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  • <img alt="\mathrm{\text { Let } f(x)=\left\{\begin{array}{cc} x^{3}-x^{2}+10 x-7, & x \leq 1 \\ -2 x+\log _{2}\left(\mathrm{~b}^{2}-4\right), & x>1. \end{array}\right.}" src="/latex-imag

\mathrm{\text { Let } f(x)=\left\{\begin{array}{cc} x^{3}-x^{2}+10 x-7, & x \leq 1 \\ -2 x+\log _{2}\left(\mathrm{~b}^{2}-4\right), & x>1. \end{array}\right.}

Then the set of all values of \mathrm{b}, for which \mathrm{f(x)} has maximum value at  \mathrm{x=1}, is :

Option: 1

(-6,-2)


Option: 2

(2,6)


Option: 3

\mathrm{[-6,-2) \cup(2,6]}


Option: 4

\mathrm{[-\sqrt{6},-2) \cup(2, \sqrt{6}]}


Answers (1)

best_answer

\mathrm{ f(1)=3}

\mathrm{for\; x<1, f^{\prime}(x)=3 x^{2}-2 x+10>0}\\

\mathrm{\Rightarrow f(x)\; is\; increasing}

\mathrm{For\; x>1, f^{\prime}(x)<0}

\Rightarrow Function is decreasing

\mathrm{\lim _{x \rightarrow 1^{+}} f(x)=-2+\log _{2}\left(b^{2}-4\right)} \\

\mathrm{\text { For maximum value at } x=1} \\

\mathrm{3 \geq-2+\log _{2}\left(b^{2}-4\right)} \\

\mathrm{32 \geq b^{2}-4>0} \\

\mathrm{b \in[-6,-2) \cup(2,6]}

Hence correct option is 3

 

Posted by

Ritika Jonwal

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