Question

∫g'(x) dx for x ∈ [-1,1] =

(a) 2g(–1) 

(b) 0 

(c) –2g(1) 

(d) 2g(1)

Answer 1

Answer is (d) 2g(1)

For x ∈ (– 2, 2),

(g(x))³ – 3g(x) + x = 0 ...(i)

(g(– x))³ – 3g(–x) – x = 0 ...(ii)

Adding Eqs. (i) and (ii), we get

{(g(x))³ + (g (x))³} – 3{g(x) + g(–x)} = 0

{g(x) + g(–x)} {(g(x))² + (g(–x))² – g(x)g(–x)) – 3} = 0

{g(x) + g(– x)} = 0

Now, ∫g'(x) dx for x ∈ [-1,1]

= (g(x)) for x ∈ [-1,1]

= g(1) – g (–1) = g(1) + g (1) = 2g(1)