AP EAPCET 20-Aug-2021 Shift 2 Paper

g(x)=

1

6

f(3x2−1)+

1

2

f(1−x2)
g′(x)=

1

6

f′(3x2−1)(6x) +

1

2

f′(1−x)2(−2x)
=x(f′(3x21)+(−1)f′(1−x2))
g(x) is increasing, when g′(x)>0
Case 1 f′(3x2−1)−f′(1−x2)>0
and x>0 ...(a)
⇒f′(3x2−1)>f′(1−x2) ...(i)
Given that, f"(x)>0
⇒f(x) is increasing.
⇒f′(x)>0
From Eq. (i), 3x2−1>1−x2
4x2>2⇒x2>

1

2

⇒x<−

1

√2

and x>

1

√2

⇒x∈(−∞,−

1

√2

)∪(

1

√2

,∞)...(b)
From (a) and (b)
⇒x∈(

1

√2

,∞)
Case 2 When x<0 and f′(3x2−1)<f′(1−x2)
⇒3x2−1<1−x2
x2<

1

2

⇒x∈(

−1

√2

,

1

√2

)
∵x<0⇒x∈(−

1

√2

,0)
using case - 1 and case - 2.
x∈(

−1

√2

,0)∪(

1

√2

,∞)