If `f(x)=(2018x-2019)/(x+lambda)` and `f(f(x))=x` then `lambda=`

If f(x)=x+(1)/(x) , such that [f(x)]^(3)=f(x)^(3)+lambdaf((1)/(x)) , then lambda=

If range of f(x)=(2x^(2)+lambda)/(x^(2)+2) is [2,6] then (lambda)/(4)=

If range of the function f(x)=(2x^(2)+lambda)/(x^(2)+2) is (2,6] then (lambda)/(2) is-

If f(x)=x+(1)/(x) , such that f^3 (x)=f(x^(3))+lambdaf((1)/(x)) , then lambda=

If f(x)=x+(1)/(x) , such that f^3 (x)=f(x^(3))+lambdaf((1)/(x)) , then lambda=

Let f(x)=(3)/(4)x+1,f^(n)(x) be defined as f^(2)(x)=f(f(x)), and for n>=2,f^(n+1)(x)=f(f^(n)(x)). If lambda=(lim)_(n rarr oo)f^(n)(x), then (a)lambda is independent of x( b) lambda is a linear polynomial in 4y=lambda touches the unit circle 0. (d)the line 4y=lambda touches the unit circle with centre at the origin.

If 2^(f(x)) = (2+x)/(2-x), x in (-2, 2) and f(x) = lambda f((8x)/(4+ x^(2))) then value 'lambda' will be

f:R rarr R is defined by f(x)={(cos3x-cos x)/(x^(2)),x!=0 lambda,x=0 and f is continuous at x=0; then lambda=

If f(x)=|(x,lambda),(2lambda,x)| , then f(lambdax)-f(x) is equal to