Mathematics I (Calculus) 2074

Given,

Since -1<0, f(-1) = -1 + 2 = 1

Since 3>0, f(3) = 1 - 3 = -2

To draw graph, calculating the points:

For f(x) = x+2 if x<0

    f(-1) = -1+2 =1  ⇒ (-1, 1)

    f(-2) = -2+2 = 0  ⇒ (-2, 0)

    f(-3) = -3+2 = -1  ⇒ (-3, -1) and so on.

For f(x) =1-x if x>0

    f(1) = 1-1 = 0  ⇒ (1, 0)

    f(2) = 1-2 = -1  ⇒ (2, -1)

    f(3)= 1-3 = -2  ⇒ (3, -2) and so on.

Plotting these points of both functions we get;

Given,

Now,

Given

For domain,   is exist only when x>0.

Thus, domain is (0, ∞).

Given

Given equation is the parabola that has the vertex (0, 0) and the line of symmetry is y = 0 with x>=0.

Given line are:

x = 0 & x =2

Sketch of the given curve is:

Area of bounded region 

Now,

Let d is any positive number with  then 

So, by Taylor’s inequality

 for 

Since  is a finite value. So

i.e.

 for all values of x.

 This implies that series converges to e^x for every value of x.

The problem of finding a function y of x when we know its derivative and its value y₀ at a particular point x₀ is called an initial value problem.

Given,

Comparing given equation with we have

P = 5 and Q = 1

Now

Applying the initial condition y(0)=2

Applying this value, we have:

The sphere of radius r can be obtained rotating the half circle graph (semi-circle) of the function  about the x-axis. 

The volume V is obtained as follows:

by the symmetry about the y-axis,

Given series is,

The general term of the series is

So, apply ratio test

The series converges if x-3<1 Þ x<4

Therefore, the series converges for x<4.

Given,

Now,

Since the function  is being a quotient of two continuous functions and   everywhere in their domain. In particular x = 4 and hence the quotient function f(x) is also continuous at x = 4.

Given,

f(x) = x³-3x+3 

Since, f(x) = x³-3x+3  is continuous on [-1, 2] and f’(x) = 3x²-3 so, differentiable on (-1, 2).

Thus f(x) = x³-3x+3 satisfy the both conditions for mean value theorem. So, there exist  such that

Clearly 

Hence, mean value theorem satisfied.

We have

Since the limit does not exist as a finite number so it divergent.

Given,

The arc length formula gives

If we substitute  then  when x = 4, u = 10.

Therefore,

Given

The characteristics equation of given differential equation is

Here the roots are real and equal.

The general solution is

Now, applying the condition y(0)=2

Again, 

Then, 

Applying the condition y’(0)=1

The particular solution of the given equation is

Given series is

The general term of the series is

Here

So, the given series is divergent by D’Alembert ratio test.

If u=(u₁, u₂, u₃) and v=(v₁, v₂, v₃) then the cross product of u and v is a vector

It is also written as

Now,

Given that,

We have,

Thus,

Let f(x, y) be a function of two variables x and y and L be a number. The we say L is the limit of f(x, y) at point (x₀, y₀) if

Now,

[This form is in  as ]

We can find its limit by rewriting it into the form   wherein L'Hospital's rule can be applied if it is applicable.

Applying L'Hospital's rule

Given

f(x, y) = y²-x²

Then

Also

For critical point,

This gives, x=0, y=0.

At point (0, 0)

Here, at point (0, 0)

and

The function has a saddle point at the (0, 0) and no local extreme values.