 2bx + a) = 0 \implies (a - 2b)x - (a - 2b) = 0$ $(a - 2b)(x - 1) = 0$ Since $a \neq 2b$ , we have $a - 2b \neq 0$ , which uniquely gives the common root $x = 1$ .
execute: Substitute the common root $x = 1$ into $f'(x) = 0$ : $1^2 + a(1) + 2b = 0 \implies 1 + a + 2b = 0 \implies a + 2b = -1$ Now evaluate the requested expression: $f'(x) = 0$
verify: Substituting $x = 1$ into $f'(x) = x^2 + ax + 2b = 0$ gives $1 + 2b + a = 0$ , which is consistent. Thus $a + 2b + 7 = 6$ holds without contradiction.

Limits, Continuity and Differentiability: Mathematics | JEE Main

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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Limits, Continuity and Differentiability: Mathematics | JEE Main

Setting f′(x)=g′(x)=0f'(x) = g'(x) = 0f′(x)=g′(x)=0 yields x=1x = 1x=1, which requires a+2b=−1a + 2b = -1a+2b=−1, making a+2b+7=6a + 2b + 7 = 6a+2b+7=6.

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Students also ask

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

Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

Let $\text{f}, \text{g}$ and $\text{h}$ be the real valued functions defined on…

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