QJEEM22Mafd06: JEE Main 2022 Mathematics - Class 12 JEEM22Mafd06 Solution: Let f : R → R be a function defined by : f ( x ) = { max { t 3…

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QJEEM22Mafd06: JEE Main 2022 Mathematics - Class 12 JEEM22Mafd06 Solution: Let f : R → R be a function defined by : f ( x ) = { max { t 3…

Let $f : R \to R$ be a function defined by :

$f (x) = {\begin{matrix} max {t^{3} - 3 t} t \leq x & ; & x \leq 2 \\ x^{2} + 2 x - 6 & ; & 2 < x < 3 \\ [x - 3] + 9 & ; & 3 \leq x \leq 5 \\ 2 x + 1 & ; & x > 5 \end{matrix}$

where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $I = \int_{- 2}^{2} f (x) d x$ . Then the ordered pair (m, I) is equal to :

Multiple Choice Options

A 3,27/4
B 3,23/4
C 4,27/4
D 4,23/4

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Accepted Answer

{\begin{cases} f (x) = x^{3} - 3 x, x \leq - 1 \\ 2, - 1 < x < 2 \\ x^{2} + 2 x - 6, 2 < x < 3 \\ 9, 3 \leq x < 4 \\ 10, 4 \leq x < 5 \\ 11, x = 5 \\ 2 x + 1, x > 5 \end{cases}

Clearly

\mathrm{f}(\mathrm{x})

is not differentiable at

\begin{aligned} x = 2, 3, 4, 5 \Rightarrow m = 4 \\ I = \int_{- 2}^{- 1} (x^{3} - 3 x) d x + \int_{- 1}^{2} 2 \cdot d x = \frac{27}{4} \end{aligned}

QJEEM22Mafd06: JEE Main 2022 Mathematics - Class 12 JEEM22Mafd06 Solution: Let f : R → R be a function defined by : f ( x ) = { max { t 3…

Multiple Choice Options

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