CGPET 2022 Solved Paper

(c) Let y=(

1

x

)x=x−x
⇒log‌y=−x‌log‌x
⇒

1

y

dy

dx

=−(1+log‌x)
⇒

dy

dx

=−y(1+log‌x)
⇒

d2y

dx2

=

−dy

dx

(1+log‌x)−

y

x

=y(1+log‌x)2−

y

x

=x−x(1+log‌x)2−

x−x

x

=x−x(1+log‌x)2−x−x−1
At points of local maximum and local minimum, we must have

dy

dx

=0
−y(1+log‌x)=0
⇒1+log‌x=0
⇒log‌x=−1
⇒x=e−1
⇒(

d2y

dx2

)x=e−1=(

1

e

)−

1

e

(1+log‌

1

e

)2−(

1

e

)−

1

e

−1
=(e−1)−

1

e

(1−log‌e)2−(e−1)−

1

e

−1
=−e

1

e

+1<0
So, x=

1

e

is a point of local maximum. The local maximum value of y is obtained by putting x=

1

e

in y and is equal to e‌

1

e

.