If `0lt=xlt=pi/3` then range of `f(x)=sec(pi/6-x)+sec(pi/6+x)` is (a)`(4/(sqrt(3)),oo)` (b) `(4/(sqrt(3)),oo)` (c)`(0,4/(sqrt(3)))` (d) `(0,4/(sqrt(3)))`

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

The value of log_((9)/(4))((1)/(2sqrt(3))sqrt(6-(1)/(2sqrt(3))sqrt(6-(1)/(2sqrt(3))sqrt(6-(1)/(2sqrt(3)))))...oo) is

The value of ("lim")_(xto2)(sqrt(1+sqrt(2+x))-sqrt(3))/(x-2)i s (a) 1/(8sqrt(3)) (b) 1/(4sqrt(3)) (c) 0 (d) none of these

The numerical value of tan(pi/3)+2tan((2pi)/3)+4tan((4pi)/3)+8tan((8pi)/3) is equal to (A) -5sqrt(3) (B) -5/(sqrt3) (C) 5sqrt(3) (D) 5/(sqrt3)

The vertices of a triangle are (0,0), (x ,cosx), and (sin^3x ,0),w h e r e0ltxltpi/2 the maximum area for such a triangle in sq. units is (a) (3sqrt(3))/(32) (b) (sqrt(3)/32) (c) 4/32 (d) (6sqrt(3)) /(32)

In each of the following cases find the period of the function if it is periodic. (i) f(x)="sin"(pi x)/(sqrt(2))+"cos"(pi x)/(sqrt(3)) " (ii) " f(x)="sin"(pi x)/(sqrt(3))+"cos"(pi x)/(2sqrt(3))

Prove that pi/6 lt int_0^1(dx) /(sqrt(4-x^2-x^3)) lt pi/(4sqrt(2))

Points on the curve f(x)=x/(1-x^2) where the tangent is inclined at an angle of pi/4 to the x-axis are (a)(0,0) (b) (sqrt(3),-(sqrt(3))/2) (-2,2/3) (d) (-sqrt(3),(sqrt(3))/2)

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