© examsiri.com
Question : 167 of 180
Marks:
+1,
-0
Solution:
‌‌Let ‌f(x)=(‌)2x2=x−2x2‌∴‌‌‌‌log‌f(x)=−2x2‌log‌x‌⇒‌‌‌f′(x)=−4x‌log‌x−‌‌⇒‌‌f′(x)=x−2x2‌‌‌‌[−4x‌log‌x−2x]‌‌ For maximum, ‌‌‌‌‌‌‌‌f′(x)=0‌⇒‌‌x−2x2[−4x‌log‌x−2x]=0‌⇒‌‌−4x‌log‌x−2x=0‌⇒‌‌log‌x=−‌‌‌ Again, ‌f′′(x)=x−2x2[−4x‌log‌x−2x]2‌‌‌‌‌‌‌‌‌‌‌x=e−‌‌∴‌‌f′′(e−‌)<0 So,
f(x) has maximum value at
x=e−‌∴‌‌fmax=f(e−‌)=(‌)2(‌)2=(√e)2(‌)=e1∕e
© examsiri.com
Go to Question: