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According to relativity theory, an observer who measures the duration of an event
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# Question : According to relativity theory, an observer who measures the duration of an event : 2151720

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Find the first four nonzero terms of the Maclaurin expansion of the given function.

1) f(x) = sin6x

A) 6x - (216x3/3!) + (7776x5/5!) - (279,936x7/7!) + ...

B) 6x - 216x3 + 7776x5 - 279,936x7 + ...

C) 6x - (6x3/3!) + (6x5/5!) - (6x7/7!) + ...

D) 6x - (36x2/2!) + (216x3/3!) - (1296x4/4!) + ...

2) f(x) = e5x

A) 1 + 5x + (5x2/2!) + (5x3/3!) + ...

B) 1 + 5x + (25x2/2!) + (125x3/3!) + ...

C) 5x + 125x3 + 3125x5 + 78,125x7 + ...

D) x + 5x + (5x2/2!) + (5x3/3!) + ...

3) f(x) = cosx5

A) 1 + 5x + (5x2/2!) + (5x3/3!) + ...

B) 1 - x5 + (2x10/3) - (x20/120) + ...

C) x + 5x + (5x2/2!) + (5x3/3!) + ...

D) 1 - (x10/2!) + (x20/4!) - (x30/6!) + ...

4) f(x) = xsin3x

A) 3x2 + (9/2)x4 + (81/40)x6 + (243/560)x8 + ...

B) 3x - (9/2)x3 + (27/8)x5 - (729/80)x7 + ...

C) 3x - (9/2)x4 + (243/40)x6 - (729/560)x8 + ...

D) 3x2 - (9/2)x4 + (81/40)x6 - (243/560)x8 + ...

5) f(x) = √(1 + x3)

A) 1 + (1/2)x3 - (1/8)x6 + (1/16)x9 - ...

B) 1 + x - (1/4)x2 + (1/8)x3 - ...

C) 1 + (3/2)x - (3/8)x2 + (3/16)x3 - ...

D) 1 + (3/2)x3 - (3/8)x6 + (3/16)x9 - ...

6) f(x) = √(1 - x6)

A) 1 + (3/2)x - (3/8)x2 + (3/16)x3 - ...

B) 1 - (1/2)x6 - (1/8)x12 - (1/16)x18 - ...

C) 1 - (1/2)x6 + (1/8)x12 - (1/16)x18 - ...

D) 1 + (3/2)x6 - (3/8)x12 + (3/16)x18 - ...

7) f(x) = ln(1 + x3)

A) x3 - (1/2)x6 + (1/6)x9 - (1/24)x12 + ...

B) (1/3)x3 - (1/6)x6 + (1/9)x9 - (1/12)x12 + ...

C) x3 + (1/2)x6 - (1/6)x9 + (1/24)x12 + ...

D) x3 - (1/2)x6 + (1/3)x9 - (1/4)x12 + ...

8) f(x) = exsinx

A) 1 + x + (1/3)x2 - (1/30)x6 - ...

B) x + x2 + (1/3)x3 - (1/30)x5 - ...

C) 1 + x2 + (1/3)x3 + (1/30)x5 + ...

D) 1 + (3/2)x + (3/8)x3 - (3/16)x5 - ...

9) f(x) = x5ln(1 - x)3

A) x6 + (1/2)x7 + (1/3)x8 + (1/4)x9 + ...

B) -3x6 - (3/2)x7 - (1/2)x8 - (1/8)x9 - ...

C) 3x5 - (3/2)x10 - x15 + (3/4)x20 + ...

D) -3x6 - (3/2)x7 - x8 - (3/4)x9 - ...

Evaluate the given integral by using three terms of the appropriate series, rounding the result to four decimal places.

