The total number of distinct x ∈ 0 ,   1 for which, ∫ 0 x t 2 1 + t 4 d t = 2 x - 1 is

The total number of distinct x∈0, 1 for which, ∫0xt21+t4dt=2x-1 is

Solution

Let fx=∫0xt21+t4dt-2x+1
⇒ f′x=x21+x4-2
As 1+x4x2 ≥2 ⇒x21+x4≤12
⇒ f′x ≤-32
⇒ fx is continuous and decreasing
f0=1 and f1=∫01t21+t4dt-1≤-12
By intermediate value theorem (IVT),fx=0possesses exactly one solution in [0, 1]. As f(0)f(1)<0 

Asked in: JEE Advanced 2016 (Paper 1)

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