Value of 125-¹/³ is (a) 5 (b)1/5 (c) 25 (d) 1/25 If P(y) - y² - y + 4 then P(2) (a) 10 (b) 4 (c) 2 (d) 6
Answers
Answer:
Chapter 1
Preliminary Test (page 3)
1. √
7. [c
2 = a
2 + b
2 − 2ab cos C.] (5 marks)
2. x
−4/3 +
y
16 = 1. [Verify that the point is on the curve. Find slope
dy
dx = 12 (at that point)
and the tangent y + 8 = 12(x + 2). (5 marks)
Rearrange the equation to get it in intercept form, or solve y = 0 for x-intercept and x = 0
for y-intercept.] (5 marks)
3. One. [Show that g
0
(t) < 0 ∀t ∈ R, hence g is strictly decreasing. Show existence. Show
uniqueness.] (5 marks)
Comment: Graphical solution is also acceptable, but arguments must be rigorous.
4. 4π(9 − 2
√
6) or 16.4π or 51.54 cm3
. [V =
R 2π
0
R Rh
0
2
√
R2 − r
2 rdrdθ, R = 3, Rh =
√
3.
Sketch domain. (5 marks)
V = 4π
R Rh
0
√
R2 − r
2 rdr, substitute R2 − r
2 = t
2
. (5 marks)
Evaluate integral V = −4π
R
√
R2−R2
h
R t
2dt.] (5 marks)
5. (a) (2,1,8). [Consider p~ − q~ = ~s − ~r.] (5 marks)
(b) p
3/5. [cosQ =
QP~ ·QR~
kQP~ k kQR~ k
.] (5 marks)
(c) 9
5
(j + 2k). [Vector projection = (QP~ · QRˆ )QRˆ .] (5 marks)
(d) 6
√
6 square units. [Vector area = QP~ × QR~ .] (5 marks)
(e) 7x + 2y − z = 8.
[Take normal n in the direction of vector area and n · (x − q).] (5 marks)
Comment: Parametric equation q + α(p − q) + β(r − q) is also acceptable.
(f) 14, 4, 2 on yz, xz and xy planes. [Take components of vector area.] (5 marks)
6. 1.625%. [A = πab ⇒ dA
A = da
a + db
b
. (5 marks)
Put a = 10, b = 16 and da = db = 0.1.] (5 marks)
7. 9/2. [Sketch domain: region inside the ellipse x
2 + 4y
2 = 9 and above the x-axis. (5 marks)
By change of order, I =
R 3
−3
R 1
2
√
9−x2
0
ydydx. (5 marks)