Right answer is (b) 3.1623
The explanation: Now f(x)=√x,
Hence, f’ (x)=\(\frac{1}{2} x^{-1/2}\)
f” (x)=\(-\frac{1}{4} x^{-3/2}\)
f”’ (x)=\(\frac{3}{8} x^{-5/2}\)
Hence,
f(x+h)=\(\sqrt{x+h}=\sqrt{x}+\frac{h}{2} x^{-1/2}-\frac{h^2}{8} x^{-3/2}+\frac{h^3}{16} x^{-5/2}+…. \) (By Taylor’s expansion)
Putting,
h=1, and x=9 we get,
f(10)=√10=3+1/6-1/216+1/3888+….=3.1623