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TS EAMCET 04 May 2019 Shift 1 Solved Paper
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© examsnet.com
Question : 69
Total: 160
The number of real roots of the equation
e
x
−
1
+ log x + x - 2 = 0 , x ≠ 0 , is
0
1
2
3
Validate
Solution:
Consider the expression,
f
(
x
)
=
e
x
−
1
+
log
x
+
x
−
2
Differentiate above w.r.t
x
f
′
(
x
)
=
e
x
−
1
+
1
x
+
1
This implies,
f
′
(
x
)
>
0
This implies,
f
(
x
)
is monotonic increasing function so,
f
(
x
)
has only one real root.
© examsnet.com
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