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Joint Entrance Examination
Mathematics
Sets, Relations and Functions
Question 128
Question 128 of 161
Medium
MCQ
If
f
(
x
)
+
2
f
(
1
x
)
=
3
x
f(x)+2f\!\left(\dfrac1x\right)=3x
f
(
x
)
+
2
f
(
x
1
)
=
3
x
,
x
≠
0
x\neq0
x
=
0
, and
S
=
{
x
∈
R
:
f
(
x
)
=
f
(
−
x
)
}
S=\{x\in\mathbb{R} : f(x)=f(-x)\}
S
=
{
x
∈
R
:
f
(
x
)
=
f
(
−
x
)}
; then
S
S
S
:
A
contains exactly two elements.
B
contains more than two elements.
C
is an empty set.
D
contains exactly one element.
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Solutions · 1
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Official solution
Given:
f
(
x
)
+
2
f
(
1
/
x
)
=
3
x
f(x)+2f(1/x)=3x
f
(
x
)
+
2
f
(
1/
x
)
=
3
x
\ldots (1). Replacing
x
→
1
/
x
x\to1/x
x
→
1/
x
:
f
(
1
/
x
)
+
2
f
(
x
)
=
3
/
x
f(1/x)+2f(x)=3/x
f
(
1/
x
)
+
2
f
(
x
)
=
3/
x
\ldots (2). Compute
2
×
(
2
)
−
(
1
)
2\times(2) - (1)
2
×
(
2
)
−
(
1
)
: \[4f(x)+2f(1/x) - f(x)-2f(1/x) = \frac6x-3x \implies 3f(x)=\frac6x-3x \implies f(x)=\frac2x-x\] Set
f
(
x
)
=
f
(
−
x
)
f(x)=f(-x)
f
(
x
)
=
f
(
−
x
)
: \[\frac2x-x = -\frac2x+x \implies \frac4x=2x \implies x^2=2 \implies x=\pm\sqrt{2}\]
S
=
{
2
,
−
2
}
S=\{\sqrt{2},-\sqrt{2}\}
S
=
{
2
,
−
2
}
--- exactly two elements.
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