JEE Mathematics 2023 Question with Solution

Given: $f(x) = \frac{x^3}{3} + 2bx + \frac{ax}{2}$ and $g(x) = \frac{x^3}{3} + ax + bx^2$ with $a \neq 2b$.

Find: The value of $a + 2b + 7$ when the two functions have a common extreme point.

For a common extreme point, the corresponding stationary point must be common. From the extracted solution:

$$f'(x) = x^2 + 2b + ax$$

and

$$g'(x) = x^2 + a + 2bx$$

Subtracting,

$$(2b-a) - x(2b-a) = 0$$

Since $a \neq 2b$, we get

$$x = 1$$

So $x=1$ is the common root.

Put $x=1$ in $f'(x)=0$ or $g'(x)=0$:

$$1 + 2b + a = 0$$

Therefore,

$$a + 2b + 7 = 6$$

Hence, the correct value is $6$.

Discrepancy note: The worked solution concludes the value is $6$, which corresponds to option D, but the solution labels the correct option as C. The numerical conclusion from the working is authoritative.

At a common extreme point, both derivatives must vanish at the same value of $x$.

Using the expressions shown in the solution:

$$f'(x) = x^2 + 2b + ax$$ $$g'(x) = x^2 + a + 2bx$$

Now subtract the second from the first:

$$f'(x) - g'(x) = (x^2 + 2b + ax) - (x^2 + a + 2bx)$$ $$= 2b - a + ax - 2bx$$ $$= (2b-a) - x(2b-a)$$

So,

$$(2b-a)(1-x) = 0$$

Given $a \neq 2b$, we must have

$$1-x = 0$$

Therefore,

$$x=1$$

Substituting into the stationary condition,

$$1 + 2b + a = 0$$

Thus,

$$a+2b = -1$$

and hence

$$a+2b+7 = 6$$

Therefore, the correct option should be D from the listed options.