# CALCULUS

###### General motors
November 18, 2018
###### Career Focus
November 18, 2018

CALCULUS

1. Let f(x) = 2x – 1
3x – 1
.
(i) Write down the domain of f. 
(ii) Find the composite functions f ? f and f ? f ? f, and their domains. 
2. For each of the following, evaluate the limit. (L’Hˆopital’s rule is not allowed.)
(a) lim
x?- 1
2
4x
3 – 3x – 1
4x
4 + 4x
3 + 5x
2 + 4x + 1
. 
(b) limx?2
v
x
2 + 1 –
v
2x + 1
v
x
3 – x
2 –
v
x + 2
. 
(c) limx?0
(b-xc + b2x + pc + b3xc). 
(d) limx?0
sin x + sin 3x + sin 5x
sin 2x + sin 4x + sin 6x
. 
3. Prove the following limits using only the precise definition of limits.
(a) limx?-2
x
3 = -8. 
(b) limx?-1
1
v
x
2 + 3
=
1
2
. 
(c) limx?8
3x + 2
2x + 3
=
3
2
. 
4. Let P 6= O be a point on the parabola y = x
2 and Q be the point where the perpendicular
bisector of OP intersects the y-axis. As P approaches the origin O along the
parabola, does Q have a limiting position? If so, find it. 

5-mark

5. Prove that the equation 
x
5 – 1102x
4 – 2015x = 0
has at least three real roots.
6. Using the definition of derivative, prove that 
d
dx
1
v
x
2 + 2
= –
x
(x
2 + 2)3/2
.
7. Let 
f(x) = (
ax3
cos(1/x) + bx + b, if x < 0, v a + bx, if x = 0, where a, b are positive real constants. Determine the values of a and b, if any, that make f differentiable everywhere. 8. Find dy dx and d 2 y dx2 at (x, y) = (1, -2) if  2x 2 + 2xy + xy2 - 3x + 3y + 7 = 0. (You may assume that the derivatives exist at (1, -2) without proof.) 9. Find the absolute maximum and minimum values of  f(x) = (x 2 - x - 5)x 2/5 , -1 = x = 2, and indicate the values of x where they occur. 10. Suppose that limx?a f(x) = -8 and limx?a g(x) = c, where c is a constant. Prove that  limx?a (f(x) + g(x)) = -8. 11. Prove that if the curve y = x 3 + px + q is tangent to the x-axis, then  4p 3 + 27q 2 = 0. 12. Let f be a nonconstant function that is continuous on [a, b], where a < b. Prove that the range of f is a finite closed interval [c, d], where c < d.  PLACE THIS ORDER OR A SIMILAR ORDER WITH US TODAY AND GET A GOOD DISCOUNT 🙂  