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Let $\mathrm{f: \mathbf{R} \rightarrow \mathbf{R}}$ be a function defined by :

$f(x)=\left\{\begin{array}{ccc} \max_{\,\,t \leq x} \left\{\mathrm{t}^{3}-3 \mathrm{t}\right\} & ; & x \leq 2 \\ x^{2}+2 x-6 & ; & 2<x<3 \\ {[x-3]+9} & ; & 3 \leq x \leq 5 \\ 2 x+1 & ; & x>5 \end{array}\right.$
where $[\mathrm{t}]$ is the greatest integer less than or equal to $\mathrm{t}.$ Let $\mathrm{m}$ be the number of points where $\mathrm{f}$ is not differentiable and $\mathrm{I}=\int_{-2}^{2} f(x) \mathrm{d} x$ . Then the ordered pair $(\mathrm{m}, \mathrm{I})$ is equal to :
Option: 1
$\left ( 3,\frac{27}{4} \right )$

Option: 2
$\left ( 3,\frac{23}{4} \right )$

Option: 3
$\left ( 4,\frac{27}{4} \right )$

Option: 4
$\left ( 4,\frac{23}{4} \right )$

Answers (1)

best_answer

For $\mathrm{x\leq 2}$

$\mathrm{\text { Let } g(t)=t^{3}-3 t} \\$

$\mathrm{g^{\prime}(t)=3 t^{2}-3=3\left(t^{2}-1\right)=3(t+1)(t-1)}$

$\mathrm{g(-1)=-1+3=2} \\$

$\mathrm{g(0)=0} \\$

$\mathrm{g(1)=1-3=-2} \\$

$\mathrm{g(2)=8-6=2 }$

$\mathrm{\therefore \max_{t \leqslant x} \left\{t^{3}-3 t\right\} \text { is }}$

$\therefore$ Differentiable in this interval

For $\mathrm{ 2<x<3} \\$

$\mathrm{y=x^{2}+2 x-6} \\$

$\mathrm{A t \: x=2^{+} ; y=4+4-6=2} \\$

$\mathrm{f^{\prime}(x)=2 x+2 }\\$

$\mathrm{f^{\prime}\left(2^{+}\right)=4+2=6} \\$

$\mathrm{\therefore f^{\prime}\left(2^{-}\right) \neq f^{\prime}\left(2^{+}\right)} \\$

$\mathrm{\therefore \text {Non-differentiable at } x=2} \\$

$\mathrm{f\left(3^{-}\right)=9+6-6=9} \\$

$\mathrm{f^{\prime}\left(3^{-}\right)=6+2=8}$

For $\mathrm{3 \leqslant x \leqslant 5} \\$

$\mathrm{y=[x-3]+9=[x]-3+9=[x]+6} \\$

$\mathrm{\text { For } 3 \leq x<4, y=3+6=9}\\$

$\mathrm{f^{\prime}\left(3^{+}\right)=0 \neq f^{\prime}\left(3^{-}\right)$ $\Rightarrow \mathrm{\text{non-differentiable at x=3}}$

For $4 \leq x<5, y = 4+6=10$

Discontinuous at $\mathrm{x=4\Rightarrow non-differentiable\: at\: x=4 }$

$\mathrm{f^{\prime}\left(5^{-1}\right)=0 }\\$

$\mathrm{\& f\left(5^{-}\right)=10 }$

For $\mathrm{x> 5 }$

$\mathrm{f\left(5^{+}\right)=10+1=10}$

$\therefore$ Discontinuous and thus non-differentiable at $\mathrm{x=5}$

$\mathrm{\therefore m =4 \quad(x=2,3,4,5)} \\$

$\mathrm{I =\int_{-2}^{2} f(x) d x} \\$

$\mathrm{=\int_{-2}^{-1}\left(x^{3}-3 x\right) d x+\int_{-1}^{2} 2 d x} \\$

$\mathrm{=\left.\frac{x^{4}}{4}\right|_{-2} ^{-1}-\left.\frac{3}{2} x^{2}\right|_{-2} ^{-1}+\left.2 x\right|_{-1} ^{2} }$

$\mathrm{=\frac{1}{4}-4-\frac{3}{2}(1-4)+2(2+1)} \\$

$\mathrm{=\frac{1}{4}-4+\frac{9}{2}+6} \\$

$\mathrm{=\frac{1-16+18+24}{4}} \\$

$\mathrm{=\frac{27}{4}}$

Hence answer is option 3

Posted by

shivangi.shekhar

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