The function `f(x)=x(x+4)e^(-x//2)` has its local maxima at `x=adot` Then (a)`a=2sqrt(2)` (b) `a=1-sqrt(3)` (c)`a=-1+sqrt(3)` (d) `a=-4`

The function f(x)=(ax+b)/((x-1)(x-4)) has a local maxima at (2,-1), then

The range of function f(x)=sqrt(x^(2)+4x)C_(2x^(2)+3) is

If mean value theorem holds good for the function f(x)=(x-1)/(x) on the interval [1,3] then the value of 'c' is 2 (b) (1)/(sqrt(3))( c) (2)/(sqrt(3))(d)sqrt(3)

The greatest value of the function f(x)=(sin2x)/(sin(x+(pi)/(4))) on the interval (0,(pi)/(2)) is (1)/(sqrt(2))(b)sqrt(2)(c)1(d)-sqrt(2)

If x^(4)+(1)/(x^(4))=623, then x+(1)/(x)=27(b)25 (c) 3sqrt(3)(d)-3sqrt(3)

The domain of the function f(x)=sqrt(2-2x-x^2) is [-sqrt(3),\ sqrt(3)] b. [-1-sqrt(3),\ -1+sqrt(3)] c. [-2,2] d. [-2-sqrt(3),-2+sqrt(3)]

If f(x)=sin^2x and the composite function g(f(x))=|sinx| , then g(x) is equal to sqrt(x-1) (b) sqrt(x) (c) sqrt(x+1) (d) -sqrt(x)

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)

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

(1+log_(2)(x-4))/(log_(sqrt(2))(sqrt(x+3)-sqrt(x-3)))=1