JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 3)
Let $$f:R \to R$$ be a function defined by :
$$f(x) = \left\{ {\matrix{ {\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \cr {{x^2} + 2x - 6} & ; & {2 < x < 3} \cr {[x - 3] + 9} & ; & {3 \le x \le 5} \cr {2x + 1} & ; & {x > 5} \cr } } \right.$$
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 $$I = \int\limits_{ - 2}^2 {f(x)\,dx} $$. Then the ordered pair (m, I) is equal to :
$$\left( {3,\,{{27} \over 4}} \right)$$
$$\left( {3,\,{{23} \over 4}} \right)$$
$$\left( {4,\,{{27} \over 4}} \right)$$
$$\left( {4,\,{{23} \over 4}} \right)$$
Explanation
$$
\left\{\begin{array}{l}
f(x)=x^3-3 x, x \leq-1 \\\\
2,-1 < x < 2 \\\\
x^2+2 x-6,2 < x < 3 \\\\
9,3 \leq x < 4 \\\\
10,4 \leq x < 5 \\\\
11, x=5 \\\\
2 x+1, x>5
\end{array}\right.
$$
Clearly $\mathrm{f}(\mathrm{x})$ is not differentiable at
$$ \begin{aligned} & x=2,3,4,5 \Rightarrow m=4 \\\\ & I=\int_{-2}^{-1}\left(x^3-3 x\right) d x+\int_{-1}^2 2 \cdot d x=\frac{27}{4} \end{aligned} $$
Clearly $\mathrm{f}(\mathrm{x})$ is not differentiable at
$$ \begin{aligned} & x=2,3,4,5 \Rightarrow m=4 \\\\ & I=\int_{-2}^{-1}\left(x^3-3 x\right) d x+\int_{-1}^2 2 \cdot d x=\frac{27}{4} \end{aligned} $$
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