Engineering Mathematics Miscellaneous Moderate Questions and Answers

Applying Laplace transform

L[teⁿ] =	n!
	Sⁿ⁺¹

L[t²] =	2
	S³

L[2t²] =	4
	S³

F(S) = L[e^-t.2t²] =	4
	(S + 1)³

Putting s = 1 we get

F(1) =	4	=	4	= 0.5
	(1 + 1)³		8

Correct Option: B

Applying Laplace transform

L[teⁿ] =	n!
	Sⁿ⁺¹

L[t²] =	2
	S³

L[2t²] =	4
	S³

F(S) = L[e^-t.2t²] =	4
	(S + 1)³

Putting s = 1 we get

F(1) =	4	=	4	= 0.5
	(1 + 1)³		8

Here, volume of solid of rotation by rotating shaded area by 360° around x-axis is,
V = ¹∫₀πy²dx

V = ¹∫₀ πx dx =	πx²		²	=	π
	2		₀		2

Correct Option: B

Here, volume of solid of rotation by rotating shaded area by 360° around x-axis is,
V = ¹∫₀πy²dx

V = ¹∫₀ πx dx =	πx²		²	=	π
	2		₀		2

I = ^e∫₁x In xdx

=	In x²	- ∫		1	×	x²		dx
	2			x		2

=	x² In x	-	x²		^e
	2		₁		4

=		e²	-	e²		+	1
		2		4			4

=	e² + 1	= 2.097
	4

Correct Option: C

I = ^e∫₁x In xdx

=	In x²	- ∫		1	×	x²		dx
	2			x		2

=	x² In x	-	x²		^e
	2		₁		4

=		e²	-	e²		+	1
		2		4			4

=	e² + 1	= 2.097
	4

= 0 + 0 + (x² + y²)k̂
n̂ = k̂
∇ × u). n̂ = (∇ × v). k̂ = x² + y²
∫∫(x² + y²)dx dy
^{2π^{∫₀^{1^{∫₀r²(rdrdθ) = π/2}}}}

Correct Option: C

= 0 + 0 + (x² + y²)k̂
n̂ = k̂
∇ × u). n̂ = (∇ × v). k̂ = x² + y²
∫∫(x² + y²)dx dy
^{2π^{∫₀^{1^{∫₀r²(rdrdθ) = π/2}}}}

y" + 2y' + y = 0
A.E is, D² + 2D + 1 = 0
⇒ (D + 1)² = 0
⇒ m = – 1
The C.F. is (C₁ + C₂ x)e^–x and P.I. = 0
Now y(0) = 0
⇒ C₁ = 0 a
nd y(1) = 0
⇒ C₂ e^–1 = 0,
⇒ C₂ = 0
⇒ C.F. = 0
⇒ y(0.5) = 0

Given a vector U =	1	(-y³î + x³ĵ + z³̂k)
	3