f(x) is a polynomial of degree 6 which decreases in the interval (0, ∞) and increases in the interval (-∞, 0) . If f'(2)=0, f'(0)=0, f''(0)=0, f(0)=1 and f(1)-f(-1)=(8/5), then -3(f(1)+f(-1)) equals

Question

f(x) is a polynomial of degree 6 which decreases in the interval (0, ∞) and increases in the interval (-∞, 0) . If f'(2)=0, f'(0)=0, f''(0)=0, f(0)=1 and f(1)-f(-1)=(8/5), then -3(f(1)+f(-1)) equals (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Since, f'(x)=k x3(x-2)2=k x3(x2-4 x+4) f'(x)=k(x5-4 x4+4 x3) ∴ f(x)=k((x6/6)-(4 x5/5)+x4)+c Also f(0)=1 ⇒ c=1. and f(1)-f(-1)=-(8/5) k=(8/5) ⇒ k=-1 . So f(1)+f(-1)=(7/3) k+2 c=-(7/3)+2=-(1/3)

Q. f(x) is a polynomial of degree 6 which decreases in the interval (0,∞) and increases in the interval (−∞,0). If f′(2)=0,f′(0)=0,f′′(0)=0,f(0)=1 and f(1)−f(−1)=58​, then −3(f(1)+f(−1)) equals

Since, f′(x)=kx3(x−2)2=kx3(x2−4x+4) f′(x)=k(x5−4x4+4x3) ∴f(x)=k(6x6​−54x5​+x4)+c Also f(0)=1⇒c=1. and f(1)−f(−1)=−58​k=58​ ⇒k=−1. So f(1)+f(−1)=37​k+2c=−37​+2=−31​

Q. $f (x)$ is a polynomial of degree $6$ which decreases in the interval $(0, \infty)$ and increases in the interval $(- \infty, 0) .$ If $f^{'} (2) = 0, f^{'} (0) = 0, f^{''} (0) = 0, f (0) = 1$ and $f (1) - f (- 1) = \frac{8}{5}$ , then $- 3 (f (1) + f (- 1))$ equals