Area lying between the curves `y=tanx, y=cotx` and x-axis, `x in [0, pi/2]` is (A) `1/2log2` (B) `log2` (C) `2log(1/sqrt(2))` (D) none of these

The area between the curves y=tanx,y=cotx and X-axis in (pi)/(6)le x le (pi)/(3) is

int_2^4 log[x]dx is (A) log2 (B) log3 (C) log5 (D) none of these

The function f is such that : f(xy)=f(x)+f(y), x, y gt 0 and f\'(1)=2 and A the area bounded by the curves y=f(x), x=2 and the x-axis, then (A) f(x)=2log_ex (B) f(x)=2 log_ex (C) A=2(2log_e2-1) (D) A=4log(2/sqrt(e))

If A is the area lying between the curve y=sin x and x-axis between x=0 and x=pi//2 . Area of the region between the curve y=sin 2x and x -axis in the same interval is given by

int_0^1 log(sqrt(1+x)+sqrt(1-x))dx= (A) 1/2(log2-pi/2+1) (B) 1/2(log2+pi/2+1) (C) 1/2(log2+pi/2-1) (D) none of these

If y=(log)_(sinx)(tanx),t h e n(((dy)/(dx)))_(pi/4)"is equal to" (a) 4/(log2) (b) -4log2 (c) (-4)/(log2) (d) none of these

Area enclosed between the curves |y|=1-x^2 and x^2+y^2=1 is (a) (3pi-8)/3 (b) (pi-8)/3 (c) (2pi-8)/3 (d) None of these

The curve x y=c ,(c >0), and the circle x^2+y^2=1 touch at two points. Then the distance between the point of contacts is 1 (b) 2 (c) 2sqrt(2) (d) none of these

The solution of dy/dx+(xy^2-x^2y^3)/(x^2y+2x^3y^2)=0 , is (A) log(y^2/x)-1/(xy)=C (B) log(x/y)+y^2/x=C (C) log(x^2y)+y^2/x=C (D) none of these

intxsecx^2\ dx is equal to 1/2log(secx^2+tanx^2)+C (b) (x^2)/2log(secx^2+tanx^2)+C (c) 2log(secx^2+tanx^2)+C (d) none of these