• # question_answer If $f(x)=\left\{ \begin{matrix} x\ln x, & x>0 \\ 0, & x=0 \\ \end{matrix} \right.$and conclusion of LMVT holds at $x=1$ in the interval [0, a] for $f(x)$, then $[{{a}^{2}}]$ is equal to [Note: [k] denotes the greatest integer less than or equal to k.] A)  2                                 B)  4 C)  7                                 D)  9

Given that, So, $f(x)=\,\left\{ \begin{matrix} x\ln \,x, & x>0 \\ 0, & x=0 \\ \end{matrix} \right.$                   So,$f'(1)x=\frac{f(a-f)(0)}{a-0}$                         ?(1) As $f'(x)\,=\left( x.\frac{1}{x}\,+\ln \,x.1 \right)\,=1+\ln \,x$ $f'(1)=1+\ln \,1=1$                               ?(2) $\therefore$ Using (2) in (1), we get $1=\frac{a\ln a-0}{a}$ $\Rightarrow \,\,\ln \,a=1\,\Rightarrow \,a=e$ $\therefore \,\,[{{a}^{2}}]\,=[{{a}^{2}}]=7$