Let f(x) be a function continuous ∀ x∈R−{0} such that f′(x)<0, ∀ x∈(−∞,0) and f′(x)>0, ∀ x∈(0,∞). If limx→0+f(x)=3, limx→0−f(x)=4 and f(0)=5, then the image of the point (0,1) about the line, y⋅limx→0f(cos3x−cos2x)=x⋅limx→0f(sin2x−sin3x), is

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

1.	Let f(x) be a function continuous ∀ x∈R−{0} such that f′(x)<0, ∀ x∈(−∞,0) and f′(x)>0, ∀ x∈(0,∞). If limx→0+f(x)=3, limx→0−f(x)=4 and f(0)=5, then the image of the point (0,1) about the line, y⋅limx→0f(cos3x−cos2x)=x⋅limx→0f(sin2x−sin3x), is
Answer» Let $f (x)$ be a function continuous $\forall x \in R - {0}$ such that $f^{'} (x) < 0, \forall x \in (- \infty, 0)$ and $f^{'} (x) > 0, \forall x \in (0, \infty) .$ If $lim x \to 0^{+} f (x) = 3, lim x \to 0^{-} f (x) = 4$ and $f (0) = 5,$ then the image of the point $(0, 1)$ about the line, $y \cdot lim x \to 0 f ({cos}^{3} x - {cos}^{2} x) = x \cdot lim x \to 0 f ({sin}^{2} x - {sin}^{3} x),$ is