From mean value theoren : `f(b)-f(a)=(b-a)f^(prime)(x_1); a lt x_1 lt b` if `f(x)=1/x` , then `x_1` is equal to

From mean value theoren :f(b)-f(a)=(b-a)f'(x_(1));a

If, from mean value theorem , f'(x_(1))=(f(b)-f(a))/(b-a), then:

If A={x:(pi)/(6) lt x lt (pi)/(3)} and f(x) =cosx-x(1+x), then f(A) is equal to :

In the Mean Value theorem (f(b)-f(a))/(b-a)=f'(c) if a=0,b=(1)/(2) and f(x)=x(x-1)(x-2) the value of c is

If from Largrange's mean value theorem, we have f(x'(1))=(f'(b)-f(a))/(b-a) then,

If f(x)=log_(e)((1-x)/(1+x)),|x| lt 1, " then " f((2x)/(1+x^(2))) is equal to

If the function f(x)={{:(,-x,x lt 1),(,a+cos^(-1)(x+b),1 le xle 2):} is differentiable at x=1, then (a)/(b) is equal to