Let `f(x)={1+(2x)/a ,0lt=x<1 and ax ,1lt=x<2` If `lim_(x->1)f(x)` exists ,then ` a` is

(a)`1`

(b)`-1`

(c) `2`

(d) `-2`

If f(x+1/2)+f(x-1/2)=f(x)fora l lx in R , then the period of f(x) is (a) 1 (b) 2 (c) 3 (d) 4

Let f(x)=x+2|x+1|+2|x-1|dot If f(x)=k has exactly one real solution, then the value of k is (a) 3 (b) 0 (c) 1 (d) 2

Let f(x)=x+1/(2x+1/(2x+1/(2x))) Compute the value of f(5) .

Let g(x)=2f(x/2)+f(x) and f''(x)lt0 " in "0lexle1 , then g(x)

Let f(x)=x^(3) + bx^(2) +cx +d, 0 lt b^(2) lt c . Then f(x)-

Let f(x)={{:(1+(2x)/(a)", "0lexlt1),(ax", "1lexlt2):}."If" lim_(xto1) "f(x) exists, then a is "

Let f (x+(1)/(x))= x^(2) +(1)/(x^(2)) , x ne 0, then the value of f (x) is-

Let f:[-1,2]->[0,oo) be a continuous function such that f(x)=f(1-x)fora l lx in [-1,2]dot Let R_1=int_(-1)^2xf(x)dx , and R_2 be the area of the region bounded by y=f(x),x=-1,x=2, and the x-axis . Then (a) R_1=2R_2 (b) R_1=3R_2 (c) 2R_1 (d) 3R_1=R_2

If f(x)=x/(x^2+1) and f(A)={y:-1/2 le y lt 0} , then set A is

Let f: R->R be a continuous function and f(x)=f(2x) is true AAx in Rdot If f(1)=3, then the value of int_(-1)^1f(f(x))dx is equal to (a)6 (b) 0 (c) 3f(3) (d) 2f(0)