Question

A function y= f (x) satisfies f "(x) = - 1/x2 - π2 sin(πx) ; f '(B) = π+ 1/2 and f (A)=0. The value of f (1/2) is

(A) ln 2

(B) 1

(C) π/2 - ln 2

(D) 1 – ln 2

Answer 1

Correct Answer is: (B) 1

f ' (x) = 1/x + π cos(πx) + C

f ' (2) = 1/2 + π + C = 1/2 + π ⇒ C =0

f (x) = ln | x | + sin(πx) + C'

f (1) = C' = 0

f (x) = ln | x | + sin(πx)