Let f(x) be a non-constant twice differentiable function on R such that f(2+x)=f(2-x) and f prime((1/2))=f prime(1)=0. Then minimum number of roots of the equation f prime prime(x)=0 in (0,4) are

Question

Let f(x) be a non-constant twice differentiable function on R such that f(2+x)=f(2-x) and f prime((1/2))=f prime(1)=0. Then minimum number of roots of the equation f prime prime(x)=0 in (0,4) are (A) 2 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

f (2+ x )= f (2- x ) f prime(2+x)=-f prime(2-x) Put x=0, f prime(2)=0 x=-1 f prime(1)=-f prime(3)=0 x=(-3/2), f prime((1/2))=-f prime((7/2))=0 ∴ f prime((1/2))=0=f prime(1)=f prime(2)=f prime(3)=f prime((7/2)) Using Rolle's theorem in y = f prime( x ). Minimum number of roots of f prime prime( x )=0 are 4 .

Q. Let f(x) be a non-constant twice differentiable function on R such that f(2+x)=f(2−x) and f′(21​)=f′(1)=0. Then minimum number of roots of the equation f′′(x)=0 in (0,4) are

2

4

5

6

f(2+x)=f(2−x) f′(2+x)=−f′(2−x) Put x=0,f′(2)=0 x=−1f′(1)=−f′(3)=0 x=2−3​,f′(21​)=−f′(27​)=0 ∴f′(21​)=0=f′(1)=f′(2)=f′(3)=f′(27​) Using Rolle's theorem in y=f′(x). Minimum number of roots of f′′(x)=0 are 4 .

Q. Let $f (x)$ be a non-constant twice differentiable function on $R$ such that $f (2 + x) = f (2 - x)$ and $f^{'} (\frac{1}{2}) = f^{'} (1) = 0$ . Then minimum number of roots of the equation $f^{''} (x) = 0$ in $(0, 4)$ are