Engineering Mathematics Miscellaneous Easy Questions and Answers | Page - 1

Engineering Mathematics Miscellaneous

At x = 0, the function f(x) = x³ + 1 has

1. a maximum value
1. a minimum value
1. a singularity
1. a point of inflection

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Function f(x) = x ³ + 1 has a point of inflection at x = 0, since in the graph sign of the curvature (i.e., concavity) is changed.

Correct Option: D

Function f(x) = x ³ + 1 has a point of inflection at x = 0, since in the graph sign of the curvature (i.e., concavity) is changed.

The area enclosed between the parabola y = x² and the straight line y = x is

1. 1
  8
1. 1
  6
1. 1
  3
1. 1
  2

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Given equations are
y = x² ...(i)
y = x ...(ii)
From equations (i) and (ii)
x ² – x = 0
⇒ x(x – 1) = 0
⇒ x = 0, 1

Area enclosed = dydx = ∫¹_{x = 0} dx ∫^{y = x²}_{_{y = x}} dy

= [y]^{y = x²}_{_{y = x}} dx = (x² - x) dx

= x³ - x² ¹
3 2 ₀

= 1 - 1 = 2 - 3 = - 1
3 2 6 6

Correct Option: B

Given equations are
y = x² ...(i)
y = x ...(ii)
From equations (i) and (ii)
x ² – x = 0
⇒ x(x – 1) = 0
⇒ x = 0, 1

Area enclosed = dydx = ∫¹_{x = 0} dx ∫^{y = x²}_{_{y = x}} dy

= [y]^{y = x²}_{_{y = x}} dx = (x² - x) dx

= x³ - x² ¹
3 2 ₀

= 1 - 1 = 2 - 3 = - 1
3 2 6 6

Consider a spatial curve in three-dimensional space given in parametric form by x(t) = cos t, y(t) = sin t,
z(t) = 2 t , 0 ≤ t ≤ π The length of the curve is ___________.
π 2

1. 0.8622
1. 1.8622
1. 11.8622
1. 1.81622

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The length of the curve

= 1.8622

Correct Option: B

The length of the curve

= 1.8622

The best approximation of the minimum value attained by e^–x sin (100 x) for x > 0 is ______.

1. 0.9844
1. 1.9844
1. 11.91844
1. None of these

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f(x) = e^{– x} sin(100x)
f '(x) = – e^{– x}sin(100x) + e^{– x}cos(100x) × 100
for minima f '(x) = 0
tan(100 x) = 100
x = 1 tan^-1(100) = 0.0156
100

f(x) = e^{– 0.0156} sin(100 × 0.0156) = 0.9844

Correct Option: A

f(x) = e^{– x} sin(100x)
f '(x) = – e^{– x}sin(100x) + e^{– x}cos(100x) × 100
for minima f '(x) = 0
tan(100 x) = 100
x = 1 tan^-1(100) = 0.0156
100

f(x) = e^{– 0.0156} sin(100 × 0.0156) = 0.9844

In the Taylor series expansion of ex about x = 2 the coefficient of (x – 2)⁴ is

1. 1 / 4!
1. 24 / 4!
1. e² / 4!
1. e⁴ / 4!

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Taylor series expansion of f(x) about a is given by
f(x) = f(a) + (x - a) f '(a) + (x - a)² f ''(a) + .......
1! 2!

coefficient of (x - a)⁴ is f ''''(a)
4!

Now f(x) = e^x
⇒ f ''''(x) = e^x
⇒ f ''''(a) = e^a
Hence for a = 2,
f ''''(a) = e²
4! 4!

Correct Option: C

Taylor series expansion of f(x) about a is given by
f(x) = f(a) + (x - a) f '(a) + (x - a)² f ''(a) + .......
1! 2!

coefficient of (x - a)⁴ is f ''''(a)
4!

Now f(x) = e^x
⇒ f ''''(x) = e^x
⇒ f ''''(a) = e^a
Hence for a = 2,
f ''''(a) = e²
4! 4!