Mathematics I (Calculus) 2074
Attempt any three questions.
1. (a) A function is defined by 
, calculate f(-1),f(3), and sketch the graph.(5)
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;
(b) Prove that 
the does not exist.
Given,
Now,
2. (a) Find the derivative of f(x) = √x and to state the domain of f℩
Given
For domain, 
is exist only when x>0.
Thus, domain is (0, ∞).
(b) Estimate the area between the curve y2 = x and the lines x=0 and x=2.
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 
3. (a) Find the Maclaurin series for ex and prove that it represents ex for all x.
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 ex for every value of x.
(b) Define initial value problem. Solve that initial value problem of y' + 5y = 1, y(0) = 2.
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:
(c) Find the volume of a sphere of radius r.
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,
4. (a) For what value of x does the series 
converge?
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.
(b) Calculate ∫ ∫ f(x, y)dA for f(x, y) = 100 − 6x2y and R: 0 ≤ x ≤ 2, −1 ≤ y ≤ 1.
5. If 
and 
, find gof and gog.
Given,
Now,
6. Use continuity to evaluate the limit , 
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.
7. Verify Mean value theorem of f(x) = x3 − 3x + 3 for [−1,2].
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.
8. Sketch the curve y = x3 + x
9. Determine whether the integer 
is convergent or divergent .
We have
Since the limit does not exist as a finite number so it
divergent.
10.Find the length of the arc of the semicubical parabola y2 = x3 between the point(1,1) and (4,8).
Given,
The arc length formula gives
If we substitute 
then 
when x = 4, u = 10.
Therefore,
11. Find the solution of y" + 6y′ + 9 = 0, y(0) = 2, y(0) = 1.
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
12. Test the convergence of the series 
Given series is
The general term of the series is
Here
So, the given series is divergent by D’Alembert ratio test.
13. Define cross product of two vectors .if a=i+3j +4k and b-= 2i+7j=5k, find the vector a × b and b × a.
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,
14. Define limit of a function . find 
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
15. Find the extreme value of f(x, y) = y2 − x2 .
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.