Question-2;

Consider the function f(x)= -cos3x -4sin3x.
(a)Find the equation of the line normal to the graph of f(x) when x= pie/6 .

(b)Find the x coordinates of the points on the graph of f(x) where the tangent to the graph is horizontal.

(c)Find the absolute extrema of the function f(x)=5+6x^3-3x^4 on the interval [-2,2] .
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Question-3: Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20.
(a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points.

(b) Compute the area enclosed by the two parabolas.

(c) Use Mathcad to draw the graphs of the parabolas and find their intersection points graphically .

(d) Solve the differential equation dy/dx = -exp(5-x)+3/5x^2 ; y(5)=3

Question-2:

(a) To find the equation of the line normal (perpendicular) to the graph of f(x) at x = pi/6, we need to find the derivative of f(x) and use it to determine the slope of the tangent line.

First, let's find the derivative of f(x) using the chain rule:

f'(x) = d/dx(-cos3x) - d/dx(4sin3x)
= 3sin3x + 12cos3x

Now, evaluate f'(x) at x = pi/6 to find the slope of the tangent line at that point:

f'(pi/6) = 3sin(pi/2) + 12cos(pi/2)
= 3(1) + 12(0)
= 3

Since the line normal to the graph is perpendicular to the tangent line, the slope of the normal line will be the negative reciprocal of 3. Therefore, the slope of the normal line is -1/3.

Now that we have the slope and a point (x = pi/6) on the normal line, we can use the point-slope form of a line to find the equation of the line:

y - y1 = m(x - x1)

Plugging in the values:
y - f(pi/6) = -1/3(x - pi/6)

Simplifying:
y + cos(pi/2) + 4sin(pi/2) = -1/3(x - pi/6)
y + 1 = -1/3(x - pi/6)

Therefore, the equation of the line normal to the graph of f(x) at x = pi/6 is y = -1/3(x - pi/6) - 1.

(b) To find the x coordinates of the points on the graph of f(x) where the tangent is horizontal, we need to find where the derivative of f(x) is equal to 0.

Setting f'(x) = 0:

3sin3x + 12cos3x = 0

Using trigonometric identities, we can rewrite this equation:

4cos3x = -sin3x

Dividing both sides by cos3x:

4 = -tan3x

Taking the inverse tangent of both sides:

3x = arctan(-4)

Solve for x:

x = (1/3) * arctan(-4)

Therefore, the x coordinate of the point where the tangent to the graph of f(x) is horizontal is x = (1/3) * arctan(-4).

(c) To find the absolute extrema of the function f(x) = 5 + 6x^3 - 3x^4 on the interval [-2, 2], we need to find the critical points of the function and evaluate the function at the critical points as well as the endpoints of the interval.

First, let's find the derivative of f(x):

f'(x) = 18x^2 - 12x^3

Setting f'(x) = 0 to find the critical points:

18x^2 - 12x^3 = 0
6x^2(3 - 2x) = 0

This equation has two solutions: x = 0 and x = 3/2.

Now, let's evaluate the function at the critical points and the endpoints:

f(-2) = 5 + 6(-2)^3 - 3(-2)^4 = -57
f(0) = 5 + 6(0)^3 - 3(0)^4 = 5
f(2) = 5 + 6(2)^3 - 3(2)^4 = -19
f(3/2) = 5 + 6(3/2)^3 - 3(3/2)^4 = 5.625

Therefore, the absolute maximum value of f(x) on the interval [-2, 2] is 5.625, which occurs at x = 3/2. The absolute minimum value of f(x) is -57, which occurs at x = -2.