1.Let fff and ggg be twice differentiable functions on R\mathbb{R}R such thatf′′(x)=g(x)+6xf''(x) = g(x) + 6xf′′(x)=g(x)+6xf′(1)=4f'(1) = 4f′(1)=4g′(1)=3g'(1) = 3g′(1)=3f(2)=3f(2) = 3f(2)=3g(2)=12g(2) = 12g(2)=12Then which of the following is NOT true?a.g(−2)−f(−2)=20g(-2) - f(-2) = 20g(−2)−f(−2)=20b.If −1<x<2-1 < x < 2−1<x<2, then ∣f(x)−g(x)∣<8|f(x) - g(x)| < 8∣f(x)−g(x)∣<8c.∣f′(x)−g′(x)∣<6⇒−1<x<1|f'(x) - g'(x)| < 6 \Rightarrow -1 < x < 1∣f′(x)−g′(x)∣<6⇒−1<x<1d.There exists x0∈(1,2)x_0 \in (1, 2)x0∈(1,2) such that f(x0)=g(x0)f(x_0) = g(x_0)f(x0)=g(x0)Login to continueOnly logged in users canattempt or see the solution.