Question to Indicate Order of Place Value with with an Example

Topic : Place Value

Question : Indicate Order of Place Value in a Number.

Solution :

A Simple Integration Problem

Topic : Integration

Problem : Integrate x^3 ln x



Solution :

Solved Differential Equation

Topic : Differential Equation

Problem : If y = 4 Sin 3x prove y" + 9y = 0

Solution :
y = 4 Sin 3x
differentiate with respect to x
y' = 4 Cos 3x . 3
y' = 12 Cos 3x
Again differentiate with respect to x
y" = 12 (- Sin 3x . 3)
y" = - 36 Sin 3x

Now consider y" + 9y = 0
Substitute the value of y and y"
-36 Sin 3x + 9(4 Sin 3x)
-36 Sin 3x + 36 Sin 3x = 0
Hence proved

Topic : Integration

Question : Solve by Integration using Trignometric Values
∫Cos³ 2x Sin² 2x dx

Solution :
∫Cos³ 2x Sin² 2x dx
∫Cos² 2x Sin² 2x(Cos 2x) dx

Now Sin 2x = 2 Sin x Cos x
Sin 4x = 2 Sin 2x Cos 2x
Sin² 4x = 4 Sin² 2x Cos² 2x
Multiplying and dividing by 4

¼ ∫4 Sin² 2x Cos² 2x(Cos 2x) dx
¼ ∫ Sin² 4x Cos 2x dx
Now Cos 2x = 1 – 2 Sin² x
2 Sin² x = Cos 2x – 1
Sin² x = (Cos 2x – 1) / 2
Sin² 4x = (Cos 8x -1) / 2

So,
¼ ∫ [(Cos 8x -1) / 2] Cos 2x dx
1/8 ∫ (Cos 8x Cos 2x - Cos 2x) dx
1/8 [ ∫ (Cos 8x Cos 2x ) dx - ∫ Cos 2x dx ]

Now 2 Cos A Cos B = Cos (A+B) + Cos (A-B)
1/8 [ ∫ Cos 10x + Cos 6x dx - ∫ Cos 2x dx ]
1/8 {[ (Sin 10x)/10 + (Sin 6x)/6] – (Sin 2x)/2}
1/8 [(3 Sin 10x + 5 Sin 6x – 15 Sin 2x) / 30]
3 Sin 10x + 5 Sin 6x – 15 Sin 2x) / 240

Problem on Multistep Equation

Topic : Multistep Equation

Problem : Solve 7x + 2 = -54

Solution :
7x + 2 = -54
7x = -54-2
7x = -56
dividing both sides by 7
7x/7 = -56/7
x=-8
So the solution is x = -8

Question on Parabola

Topic : Parabola

Question : Natural bridge is a stone arch formed over a river or stream. The longest natural bridge in the world is rainbow bridge in Utah. if the origin is placed at one end of the arch, the cure of the arch can be modeled by the equation h=-0.042d^2+3.57d where h meters represents the height and d meters represents the horizontal distance.
a. what is the max height of the arch, to the nearest meter?
b. at a horizontal distance of 10 m from the vertex, what is the height of the arch, to the nearest meter

Answer :

H(d) = -0.042d^2 + 3.57d
where h(d) represents the height of the arch and d represents the horizontal distance from one end of the base of the arch in meters. Find the maximum height of the arch.
-------------------
Max and min problems use the 1st derivative of the function..
h(d) = -0.042d^2 + 3.57d
Set the 1st derivative to zero.
0 = -0.084d + 3.57
0.084d = 3d = 42.5 meters
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h(42.5) = -0.042*(42.5^2 )+ 3.57*(42.5)= -75 .8625+ 151.725h = 75.8625 meters = 76 appx..
At a horizontal distance of 10 m from the vertex, what is the height of the arch, to the nearest meter
h(d) = -0.042d^2 + 3.57d
h(10) = -0.042*10^2 + 3.57*10
=-4.2+35.7=31.5 meters

Question on Factorisation using identities:

Topic : Factorisation using identities:

Question : Factorise 4y^2 – 12y + 9

Solution :
We know that (a + b)^ 2 = a^2 + 2ab + b^2 …………… (I)
(a – b)^2= a^2 – 2ab + b^2………….. (II)
(a + b) (a – b) = a^2 – b^2………… (III)
let us use these identities for factorisation

Observe 4 y^2 = (2y) 2, 9 = 3^2 and 12y = 2 × 3 × (2y)
Therefore, 4 y^2 – 12y + 9 = (2y)^ 2 – 2 × 3 × (2y) + (3)^2
= (2y – 3)2 (required factorisation)