Find f (x) if `f(x)=lambdaint_(0)^(pi//2)sinxcostf(t)dt+sinx` where `lambda` is constant `ne 2`.

f(x) satisfies the relation f(x)-lambdaint_0^(pi//2)sinx*costf(t)dt=sinx If lambda > 2 then f(x) decreases in

f(x) satisfies the relation f(x)-lambdaint_0^(pi//2)sinx*costf(t)dt=sinx If lambda > 2 then f(x) decreases in

f(x) satisfies the relation f(x)-lamda int_(0)^(pi//2)sinxcostf(t)dt=sinx If f(x)=2 has the least one real root, then

f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt The range of f(x) is

f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt f(x) is not invertible for

Let A={x: 0 le x lt pi//2} and f:R to A be an onto function given by f(x)=tan^(-1)(x^(2)+x+lambda) , where lambda is a constant. Then,

f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt The value of int_(0)^(pi//2) f(x)dx is

If f(x) =int_(0)^(x) sin^(4)t dt , then f(x+2pi) is equal to

Find the points of minima for f(x)=int_0^x t(t-1)(t-2)dt

Find the points of minima for f(x)=int_0^x t(t-1)(t-2)dt