Engineering Mathematics Miscellaneous
- The best approximation of the minimum value attained by e–x sin (100 x) for x > 0 is ______.
-
View Hint View Answer Discuss in Forum
f(x) = e– x sin(100x)
f '(x) = – e– xsin(100x) + e– xcos(100x) × 100
for minima f '(x) = 0
tan(100 x) = 100x = 1 tan-1(100) = 0.0156 100
f(x) = e– 0.0156 sin(100 × 0.0156) = 0.9844Correct Option: A
f(x) = e– x sin(100x)
f '(x) = – e– xsin(100x) + e– xcos(100x) × 100
for minima f '(x) = 0
tan(100 x) = 100x = 1 tan-1(100) = 0.0156 100
f(x) = e– 0.0156 sin(100 × 0.0156) = 0.9844
- 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
-
View Hint View Answer Discuss in Forum
The length of the curve
= 1.8622Correct Option: B
The length of the curve
= 1.8622
- The area enclosed between the parabola y = x2 and the straight line y = x is
-
View Hint View Answer Discuss in Forum
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 = ∫1x = 0 dx ∫y = x²y = x dy = [y]y = x²y = 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 = ∫1x = 0 dx ∫y = x²y = x dy = [y]y = x²y = x dx = (x² - x) dx = x3 - x2 1 3 2 0 = 1 - 1 = 2 - 3 = - 1 3 2 6 6
- Consider the shaded triangular region P shown in the figure. What is ∬P xy dxdy ?
-
View Hint View Answer Discuss in Forum
I = ∬ xy .dxdy
The limit of y is from 0 to 
and limit of x from 0 to 2. 
= 
x 
1 - x 
2 dx 2 2 = 1 x(x² + 4 - 4x).dx 8 = 1 (x³ + 4x - 4x²).dx 8 
Correct Option: A
I = ∬ xy .dxdy
The limit of y is from 0 to and limit of x from 0 to 2. = x 1 - x 2 dx 2 2 = 1 x(x² + 4 - 4x).dx 8 = 1 (x³ + 4x - 4x²).dx 8
- The right circular cone of largest volume that can be enclosed by a sphere of 1 m radius has a height of
-
View Hint View Answer Discuss in Forum
Given R = 1, radins of sphere.
Let height of cone is H = h + RVolume , V = 1 π × (√R² - h²)2(R + h) 3
for maximum value ,⇒ dV = 0 dh ⇒ d 1 (R² - h²)(R + h) dh 3
⇒ -2h(R + h) + (R² - h²) = 0
⇒ (R + h)(R - 3h) = 0h = -R , R 3 Height of the come = R + R = 4R 3 3 = 4 × 1 m = 4 m 3 3
Correct Option: D
Given R = 1, radins of sphere.
Let height of cone is H = h + RVolume , V = 1 π × (√R² - h²)2(R + h) 3
for maximum value ,⇒ dV = 0 dh ⇒ d 1 (R² - h²)(R + h) dh 3
⇒ -2h(R + h) + (R² - h²) = 0
⇒ (R + h)(R - 3h) = 0h = -R , R 3 Height of the come = R + R = 4R 3 3 = 4 × 1 m = 4 m 3 3
