2017-2-PEN-JIT SIN
Section A [45 marks]
Answer all questions in this section.
1. Evaluate
(a) lim
x →
( x + 6x − x )
[3 marks]
x2 + 4
x−2
[3 marks]
(b) lim
x →−
2
2. The point P(x, y) moves along a curve of which equation is given by (2 x + y)2 + 3( x − y)2 = 16 .
Find
(a) the gradient of the curve at P.
(b) the relationship between x and y if the normal at P is parallel to the x-axis.
[3 marks]
[3 marks]
(c) the equation of the normal at P where the curve cuts the negative y-axis.
[3 marks]
3. (a) By using the substitution u = 2 + x , find
e 2+ x
dx .
2+ x
[3 marks]
Hence, using integration by parts, show that
e
2+ x
dx = 2e 2+ x
( 2 + x −1) + C.
[4 marks]
1
. The region R is bounded by
4
the curve, the y-axis and the lines y = −2 , y = 2 . Using the result in (a), find the area of R
in exact form.
[3 marks]
y
(b) The diagram below shows the graph of y = ( ln 4 x ) − 2, x
2
x
0
1
, −2
4
4. (a) Find the general solution of the differential equation
d2 y
(4 − 2 x)3 2 = 1 ,
dx
giving your answer in the form y = f ( x) .
(b) The variables x and y are connected by the differential equation
By using the substitution u = y x , show that
[3 marks]
dy y 1 3
+
= +
.
dx 2 x 2 2 x
du 1
3
.
=
x+
dx 2
2 x
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[3 marks]
5. Given that y = (sin −1 x)2 , show that
2
dy
[3 marks]
(1 − x 2 ) = 4 y
dx
d2 y
dy
and (1 − x 2 ) 2 − x = 2 .
[2 marks]
dx
dx
By further differentiation of these results, find the Maclaurin series of y up to including the term
in x 4 .
[4 marks]
6. Use the trapezium rule with five ordinates to estimate, to three decimal places, the value of
3
0
0.75 + cos 2 x dx .
[5 marks]
Section B [15 marks]
Answer only one question in this section.
7. The function f is defined by f ( x) = x 4 e 2 x for xR.
(a) Find the stationary points and determine the nature of the stationary points.
[7 marks]
(b) Determine the interval(s) on the domain of f where f is concave upwards.
[5 marks]
[3 marks]
(c) Sketch the curve y = f(x).
8. (a) Solve the differential equation e x cos x −
dy
= y tan x . Express y in term of x.
dx
[7 marks]
(b) In a fragmentation process, the rate of disintegration of a substance at any time t varies
proportionately to the amount of a substance y and also to the amount of active yeast x. If the
constant of proportion is ½ , the value of x at any t is
4
and y = e3 initially. Find y in
(1 + 2t )2
terms of t. Hence, find the amount of a substance remaining when 1 approaches zero.
t
[8 marks]
For More Info: Dr. Math (Shopee) #All the HARDWORK, SACRIFICES, SLEEPLESS NIGHT, DOWNFALLS definitely PAYS OFF!
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