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) ≥ 0, I. x (-∞,1) (1,8/5) (2,∞), B., f(x) < 0, II. x ≥ 1, C., f(x) is increasing if, III. x < 1, D., f(x) is decreasing if, IV. x (8/5,2), Choose the correct answer from the options given below: Option 1: A - I, B - IV, C - II, D - III Option 2: A - II, B - III, C - I, D - IV Option 3: A - II, B - III, C - IV, D - I Option 4: A - I, B - IV, C - III, D - II

Answer: A - II, B - III, C - I, D - IV. Solution: (x-2)^2≥0 always, so the sign of f follows (x-1)^3: f≥0 for x≥1 (A-II) and f 0, i.e. on (-∞,1)(1,85)(2,∞) (C-I), and negative on (85,2) (D-IV).

Calculus

Application of Derivatives

Hard

common

For the function $f(x) = (x-1)^3(x-2)^2$

Match the LIST-I with LIST-II

LIST-I LIST-II
A. $f(x) \geq 0$ I. $x \in (-\infty,1) \cup \left(1,\frac{8}{5}\right) \cup (2,\infty)$
B. $f(x) < 0$ II. $x \geq 1$
C. $f(x)$ is increasing if III. $x < 1$
D. $f(x)$ is decreasing if IV. $x \in \left(\frac{8}{5},2\right)$

Choose the correct answer from the options given below:

LIST-I		LIST-II
A.	$f(x) \geq 0$	I. $x \in (-\infty,1) \cup \left(1,\frac{8}{5}\right) \cup (2,\infty)$
B.	$f(x) < 0$	II. $x \geq 1$
C.	$f(x)$ is increasing if	III. $x < 1$
D.	$f(x)$ is decreasing if	IV. $x \in \left(\frac{8}{5},2\right)$

A - I, B - IV, C - II, D - III

A - II, B - III, C - I, D - IV

A - II, B - III, C - IV, D - I

A - I, B - IV, C - III, D - II

✅ Correct Option: 2

$(x-2)^2\ge0$ always, so the sign of $f$ follows $(x-1)^3$ : $f\ge0$ for $x\ge1$ (A-II) and $f<0$ for $x<1$ (B-III). Also $f'(x)=(x-1)^2(x-2)(5x-8)$ , positive when $(x-2)(5x-8)>0$ , i.e. on $(-\infty,1)\cup(1,\frac85)\cup(2,\infty)$ (C-I), and negative on $\left(\frac85,2\right)$ (D-IV).

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