JEE Main 2022 — Application Of Derivatives Question with Solution

$$f(x)=\left|(x-1)(x+1)(x-3)\right|+(x-3)$$

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.

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.

$$f^{\prime }(3^{+})>0f^{\prime }(3^{-})<0\to$$ Minimum

$$f^{\prime }(1^{+})>0f^{\prime }(1^{-})<0\to$$ Minimum

$$x\in (1,3)f^{\prime }(x)=0$$ at one point $$\to$$ Maximum

$$x\in (3,4)f^{\prime }(x)\ne 0$$

$$x\in (0,1)f^{\prime }(x)\ne 0$$

So, 3 points.