Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

Let $f(x) = |(x - 1)({x^2} - 2x - 3)| + x - 3,\,x \in R$. 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 ____________.

Answer (integer) 3

Solution

<p>$f(x) = \left| {(x - 1)(x + 1)(x - 3)} \right| + (x - 3)$</p> <p>$$f(x) = \left\{ {\matrix{ {(x - 3)({x^2})} & {3 \le x \le 4} \cr {(x - 3)(2 - {x^2})} & {1 \le x < 3} \cr {(x - 3)({x^2})} & {0 < x < 1} \cr } } \right.$$</p> <p>$$f'(x) = \left\{ {\matrix{ {3{x^2} - 6x} & {3 < x < 4} \cr { - 3{x^2} + 6x + 2} & {1 < x < 3} \cr {3{x^2} - 6x} & {0 < x < 1} \cr } } \right.$$</p> <p>$f'({3^ + }) > 0\,\,\,f'({3^ - }) < 0 \to$ Minimum</p> <p>$f'({1^ + }) > 0\,\,\,f'({1^ - }) < 0 \to$ Minimum</p> <p>$x \in (1,3)\,\,f'(x) = 0$ at one point $\to$ Maximum</p> <p>$x \in (3,4)\,\,f'(x) \ne 0$</p> <p>$x \in (0,1)\,\,f'(x) \ne 0$</p> <p>So, 3 points.</p>

About this question

Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals

This question is part of PrepWiser's free JEE Main question bank. 99 more solved questions on Application of Derivatives are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →