JEE Mathematics 2023 Question with Solution

Given: $f(x) = 2x^3 + (2p-7)x^2 + 3(2p-9)x - 6$

Find: All values of $p$ such that the function has a maximum at some $x<0$ and a minimum at some $x>0$.

For a cubic polynomial, extrema occur at the roots of $f'(x)$. So we first differentiate:

For the function to have both a maximum and a minimum, $f'(x)=0$ must have two distinct real roots. Also, since the maximum is at some $x<0$ and the minimum is at some $x>0$, the two critical points must lie on opposite sides of $0$.

Using the product of roots of the quadratic equation

we get

For one root to be negative and the other positive, their product must be negative. Hence,

so

The solution also checks the discriminant condition for two distinct real roots:

Combining with the sign condition, the solution concludes

Therefore, the interval obtained from the working is $\left(-\infty, \frac{9}{2}\right)$.

The solution explicitly states The Correct Option is D, and also says the correct answer is option (4), even though the interval found in the working matches option C in the given options list. the correct option is D.

Given: $f(x) = 2x^3 + (2p-7)x^2 + 3(2p-9)x - 6$

Find: The condition on $p$ for one critical point to lie at $x<0$ and the other at $x>0$.

A quicker observation from the provided solution is to evaluate

If the two turning points lie on opposite sides of $0$ for an upward-opening quadratic $f'(x)$, then $0$ must lie between its roots, so

Thus,

which gives

Hence the interval from the working is $\left(-\infty, \frac{9}{2}\right)$. The solution, however, marks the option as D despite this interval corresponding to option C in the listed choices.