If `f(x)=sqrt(1+cos^2(x^2)),t h e nf^(prime)((sqrt(pi))/2)` is `(sqrt(pi))/6` (b) `-sqrt(pi//6)` `1//sqrt(6)` (d) `pi//sqrt(6)`

If f(x)=cot^(-1)sqrt(cos2x)," then: "f'((pi)/(6))=

sin(pi-6x)+sqrt3 sin(pi/2+6x)=sqrt3

If int_ (0)^(oo) e^(-x^(2)) dx = (sqrt (pi))/(2), and int_ (0)^(oo) e^(-ax^(2) ) dx, a> 0 is (i) (sqrt (pi))/(2) (ii) (sqrt (pi))/(2a) (iii) 2 (sqrt (pi))/(a) (iv) (1)/(2) sqrt ((pi)/(a))

If int_(0)^(oo)e^(-x^(2))dx=(sqrt(pi))/(2), then int_(0)^(oo)e^(-ax^(2))dx where a>0 is: (A)(sqrt(pi))/(2) (B) (sqrt(pi))/(2a)(C)2(sqrt(pi))/(a) (D) (1)/(2)(sqrt((pi)/(a)))

Evaluate cos [(pi)/(6) + cos^(-1) (-(sqrt(3))/(2))]

the greater number between sqrt(pi)-sqrt(12) and sqrt(pi)-sqrt(6)

Let f(x) be a non-negative continuous function such that the area bounded by the curve y=f(x), the x -axis,and the ordinates x=(pi)/(4) and x=beta>(pi)/(4) is beta sin beta+(pi)/(4)cos beta+sqrt(2)beta Then f'((pi)/(2)) is ((pi)/(2)-sqrt(2)-1)( b) ((pi)/(4)+sqrt(2)-1)-(pi)/(2)( d) (1-(pi)/(4)-sqrt(2))

The amplitude of (1+i sqrt(3))/(sqrt(3)+i) is (pi)/(3) b.-(pi)/(6) c.(pi)/(3) d.(pi)/(6)

int_(0)^(3)(3x+1)/(x^(2)+9)dx=(pi)/(12)+log(2sqrt(2))(b)(pi)/(2)+log(2sqrt(2))quad (c)(pi)/(6)+log(2sqrt(2))(d)(pi)/(3)+log(2sqrt(2))

If (dy)/(dx) +y sec x=tan x," then (sqrt(2) +1) y((pi)/(4)) - y(0)=, (A) sqrt(2) - (pi)/(4) (B) sqrt(2) + (pi)/(4) (C) sqrt(2) - (pi)/(2) (D) sqrt(2) + (pi)/(2)