TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Given, ‌‌2f(x)+f(‌

1

x

)=4x . . . (i)
Replace x by 1∕x, we get,
2f(1∕x)+f(x)=‌

4

x

. . . (ii)
On multiply by 2 in Eq. (i) and then subtracting Eq. (ii), we get
4f(x)+2f(‌

1

x

)=8x
f(x)+2f(1∕x)=‌

4

x

3f(x)=8x−‌

4

x

=‌

8x2−4

x

∴‌‌f(x)=‌

8x2−4

3x

=‌

4(2x2−1)

3x

and ‌‌f(−x)=‌

4(2x2−1)

−3x

Again given, ‌‌f(x)=f(−x)
‌

4(2x2−1)

3x

=−[‌

4(2x2−1)

3x

]
⇒‌‌2[‌

4(2x2−1)

3x

]=0
⇒‌‌2x2−1=0 but x≠0
⇒‌‌x=±‌

1

√2

but x≠0
∴‌‌S=(−1∕√2,1∕√2}