f′(1−x)2(−2x) =x(f′(3x21)+(−1)f′(1−x2)) g(x) is increasing, when g′(x)>0 Case 1f′(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<