Let f(h) be a function continuous ∀ h∈R−{0} such that f′(h)<0, ∀ h∈(−∞,0) and f′(h)>0, ∀ h∈(0,∞). If limh→0+f(h)=3, limh→0−f(h)=4 and f(0)=5, then the image of the point (0,1) about the line, y⋅limh→0f(cos3h−cos2h)=x⋅limh→0f(sin2h−sin3h), is

Let f(h) be a function continuous ∀ h∈R−{0} such that f′(h)<0,

1.	Let f(h) be a function continuous ∀ h∈R−{0} such that f′(h)<0, ∀ h∈(−∞,0) and f′(h)>0, ∀ h∈(0,∞). If limh→0+f(h)=3, limh→0−f(h)=4 and f(0)=5, then the image of the point (0,1) about the line, y⋅limh→0f(cos3h−cos2h)=x⋅limh→0f(sin2h−sin3h), is
Answer» Let $f (h)$ be a function continuous $\forall h \in R - {0}$ such that $f^{'} (h) < 0, \forall h \in (- \infty, 0)$ and $f^{'} (h) > 0, \forall h \in (0, \infty) .$ If $lim h \to 0^{+} f (h) = 3, lim h \to 0^{-} f (h) = 4$ and $f (0) = 5,$ then the image of the point $(0, 1)$ about the line, $y \cdot lim h \to 0 f ({cos}^{3} h - {cos}^{2} h) = x \cdot lim h \to 0 f ({sin}^{2} h - {sin}^{3} h),$ is