(A) Given f(x)=x2+10sinx∀x∈R
Also, f(0)=0 and f(x) is continuous on R. x→±∞Limf(x)→∞⇒ there exists some real number c such that f(x)=1000
(Using intermediate value theorem).
(B) Let f(x)=∣∣x2+x∣∣=∣x(x+1)∣
So, f(x) is non-derivable at two points viz x=−1,0.
(C) If y=f(x) and x=g(y) where g=f−1, then dy2d2x=−(dxdy)3dx2d2y,dxdy=0.
(D) As f(x) is continuous on (0,5) and f(x) takes only irrational values such that f(2)=π, so f(π)=π( As f(x) must be only constant function).