A function `f` is defined by `f(x)=int_(0)^(x)(dt)/(1+t^(2))` The normal line to `y=f(x)` at `x=1` has `x` -intercept equal to `X` and `y` -intercept equal to `Y` then

If f(x)=int_(-1)^(x)|t|dt , then for any x ge0,f(x) is equal to

A Function f(x) satisfies the relation f(x)=e^(x)+int_(0)^(1)e^(x)f(t)dt* Then (a)f(0) 0

The intervals of increase of f(x) defined by f(x)=int_(-1)^(x)(t^(2)+2t)(t^(2)-1)dt is equal to

If f(x) =int_(x)^(-1) |t|dt , then for any x ge 0 , f(x) equals

Let f:(0,oo)R be a continuous,strictly increasing function such that f^(3)(x)=int_(0)^(0)tf^(2)(t)dt. If a normal is drawn to the curve y=f(x) with gradient (-1)/(2), then find the intercept made by it on the y-axis is 5(b)7(c)9(d)11

If f(x)=int_(-1)^(x)|t|dt , then for any xge0,f(x) equals

If f(x)=cos x-int_(0)^(x)(x-t)f(t)dt, then f'(x)+f(x) equals