It is given that f(x) = x3 + 3qx2 + 2, where q is a real constant.
The equation f(x) = 0 has 3 distinct real roots.
Which of the following statements is/are necessarily true?
I The equation f(x) + 1 = 0 has 3 distinct real roots.
II The equation f(x+1) = 0 has 3 distinct real roots.
III The equation f(−x) − 1 = 0 has 3 distinct real roots.
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Show the answer and worked solution
answer · G
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Statement II is free: replacing x by x+1 slides the graph sideways, which moves the roots but never changes how many there are. For the other two, locate the turning points. f'(x) = 3x(x + 2q), so the stationary values are f(0) = 2 and f(−2q) = 4q3 + 2. Three distinct roots requires these to straddle zero, so 4q3 + 2 < 0, and then the local maximum is 2 and the local minimum is negative. Statement III asks for f = 1 three times, and 1 always lies strictly between a negative minimum and the maximum 2, so III holds. Statement I asks for f = −1 three times, which needs 4q3 + 2 < −1; taking q = −0.85 satisfies the original condition but gives a minimum of about −0.46, so I can fail.