Solved Math Problems for Students | Desklib

Q 22.
f (x) = x − 15 and g(x) = √x − 3
a)
( f
g
)
(x) = x − 15
√x − 3
[Put x=12]
∴
( f
g
)
(12) = 12 − 15
√12 − 3
= −3
√9 = −1
b)
Domain: It is the interval over which the function is defined.
The function f
g has a square root in the denominator which is defined only
non-negative numbers. Also the function is undefined when the denominator
is zero, therefore
x − 3 > 0 =⇒ x > 3
The domain of the function f
g is
{x ∈ R | x > 3}
Q 23.
Answer
a)
The end points of the diameter are:
(x1, y1) = (−6, 5) and (x2, y2) = (4, 3)
Diameter of the circle is given by
d = √(∆y)2 + (∆x)2 = √(y2 − y1)2 + (x2 − x1)2 = √(3 − 5)2 + (4 − (−6))2 = √104
b)
Center C of the circle is the midpoint of diameter, which is
C =
( x1 + x2
2 , y1 + y2
2
)
=
( −6 + 4
2 , 5 + 3
2
)
= (−1, 4)
1

c)
Symmetry about y- axis is obtained by changing the sign of x- coordinate.
Therefore, the point (1, 4) is symmetric to C about y- axis.
Q 24.
Answer
Line passes through points: (x1, y1) = (1,-7) and (x2, y2) =(3,-9).
Slope of the line is given by
m = ∆y
∆x = y2 − y1
x2 − x1
= −9 − (−7)
3 − 1 = −1
Equation of line passing through point (x1, y1) =(1,-7), having slope m =
−1, is
y − y1 = m(x − x1) =⇒ y + 7 = −1(x − 1)
Re-arranging the equation in slope-intercept form:
y = −x − 6
Q 25.
Answer
Let the annual income be x $.
Tax paid is a fixed amount of 90 $ and 1.4 % of x. Therefore,
Total tax paid = 90 + 1.4
100 x
The total tax paid for the year was 1035 $. Implies,
90 + 1.4
100 x = 1035 =⇒ 1.4
100 x = 945 =⇒ x = 67500
Therefore, the annual income is 67, 500 $
Q 26.
f (x) = 4x2 − 20 and g(x) = x − 3
Answer
a)
2

(f ◦ g)(x) = 4(g(x))2 − 20
= 4(x − 3)2 − 20
= 4(x2 − 6x + 9) − 20
∴ (f ◦ g)(x) = 4x2 − 24x + 16
b)
(f ◦ g)(x) = 4x2 − 24x + 16
[Put x= -2]
(f ◦ g)(−2) = 4(−2)2 − 24(−2) + 16
= 80
Q 27.
5x2 + 12 = 20x
Answer
5x2 + 12 = 20x
[Subtract 20x from both sides]
∴ 5x2 − 20x + 12 = 0
[Divide throughout by 5]
x2 − 4x + 2.4 = 0
The quadratic equation ax2 + bx + c = 0, has roots
x = −b ± √b2 − 4ac
2a
Comparing,
ax2 + bx + c ' x2 − 4x + 2.4
3