`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)`

" (vii) f(x)=3x^(2)-1;x=(1)/(sqrt(3)),(2)/(sqrt(3))

The value of lim_(x rarr2)(sqrt(1+sqrt(2+x))-sqrt(3))/(x-2) is (1)/(8sqrt(3))(b)(1)/(4sqrt(3)) (c) 0 (d) none of these

int((1)/(sqrt(x))+(2)/(sqrt(x^(2)-1))-(3)/(2x^(2)))dx on (1,oo)

If tan^(-1)x+2cot^(-1)x=(2 pi)/(3), then x, is equal to (a) (sqrt(3)-1)/(sqrt(3)+1) (b) 3 (c) sqrt(3)(d)sqrt(2)

Let f(x)=(x+1)^(2)-1,x>=-1. Then the set {x:f(x)=f^(-1)(x)} is (a){0,1,(-3+i sqrt(3))/(2),(-3-i sqrt(3))/(2)}(b){0,-1}(c){0,1}(d) empty

If f(x)=sin^(-1)(3x-4x^(3)). Then answer the following The value of f'((1)/(sqrt2)) , 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)

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)

The tangent to the graph of the function y=f(x) at the point with abscissae x=1, x=2, x=3 make angles pi/6,pi/3 and pi/4 respectively. The value of int_1^3f\'(x)f\'\'(x)dx+int_2^3f\'\'(x)dx is (A) (4-3sqrt(3))/3 (B) (4sqrt(3)-1)/(3sqrt(3)) (C) (4-3sqrt(3))/2 (D) (3sqrt(3)-1)/2

If I=int(sin x+sin^(3)x)/(cos2x)dx=P cos x+Q log|f(x)|+R then (a)P=(1)/(2),Q=-(3)/(4sqrt(2))(b)P=(1)/(4),Q=(1)/(sqrt(2)) (c) f(x)=(sqrt(2)cos x+1)/(sqrt(2)cos x-1)(d)f(x)=(sqrt(2)cos x-1)/(sqrt(2)cos x+1)