`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)-lambda int_(0)^( pi/2)sin x*cos tf(t)dt=sin x If lambda>2 then f(x) decreases in

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

Find f(x) which satisfies f(x)=x+int_0^(pi/2) sinx*cosy*f(y)dy

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 function f(x), that satisfies the condition f(x) = x + int_0^(pi//2) sinx . Cosyf(y) dy, is :

The function f(x), that satisfies the condition f(x) = x + int_0^(pi//2) sinx . Cosyf(y) dy, is :

y=f(x) satisfies the relation int_(2)^(x)f(t)dt=(x^(2))/2+int_(x)^(2)t^(2)f(t)dt The value of x for which f(x) is increasing is

A derivable function f(x) satisfies the relation f(x)=int_(0)^(1)xf(t)dt+int_(0)^(x)x^(2)f(t)dt. The value of (2f'(1))/(f(1)) is

y=f(x) satisfies the relation int_(2)^(x)f(t)dt=(x^(2))/2+int_(x)^(2)t^(2)f(t)dt The range of y=f(x) is