For the function f(x)= cases x+4 2 ,&x 0 x+1 ,&x>0 cases the range of is

MATHEMATICS · Single correct · CBSE · Mathematics · Relations and Functions · Relations and Functions

Problem

For the function f(x)={x+42,x≤0x+1,x>0f(x)=\begin{cases}\dfrac{x+4}{2},&x\leq0\\\sqrt{x+1},&x>0\end{cases} the range of ff is (a) (−∞,2](-\infty,2] (b) (1,2](1,2] (c) R\mathbb{R} (d) [1,∞)[1,\infty)

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