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TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Physics

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Question : 107 of 160

Marks: +1, -0

The approximate value of y =

(1.01)^3

(1.01)^{\frac{3}{2}}

+ 5 is

8.06
8.04
8.02
8.16

Solution:

Assume the function be,

y = f(x)

=x^{3}+2 x^{\frac{3}{2}}+5

Differentiate the above equation with respect to x .

f'(x) = 3x^{2}+2 \cdot \frac{3}{2} x^{\frac{1}{2}}

=3x^{2}+3x^{\frac{1}{2}}

\text{Let } x = 1

and

\Delta x = 0.01

f(1) = 1^{3}+2(1)^{\frac{3}{2}}+5

=8

And,

\Delta y = f'(x) \cdot \Delta x

= \left(3(1)^{2}+3(1)^{\frac{1}{2}}\right) \cdot (0.01)

=0.06

The given function is calculated as,

f(x + \Delta x) = f(x) + \Delta y

= f(1) + \Delta y

=8+0.06

=8.06

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