If \( f(x)=x^3+px^2+qx-3 \) has a local maxima at \( x=0 \) and local minima at \( x=1 \) , then the values of \( p \) and \( q \) respectively, are:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −3/2 and 0
At a local maximum or minimum of a differentiable function, f′(x) = 0.
\( \displaystyle f'(x)=3x^2+2px+q \)
\( \displaystyle f'(0)=0\ \Rightarrow\ q=0 \)
\( \displaystyle f'(1)=0\ \Rightarrow\ 3+2p=0 \)
\( \displaystyle p=-\frac{3}{2} \)
Check with the second derivative, \( f''(x)=6x+2p=6x-3 \):
\( \displaystyle f''(0)=-3\lt 0\ \text{(local maximum)} \)
\( \displaystyle f''(1)=3\gt 0\ \text{(local minimum)} \)
So p = −3/2 and q = 0.
Why the other options are incorrect
Option (1) 3 and −3 — Incorrect. With p = 3 and q = −3, f′(0) = −3 ≠ 0.
Option (3) −3 and 3 — Incorrect. With p = −3 and q = 3, f′(x) = 3(x − 1)2, which does not change sign, so there is no local maximum or minimum at all.
Option (4) 0 and −3/2 — Incorrect. These are the right values in the wrong order; the question asks for p and q respectively.