1.

The approximate value of `(1.0002)^3000`, isA. `1.2`B. `1.4`C. `1.6`D. `1.8`

Answer» Correct Answer - C
Let `y=x^(3000), x=1 and x+Delta x = 1.0002`
Then, `Delta x=1.0002-1=0.0002`
Also,y=1 when x=1
Now,
`y=x^(3000)`
`rArr" "(dy)/dx=3000" "x^(2999) rArr((dy)/dx)_(x=1)=3000`
`:." "Deltay=(dy)/dxDeltax`
`rArr" "Deltay= 3000 xx0.0002 = 6/10=0.6`
Hence
`(1.0002)^(3000)= y+Deltay=1+0.6=1.6`


Discussion

No Comment Found