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Manipal Entrance Test 2018 Math Paper
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© examsnet.com
Question : 55
Total: 70
The minimum value of
x
log
x
is equal to
e
1
e
−
1
e
2
e
Validate
Solution:
Let
y
=
x
log
e
x
On differentiating w.r.t.
x
, we get
d
y
d
x
=
x
.
1
x
+
log
x
=
(
1
+
log
x
)
Again, differentiating, we get
d
2
y
d
x
2
=
1
x
Put,
d
y
d
x
=
0
for maxima or minima.
⇒
1
+
log
x
=
0
⇒
x
=
1
e
∴
(
d
2
y
d
x
2
)
(
x
=
1
c
)
=
e
∴
y
is minimum at
x
=
1
e
∴
y
min
=
1
e
log
e
(
1
e
)
=
−
1
e
© examsnet.com
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