10) A) 0.0721

B) 0.0368

C) 0.1921

D) 0.0768

11) A) 4.7034

B) 0.0034

C) 2.4034

D) 0.2034

12) A) 0.2072

B) 0.2859

C) 0.2318

D) 0.2546

13) A) 0.3208

B) 0.7608

C) 0.3809

D) 0.6428

14) A) 0.5187

B) 0.4013

C) 0.3791

D) 0.3815

15) A) 0.5759

B) 0.2215

C) 0.5725

D) 0.1795

16) A) 0.8835

B) 0.6653

C) 0.4705

D) 0.3069

Find the indicated series by the given operation.

17) Find the first four nonzero terms of the expansion of the function f(x) = (1/2)(ex + e-x) by adding the terms of the appropriate series. The result is the series for cosh x.

A) x + (1/2)x2 + (1/24)x4 + (1/720)x6 + . . .

B) 1 + (1/2)x2 + (1/24)x3 + (1/720)x4 + . . .

C) x + (1/2)x2 + (1/24)x3 + (1/720)x5 + . . .

D) 1 + (1/2)x2 + (1/24)x4 + (1/720)x6 + . . .

18) Find the first four nonzero terms of the expansion of the function f(x) = (7 + 6x/1 + x) by rearranging the expression to get f(x) = (1/1 + x) + k, where k is a constant, and adding k to the series for (1/1 + x).

A) 6 + x - x2 + x3 - . . .

B) 1 - x + x2 - x3 + . . .

C) 7 - x + x2 - x3 + . . .

D) 6 - x + x2 - x3 + . . .

19) Find the first three terms of the expansion for f(x) = e-xsin(-x) by multiplying the proper expansions together, term by term.

A) -x + x2 - (1/3) x3 + . . .

B) 1 - x + x2 - . . .

C) x - x2 + (1/3) x3 - . . .

D) x + x2 - (1/3) x3 + . . .

20) Find the first three terms of the expansion for f(x) = excosx by multiplying the proper expansions together, term by term.

A) 1 + x - (1/3) x3 - . . .

B) x - x2 + (1/3) x3 - . . .

C) 1 + x - (1/2) x2 - . . .

D) x + x2 - (1/3) x3 + . . .

21) Find the first three nonzero terms of the expansion for f(x) = secx by dividing 1 by the series for cosx.

A) 1 + (1/2)x2 - (5/24)x4 + . . .

B) 1 + (1/2)x2 + (5/24)x4 + . . .

C) x + (1/3)x3 - (1/120)x5 + . . .

D) -x + (1/3)x3 - (1/120)x5 + . . .

22) Show that by differentiating term by term the expansion of f(x) = (1/(1 + x)2), the result is the expansion for f(x) = - (2/(1 + x)3).

A) (d/dx)(2x + 3x2 - 4x3 + . . .) = -2(1 - 3x + 6x2 - . . .)

B) (d/dx)(1 + 2x - 3x2 + 4x3 - . . .) = -2(1 + 3x - 6x2 + . . . )

C) (d/dx)(1 - 4x + 6x2 - 8x3 + . . .) = -2( 2 - 6x + 12x2 - . . .)

D) (d/dx)(1 - 2x + 3x2 - 4x3 + . . .) = -2(1 - 3x + 6x2 - . . .)

23) Show that by differentiating term by term the expansion of ln(1 + x2), the result is the expansion for (2x/1 + x2).

A) (d/dx)(2x - x2 + x4 - (1/4)x6 + . . .) = 2(1 - x + 2x3 - (3/4)x5 + . . .)

B) (d/dx)(x2 - (1/2)x4 + (1/3)x6 - (1/4)x8 + . . .) = 2x(1 - x2 + x4 - x6 + . . . )

C) (d/dx)(2x + x2 - x4 + (1/4)x6 - . . .) = 2(1 + x - 2x3 + (3/4)x5 - . . .)

D) (d/dx)(x2 + (1/2)x4 - (1/3)x6 + (1/4)x8 - . . .) = 2x(1 + x2 - x4 + x6 - . . .)

24) Show that by integrating term by term the expansion of - (2x/1 - x2), the result is the expansion for ln (1 - x2).

A) = -x2 + (1/2)x4 - (1/3)x6 + (1/4)x8 - . . .

