If `f(x) =(2x+1)/(2x^3+3x^2+x)` and `S={x:f(x)>0}` then S contains (a) `(-oo,-3/2)` (b) `(-3/2,-1/4)` (c)`(-1/4,1/2)` (d) `(1/2,3)`

If f(x)=(2x+1)/(2x^(3)+3x^(2)+x) and S={x:f(x)>0} then s contains (a) (-oo,-(3)/(2))(b)(-(3)/(2),-(1)/(4))(c)(-(1)/(4),(1)/(2)) (d) ((1)/(2),3)

If S is the set of all real x such that (2x-1)/(2x^3+3x^2+x) is positive (-oo,-3/2) b. (-3/2,1/4) c. (-1/4,1/2) d. (1/2,3) e. None of these

If S is the set of all real x such that (2x-1)/(2x^3+3x^2+x) is positive (-oo,-3/2) b. (-3/2,1/4) c. (-1/4,1/2) d. (1/2,3) e. None of these

If S is the set of all real x such that (2x-1)/(2x^(3)+3x^(2)+x) is positive (-oo,-(3)/(2)) b.(-(3)/(2),(1)/(4)) c.(-(1)/(4),(1)/(2)) d.((1)/(2),3) e. None of these

f(x)=(x-1)|(x-2)(x-3)|dot Then f decreases in (a) (2-1/(sqrt(3)),2) (b) (2,2+1/(sqrt(3))) (c) (2+1/(sqrt(3)),4) (d) (3,oo)

f(x)=(x-1)|(x-2)(x-3)|dot Then f decreases in (a) (2-1/(sqrt(3)),2) (b) (2,2+1/(sqrt(3))) (c) (2+1/(sqrt(3)),4) (d) (3,oo)

If f'(x)=4x^3-3x^2+2x+k and f(0)=1, f(1)=4 find f(x)

If f(x)={x^3,if x^2 =1 then f(x) is differentiable at (a) (-oo,oo)-{1} (b) (-oo,oo)-{1,-1} (c) (-oo,oo)-{1,-1,0} (d) (-oo,oo)-{-1}

f(x)=(x-1)|(x-2)(x-3)|* Then f decreases in (a) (2-(1)/(sqrt(3)),2) (b) (2,2+(1)/(sqrt(3)))(c)(2+(1)/(sqrt(3)),4) (d) (3,oo)