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 ]
$=16\times \frac{3.005}{3}+64\times 1.0025$
$=80.186$