Let f(x) = cases a ( (x-[x])) x-[x] , & x (-1, 0) (bx, x^2), & x [0, 1) c x + d, & otherwise cases where [t] denotes the greatest integer less than or equal to t . It is given that f(x) is continuous on R , differentiable at x=1 , and f(2) = 5 . The value of 2(b + c + d - a ) is:

Step-by-step solution

For x (-1, 0) , [x] = -1 x - [x] = x + 1 . Thus, f(x) = a ( (x+1)) x+1 for x (-1, 0) . Continuity at x = -1 : _ x -1^- f(x) = _ x -1^- (cx + d) = -c + d . _ x -1^+ f(x) = _ x -1^+ a ( (x+1)) x+1 = a . Thus, -c + d = a d - c = a . Continuity at x = 1 : _ x 1^- f(x) = _ x 1^- (bx, x^2) = (b, 1) . _ x 1^+ f(x) = _ x 1^+ (cx + d) = c + d . Thus, (b, 1) = c + d . Differentiability at x = 1 : LHD at x = 1 : If b > 1 , near x=1^- , bx > x^2 f(x) = bx LHD = b . If b 1 , near x=1^- , x^2 bx f(x) = x^2 LHD = 2 . RHD at x = 1