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  • Let  <img alt="\mathrm{f(x)=\left|(x-1)\left(x^{2}-2 x-3\right)\right|+x-3, x \in \mathbb{R}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Cleft%7C%28x-1%29%5

Let  \mathrm{f(x)=\left|(x-1)\left(x^{2}-2 x-3\right)\right|+x-3, x \in \mathbb{R}}. If \mathrm{m} and \mathrm{M} are respectively the number of points of local minimum and local maximum of \mathrm{f} in the interval \mathrm{\left ( 0,4 \right )}, then \mathrm{m+M} is equal to ________.

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\mathrm{f(x)=\left\{\begin{array}{l} \left(x^{2}-1\right)(x-3)+(x-3), x \in(0,1] \cup[3,4) \\ -\left(x^{2}-1\right)(x-3)+(x-3), x \in[1,3] \end{array}\right.}

\mathrm{f^{\prime}(x)=\left\{\begin{array}{l} 3 x^{2}-6 x, x \in(0,1) \cup(3,4) \\ -3 x^{2}+6 x+2 ; x \in(1,3) \end{array}\right.}

\mathrm{f(x)} is non-derivable at \mathrm{x=1\: and \: x=3}

Also   \mathrm{f'(x)=0,at \: x=1+\sqrt{\frac{5}{3}}}

\mathrm{\Rightarrow m+M=3 }

Hence the correct answer is \mathrm{3 }

Posted by

Ramraj Saini

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