MHT CET 2017 Math Solved Paper

(b) since, f(x)=

log‌x

x

After differentiating on both sides w.r.t. x, we get
f′(x)=

x⋅ 1 x −log‌x⋅1

x2

=

1−log‌x

x2

For maximum or minimum value of f(x), put f′(x)=0
⇒

1−log‌x

x2

=0 ⇒ logx=1⇒x=e
Now, f′′(x)=

3+2‌log‌x

x3

∴ f′′(e)=−

1

e3

<0
After substituting x = e in eq. (i), we get
f(e)=

log‌e

e

=

1

e

Hence, maximum value of f(x) is

1

e

at x = e