B) = -x2 - (1/2)x4 - (1/3)x6 - (1/4)x8 - . . .

C) = x2 + (1/2)x4 + (1/3)x6 + (1/4)x8 + . . .

D) = x2 - (1/2)x4 + (1/3)x6 - (1/4)x8 + . . .

Solve the problem.

25) According to relativity theory, an observer who measures the duration of an event, such as the decay of an unstable subatomic particle, will obtain a reading that is affected by any motion of the observer relative to the location of the event. The time, Δtr, measured by an observer moving at a speed of v relative to the location of the event will be larger than the time, Δto, measured by an observer who is not in relative motion. This phenomenon is described by the equation (Δtr/Δto) = (1/√(1 - (v/c) 2)), where c is the speed of light. If v/c = 0.30, compute Δtr/Δto using the first four nonzero terms of the series expansion for Δtr/Δto.

A) 1.18327

B) 1.04827

C) 1.09327

D) 1.04781

26) Evaluate (x→0 is under lim) (1 - cosx/x) by using the series expansion for cosx.

A) (1/2)

B) 1

C) 0

D) -1

27) Find the approximate value of the area bounded by y = x2ex2, x = 0.5, and the x-axis by using three terms of the appropriate Maclaurin series.

A) 0.14100

B) 0.04847

C) 0.04484

D) 0.03597

28) Find the approximate area bounded by y = exsin4x, x = 0, x = 0.4, and the x-axis by using the first three nonzero terms of the appropriate series.

A) 1.36533

B) 2.79467

C) 0.34987

D) 1.13067

29) The dome of a sports arena is designed as the surface generated by revolving the curve of y = 20.0cos0.00183x (0 ≤ x ≤ 75.0 m) about the y-axis. Find the volume within the dome by using the first three nonzero terms of the appropriate series.

A) 352,000 m3

B) 355,000 m3

C) 357,000 m3

D) 705,000 m3

30) The function f(x) = e-bx occurs frequently in the descriptions of physical processes. It is often present as a damping factor. For example, a sinusoidally varying quantity, such as f(t) = cos(at), might be damped with the passing of time, t, according to the relation f(t) = e-bt cosat. Write three terms of the series expansion for this function, taking b = 0.2 and a = 4.

A) 1 + 0.8t - 2.96t2 + ...

B) 1 - 0.2t - 7.98t2 + ...

C) 1 + 0.2t - 7.98t2 + ...

D) 1 - 0.8t + 2.96t2 + ...

31) The addition of sinusoidal functions is one way of describing the interaction of some physical entities, such as light beams, sounds, and radio signals. Write the first four terms of the series expansion for the sum 6cos3x + sin3x.

A) 6x - 3x2 + 27x3 + (9/2)x4 + ...

B) 6 + 3x - 27x2 - (9/2)x3 + ...

C) 6x + 6x2 - 27x3 - (9/2)x4 + ...

D) 6x + 3x2 - 27x3 - (9/2)x4 + ...

32) The displacement y (in cm) of an object hung vertically from a spring and allowed to oscillate is given by the equation y = 9e-0.4tsint, where t is the time (in s). By multiplication of series, find the first three terms of the expansion for the displacement of the oscillating object.

A) 9t + 1.8t2 - 1.02t3 + . . .

B) 9t + 3.6t2 - 0.78t3 + . . .

C) 9t + 7.2t2 - 1.56t3 + . . .

D) 9t + 3.6t2 + 0.72t3 + . . .

33) A resistor and inductor are connected in series to a battery. The current in the circuit (in A) is given by I = 2.4(1 - e-0.4), where t is the time since the circuit was closed. By using the series for ex, find the first three terms of the expansion of this function.

A) 2.4t + 0.192t2 + 0.0256t3 + . . .

B) 0.96t - 0.192t2 + 0.0512t3 + . . .

C) 0.96t - 0.48t2 + 0.16t3 + . . .

D) 0.96t - 0.192t2 + 0.0256t3 + . . .

## Solution 5 (1 Ratings )

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