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`

A line tangent to the graph of the function y=f(x) at the point x= a forms an angle (pi)/(3) with the axis of Abscissa and angle (pi)/(4) at the point x=b then int_(a)^(b)f'(x)dx is

int_(sqrt(2)//3)^(sqrt(3)//3)(1)/(sqrt(4-9x^(2)))dx=

The tangent, represented by the graph of the function y=f(x), at the point with abscissa x = 1 form an angle of pi//6 , at the point x = 2 form an angle of pi//3 and at the point x = 3 form and angle of pi//4 . Then, find the value of, int_(1)^(3)f'(x)f''(x)dx+int_(2)^(3)f''(x)dx.

The equation of the curve is y=f(x)dot The tangents at [1,f(1),[2,f(2)],a n d[3,f(3)] make angles pi/6,pi/3,a n dpi/4, respectively, with the positive direction of x-axis. Then the value of int_2^3f^(prime)(x)f^(x)dx+int_1^3f^(x)dx is equal to (a)-1/(sqrt(3)) (b) 1/(sqrt(3)) (c) 0 (d) none of these

int_0^(4/pi) (3x^2sin(1/x)-xcos(1/x))dx= (A) (8sqrt(2))/pi^3 (B) (32sqrt(2))/pi^3 (C) (24sqrt(2))/pi^3 (D) sqrt(2048)/pi^3

f(x) is double differentiable function. Tangent at (1, f(1)) and (3, (f(3)) cuts positive x-axis at an angle pi/6 and pi/4 then value of int_1^3(f'(x)^2+1)f''(x)dx is:

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)))

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)

The area of the circle x^2+y^2=16 exterior to the parabola y^2=6x is (A) 4/3(4pi-sqrt(3)) (B) 4/3(4pi+sqrt(3)) (C) 4/3(8pi-sqrt(3)) (D) 4/3(8pi+sqrt(3))