For the function \( f(x)=(x-1)^3(x-2)^2 \)
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( f(x)\ge 0 \) | I. | \( x\in(-\infty,1)\cup\left(1,\displaystyle\frac{8}{5}\right)\cup(2,\infty) \) |
| B. | \( f(x)\lt 0 \) | II. | \( x\ge 1 \) |
| C. | \( f(x) \) is increasing if | III. | \( x\lt 1 \) |
| D. | \( f(x) \) is decreasing if | IV. | \( x\in\left(\displaystyle\frac{8}{5},2\right) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A - II, B - III, C - I, D - IV
\( (x-2)^2\ge 0 \) always, so the sign of f(x) is decided by \( (x-1)^3 \), which has the sign of x − 1.
For C and D, the product rule gives f'(x); taking out the common factors:
\( \displaystyle f'(x)=3(x-1)^2(x-2)^2+2(x-1)^3(x-2) \)
\( \displaystyle =(x-1)^2(x-2)\left[3(x-2)+2(x-1)\right] \)
\( \displaystyle =(x-1)^2(x-2)(5x-8) \)
\( (x-1)^2\ge 0 \), so the sign of f'(x) is the sign of \( (x-2)(5x-8) \), which is positive outside \( \left[\frac{8}{5},2\right] \) and negative inside it.
(A) \( f(x)\ge 0 \)
This matches (II) x ≥ 1.
\( (x-1)^3\ge 0 \) exactly when x ≥ 1 (at x = 2, f is 0).
(B) \( f(x)\lt 0 \)
This matches (III) x < 1.
\( (x-1)^3\lt 0 \) when x < 1, and then \( (x-2)^2\gt 0 \).
(C) \( f(x) \) is increasing if
This matches (I) \( x\in(-\infty,1)\cup\left(1,\frac{8}{5}\right)\cup(2,\infty) \).
f'(x) > 0 for x < 8/5 (except x = 1, where f'(1) = 0) and for x > 2.
(D) \( f(x) \) is decreasing if
This matches (IV) \( x\in\left(\frac{8}{5},2\right) \).
f'(x) < 0 when \( \frac{8}{5}\lt x\lt 2 \).
So the matching is A - II, B - III, C - I, D - IV.