Question

The approximate value of $(8.01)^{\frac{4}{3}}+(8.01)^2$ upto 3 decimal places is

• A. $80.116$
• B. $80.216$
• C. 80
• D. $80.186$

Answer 1

Answer: (D)

$ (8.01)^{4 / 3}+(8.01)^2$
$=(8+0.01)^{4 / 3}+(8+0.01)^2$
$ =8^{4 / 3}\left(1+\frac{0.01}{8}\right)^{4 / 3}+8^2\left(1+\frac{0.01}{8}\right)^2 $
$=16(1+0.00125)^{4 / 3}+64(1+0.00125)^2 $
$ =16\left(1+\frac{4}{3} \times 0.00125\right)+64(1+2 \times 0.00125)$
[Using binomial expansion and neglecting higher order terms ]
$ =16 \times \frac{3.005}{3}+64 \times 1.0025$
$=80.186$