Let f: R R be a function defined by f(x)= l t^ 3 -3 t ; x 2 t x x…

MCQ

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

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

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

A
$\left(3, \frac{27}{4}\right)$
B
$\left(3, \frac{23}{4}\right)$
✓
$\left(4, \frac{27}{4}\right)$
D
$\left(4, \frac{23}{4}\right)$

✓

Answer

Correct option: C.

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

c
$\left\{\begin{array}{l} f ( x )= x ^{3}-3 x , x \leq-1 \\ 2,-1< x <2 \\ x ^{2}+2 x -6,2< x <3 \\ 9,3 \leq x <4 \\ 10,4 \leq x <5 \\ 11, x =5 \\ 2 x +1, x >5\end{array}\right.$

Clearly $f ( x )$ is not differentiable at

$x =2,3,4,5 \Rightarrow m =4$

$I =\int\limits_{-2}^{-1}\left( x ^{3}-3 x \right) dx +\int\limits_{-1}^{2} 2 \cdot dx =\frac{27}{4}$

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$List-I$	$List-II$
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($Q$) The number of symmetric matrices $M=\left(a_{i j}\right) 3 \times 3$ with all entries in $T$ such that $C_j=0$ for all $j$ is	($2$) ($2$)
($R$) Let $M=\left(a_{i j}\right) 3 \times 3$ be a skew symmetric matrix such that $a_{i j} \in T$ for $i>j$. Then the number of elements in the set $\left\{\left(\begin{array}{l}x \\ y \\ z\end{array}\right): x, y \cdot z \in R, M\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}a_{12} \\ 0 \\ -a_{23}\end{array}\right)\right\}$ is is	($3$) Infinite
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	($5$) ($0$)