Mathematics Study Material

Find a wide range of study material for Mathematics including solved assignments, essays, dissertations, and more at Desklib. Get access to high-quality content for your Mathematics course.

Mathematics

Question 1:
Solve the inequality
2x2 + 3x > 2
2x2 + 3x – 2 > 0
(x + 2) (2x – 1) > 0
or, x + 2 > 0 or, 2x – 1 > 0
x > -2 or, x > ½
so, -2 < x < ½
1

Question 3
Solve the quadratic equation
(a) Given,
f(x) = ½ x2 – 4x + 6
(i) Completing square -
f(x) = 0
½ x2 – 4x + 6 = 0
x2 – 8x + 12
x2 – 8x + (4)2 + 12 – (4)2
(x – 4)2 + 12 – 16
(x – 4)2 – 4
or, ½ (x – 4)2 – 2
(ii) To draw the graph, above obtained equation can be represented as –
y = ½ (x – 4)2 – 2
(as compared with parabolic equation y = ax2)
here, a > 0 therefore, parabola of this equation will open up in upward direction.
(iii) Interval form of given equation is
(x – 4)2 – 4 ≥ 0
(x – 4)2 ≥ 4
0 ≤ x ≤ 4
(b) g = ½ x2 – 4x + 6
(i) Given equation, ½x2 – 4x + 6
can be written in interval form as 0 ≤ x ≤ 4
therefore,
0 1 2 3 4 5 6
(ii) Image set of g can be identified in following way –
x2 – 8x + 12
2

Firstly, coordinate vertex of parabola –
xw = -b/2a
= - (-8)/2.1
= 4
now, Δ = b2 – 4ac
= (-8)2 – 4. 12. 1
= 64 – 48 = 16
so, yw = -Δ/4a
= -16/ 4.1 = -4
so, coordinate vertex (-4, 4)
Therefore, image of function will be [-4, ∞).
(iii) To find inverse function of g
Let, y = ½ x2 – 4x + 6 = 0
= ½ (x – 4)2 – 2
then,
x = ½ (y – 4)2 – 2
(y – 4)2 – 4 = 2x
(y – 4)2 = 2x + 4
y – 4 = ±√2x+4
y = (4 – √2x+4)
so, g-1 = (4 – √2x+4)
3

Question 4
(a) Find all solutions
sinϴ = 1/√2
then, sinϴ = sin π/4
so, ϴ = π/4
(b) Given, in ΔABC, A = 36 , b = 10 and c = 6⁰
using Cosine rule –
a2 = b2 + c2 – 2bc Cos (A)
= (10)2 + (6)2 – 2.10.6 Cos (36 )⁰
= 100 + 36 – 120 x 0.81
= 136 – 97.2
= 38.8
then a ≈ 6.24
Cos (B) = c2 + a2 – b2
2ac
= (6)2 + (6.24)2 – (10)2
2 (6.24) (6)
= 36 + 38.9 – 100
74.88
= -0.335
= 70.42⁰
Cos (C) = a2 + b2 – c2
2ab
= (6.24)2 + (10)2 – (6)2
2 (6.24) (10)
= 38.9 + 100 – 36
124.8
= 0.824
= 34.46⁰
4

Question 5
(a) Evaluate value of sin (π/12)
sin (π/12) = sin (π/3 – π/4)
using trigonometry rule –
sin (A – B) = sin A. cos B – Cos A. Sin B
then,
sin (π/12) = sin π/3. cos π/4 – Cos π/3. Sin π/4
= √3 x 1 - 1 x 1
2 √2 2 √2
= √3 – 1
2√2
rationalise the given term by √2
= √3 – 1 x √2
2√2 √2
= √6 - √2
4
(b) Evaluate value of cos (π/12)
Using half angle identity of cos –
Cos A = ±√(1 + cos A)
2 2
cos π/12 = cos (π/6)
2
= ± √ 1 + cos (π/6)
2
= ± √ 1 + √3/2
2
= √(2 + √3)
2
5

(C) To prove – sin2ϴ + cos2ϴ = 1
LHS –
sin2(π/12) + cos2(π/12)
√6 - √2 2 + √(2 + √3) 2
4 2
(√6)2 + (√2)2 – 2 (√6) (√2) +(2) + (√3)
16 4
= 6 + 2 – 4√3 + 2 + √3
16 4
= 8 – 4√3 + 2 + √3
16 4
= 2 – √3 + 2 + √3
4
= 1 Hence proved.
6

Question 6
Solve the following –
Given –
eqn of circle –
x2 + y2 – 2x – 6y – 6 = 0
Rearrange –
x2 – 2x + y2 – 6y – 6 = 0
Using completing the square method –
(x – 1)2 + (y – 3)2 = 22 (i)
here, coordinates of centre (1,3) and radius = 2
eqn of ellipse –
25x2 + 4y2 – 100x – 32y + 64 = 0
Rearrange
25x2 – 100x + 4y2 – 32y + 64 = 0
25 (x – 2)2 + 4 (y – 4)2 = 100
(x – 2)2 + (y – 4)2 = 1 (ii)
4 25
so, radius of ellipse will be (2,5)
Now, point of intersection of both equations will be (2, 0)
7

Question 7
a) Given
v
vy
ϴ
vx
vx = velocity of wind
vy = velocity of drone
then, vx = speed of wind x direction
= 7 x cos 125⁰
= 7 x (-0.57)
=- 4.01 m/sec-1
vy = speed of drone x direction
= 20 x sin 125⁰
= 20 x (0.82)
= 16.38 m/sec-1
(b) Resultant velocity v =
v = √vx2 + vy2
= √(-4.01)2 + 16.382
= 16.86 m/sec-1
(c) Magnitude of resultant vector –
ϴ = tan-1 (vy/vx)
= tan-1 (16.38/-4.01)
= 76⁰
Magnitude of Resultant velocity v =
v = √vx2 + vy2
= √(-4.01)2 + 16.382
= 16.86 m/sec-1
8

Question 8
f(x) = 2x3 – 9x2 + 12x – 3
(a) Exact x and y coordinates of above equation -
Differentiate given equation with respect to x
f’(x) = 6x2 – 18x + 12
at stationery points f’(x) = 0
then,
6x2 – 18x + 12 = 0
x2 – 3x + 2 = 0
(x – 1) (x – 2) = 0
so, x = 1 or, 2
at x = 1
y = 2(1)3 – 9(1)2 + 12(1) – 3
= 2 – 9 + 12 – 3
= -7 + 9 = 2
while at x = 2
y = 2(2)3 – 9(2)2 + 12(2) – 3
= 16 – 36 + 24 – 3
= 1
therefore, at stationary coordinates will be – (1,2) and (2,1)
(c)
y
x
9

(d) Greatest value of f(x) in the interval [1,3] –
f(1) = 2(1)3 – 9(1)2 + 12(1) – 3
= 2 – 9 + 12 – 3
= -7 + 9 = 2 (least value)
f(3) = 2(3)3 – 9(3)2 + 12(3) – 3
= 54 – 81 + 36 – 3
= 6 (greatest value)
10

Question 9
Given, displacement (s) as –
s = 6t3 – 21t2 + 12t
(a) Expression for velocity
v = ds/dt
= d (6t3 – 21t2 + 12t)
dt
= 18t2 – 42t + 12
Expression for acceleration
a = dv/dt
= d (18t2 – 42t + 12)
dt
= 36t – 42
(b) when velocity becomes zero then t will be –
18t2 – 42t + 12 = 0
3t2 – 7t + 2 = 0
(t – 2) (3t – 1) = 0
or, t = 2 unit or, 1/3 unit
11