KCET 2016 Math Solved Paper

Let y=(

1

x

)x
Taking log on both the sides, we get
log‌y=x‌log(‌

1

x

)=−x‌log‌x
log‌y=−x‌log‌x → (1)
Now, differentiating Eq. (1) with respect to x, we get
‌

1

y

‌

dy

dx

=−(x⋅‌

1

x

+log‌x)
⇒‌

1

y

‌

dy

dx

=−(1+log‌x)
⇒‌

dy

dx

=−y(1+log‌x) → (2)
⇒‌

dy

dx

=−(‌

1

x

)x(1+log‌x)
Now, ‌

dy

dx

=0
−y(1+log‌x)=0
⇒log‌x=−1⇒x=e−1
Now, differentiating Eq. (2) with respect to x, we get
‌

d2y

dx2

=−(1+log‌x)‌

dy

dx

−‌

y

x

‌

d2y

dx2

=y(1+log‌x)2−‌

y

x

At x=e−1, we get
⇒

d2y

dx2

|x=e−1<0
Therefore, maximum value of y is,ymax=e

1

e