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  • For the function, <img alt="\mathrm{f(x)=4 \log _{\mathrm{e}}(x-1)-2 x^{2}+4 x+5, x>1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D4%20%5Clog%20_%7B%5Cmathr

For the function, \mathrm{f(x)=4 \log _{\mathrm{e}}(x-1)-2 x^{2}+4 x+5, x>1}, which one of the following is NOT correct?

Option: 1

\mathrm{f} is increasing in \mathrm{(1,2)} and decreasing in \mathrm{(2, \infty)}


Option: 2

\mathrm{f(x)=-1}  has exactly two solutions


Option: 3

\mathrm{f^{\prime}(\mathrm{e})-f^{\prime \prime}(2)<0}


Option: 4

\mathrm{f(x)=0} has a root in the interval \mathrm{(\mathrm{e}, \mathrm{e}+1)}


Answers (1)

best_answer

\mathrm{f(x)=4 \log _{\mathrm{e}}(x-1)-2 x^{2}+4 x+5, x>1}

\mathrm{{f}'(x)=\frac{4}{x-1}-4x+4= 4\frac{\left [1-x^{2}+x+x-1 \right ]}{x-1}}
                                                  \mathrm{= -4x\frac{\left ( x-2 \right )}{x-1}}

increasing in \mathrm{\left ( 1,2 \right )} & decreasing in \mathrm{\left ( 2,\infty \right )}

\mathrm{4\log_{e}\left ( x-1 \right )-2x^{2}+4x+5= -1}
\mathrm{4\log_{e}\left ( x-1 \right )= 2x^{2}-4x-6}
\mathrm{2\log_{e}\left ( x-1 \right )= x^{2}-2x-3}



Hence, two solutions for \mathrm{f\left ( x \right )= -1}

Now, 

\mathrm{{f}''\left ( x \right )= \frac{-4}{\left ( x-1 \right )^{2}}-4}
\mathrm{{f}'\left ( e \right )-{f}''\left ( 2 \right )= \frac{4}{\left ( e-1 \right )^{2}}-4e+4+4+4= \, '+' ve}
Now
       \mathrm{f\left ( e \right )= \, '+' ve}
        \mathrm{f\left ( e +1\right )= \, '-' ve}
Hence a root lies in \mathrm{\left ( e,e+1 \right )}

Hence, option (C) 

Posted by

Rishi

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