Let f and g be twice differentiable functions on R such that f′′(x)=g′′(x)+6x f′(1)=4g′(1)−3=9 f(2)=3g(2)=12 Then which of the following is NOT true?
+C ‌ At ‌‌x=1 ‌f′(1)=g′(1)+3+C ⇒9=4+3+C⇒C=3 ∴f′(x)=g′(x)+3x2+3 Again by integrating, ‌f(x)=g(x)+‌
3x3
3
+3x+D ‌‌ At ‌x=2 ‌f(2)=g(2)+8+3(2)+D ‌⇒12=4+8+6+D⇒D=−6 So, f(x)=g(x)+x3+3x−6 ‌⇒f(x)−g(x)=x3+3x−6 ‌‌ At ‌x=−2 ‌⇒g(−2)−f(−2)=20‌‌‌ (Option (1) is true) ‌ Now, for −1<x,2 ‌h(x)=f(x)−g(x)=x3+3x−6 ‌⇒h′(x)=3x2+3 ‌⇒h(x)↑ ‌‌ So, ‌h(−1)<h(x)<h(2) ‌⇒−10<h(x)<8 ‌⇒|h(x)|<10‌‌‌ (option (2) is NOT true) ‌ Now, h′(x)=f′(x)−g′(x)=3x2+3 If |h′(x)|<6⇒|3x2+3|<6 ⇒3x2+3<6 ⇒x2<1 ⇒−1<x<1‌‌ (option (3) is True) If x∈(−1,1)|f′(x)−g′(x)|<6 option (3) is true and now to solve ‌f(x)−g(x)=0 ‌⇒x3+3x−6=0 ‌h(x)=x3+3x−6 here, h(1)=−ve and h(‌