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Answer :

`ararrp;s;brarrp,s;crarrq,r;d;drarrq`Solution :

f(x)=`x^(2)logx` <br> For `f(X) =x(2 logx +1)=0 x=(1)/sqrt(e )` which is the point of minima as derivative change sign form negative to positive <br> Also the function decreases in `0,(1)/sqrt(e )` <br> (b) y=x logx <br> `therefore (dy)/(dx)=xxx1/x+logxxx1` <br> `=1+logx and (d^(2)y)/(dx^(2))=1/x` <br> for `(dy)/(dx)=0 logx =-1 or x =1/x` <br> `(d^(2)y)/(dx^(2))=(1)/(1//e) =egt0 at x=1/e` <br> Thus y is minimum for `x =1/e` <br> Thus y is minimum for `x =1/e` <br> (c ) `f(X) =(logx)/(x)` <br> For `f(x) =(1-logx)/(x^(2)) =0 x=e` Also derevative changes singn form positive to negative at x =e hence it is the point of maxima <br> (d) `f(X)=x^(-x)` <br> `f(x) =-x^(-x)(1+logx)=0 or x=1//e` <br> Which is clearly point of maxima