# Errors and Approximations || V.S.A.Q’S||

## Errors and Approximations || V.S.A.Q’S||

These solutions designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

## Errors and Approximations

Question 1

Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions

(i) y = f(x) = x2 + x at x = 10 when ∆x = 0.1.

Sol:

Given y = f(x) = x2 + x at x = 10, ∆x = 0.1

∆y = f (x + ∆x) – f (x)

= f (10 + 0.1) – f (10)

= f (10.1) – f (10)

= (10. 1)2 + 10.1 – (102 + 10)

= 102.01 + 10.1 – 100 – 10

= 112.11 – 110

= 2.11

dy = f’ (x) ∆x

= (2x + 1) (0.1)

= [2(10) + 1] (0.1)

= 21 × 0.1

= 2.1

(ii)  y = cos x at x = 600 with ∆x = 10 (10 = 0.0174 radians)

Sol:

Given y = cos x, x = 600 and ∆x = 10

∆y = f (x + ∆x) – f (x)

= cos (600 + 10) – cos 600

= cos (610) – 0.5

= 0.4848 – 0.5

= – 0.0152

dy = f’ (x) ∆x

= – sin x (10)

= – sin 600 × 0.0174

=– 0.8660 × 0.0174

= – 0.0150

(iii)  y = x2 + 3x + 6, x = 10 with ∆x = 0.01

Sol:

y = x2 + 3x + 6

∆y = f (x + ∆x) – f (x)

= f (10 + 0.01) – f (10)

= f (10.01) – f (10)

= (10.01)2 + 3 (10.01) + 6 – (102 + 3 (10) + 6)

= 100. 2001 + 30.03 + 6 – 100 – 30 – 6

=130. 2301 – 130

= 0.2301

dy = f’ (x) ∆x

= (2x + 3 + 0) (0.01)

= (2× 10 + 3) (0.01)

= 23 × 0.01

= 0.23

Sol: Ts Inter Maths IA Concept

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