JEE Main 2022 (Online) 25th June Evening ShiftQuestion 1
Let $f(x)=|(x-1)(x^{2}-2x-3)|+x-3$, $x\in\mathbf{R}$. If $m$ and $M$ are respectively the number of points of local minimum and local maximum of $f$ in the interval $(0,4)$, then $m+M$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$3$
Solution
Concept. $p(x)=(x-1)(x-3)(x+1)$ is $\ge0$ on $(0,1)$ and $(3,4)$, $\le0$ on $(1,3)$. Write $f$ on each piece and check the joins. Formulas used. (1) $p(x)=x^3-3x^2-x+3$Step 1 - Pieces. By (1): $$(0,1),\ (3,4):\ f=x^3-3x^2,\qquad(1,3):\ f=-x^3+3x^2+2x-6.$$ Step 2 - $(1,3)$. $f'=-3x^2+6x+2=0$ at $x=1+\frac{\sqrt{15}}{3}\approx2.29$, where $f'$ goes $+\to-$: a local maximum. Step 3 - Joins. At $x=1$: slope $3-6=-3$ on the left, $-3+6+2=5$ on the right: local minimum. At $x=3$: slope $-27+18+2=-7$ on the left, $27-18=9$ on the right: local minimum. Step 4 - Rest. On $(0,1)$, $f'=3x(x-2)<0$; on $(3,4)$, $f'>0$. No other extrema.Step 5 - Count. $m=2$, $M=1$, so $m+M=3$. Answer: 3