f4 maths3

1.
f(x)=x3+2x2+ax+b

f(x) is divisible by x+1:
f(-1) = 0
(-1)3+2(-1)2+a(-1)+b = 0
1 -a + b = 0
a - b = 1 ...... (1)

f(x) is divisible by x-2:
f(2) = 0
(2)3+2(2)2+a(2)+b = 0
16 + 2a + b = 0
2a + b = -16 ...... (2)

(1)+(2):
3a = -15
a = -5

Put a = -5 into (2)
2(-5) + b = -16
b = -6

f(x) = x3 + 2x2 -5x - 6

Both x+1 and x-2 are the factors of f(x):
x3 + 2x2 -5x - 6 = (x+1)(x-2)(x+c)
Compare the constant term:
-6 = (1)(-2)(c)
c = 3

Ans: f(x) = (x+1)(x-2)(x+3)

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2.
f(x)=x2-x-3

f(k) = k
k2-k-3 = k
k2-2k-3 = 0
(k-3)(k+1) = 0
k-3 = 0 ooro k+1 = 0
k = 3 ooro k = -1

Ans: k = 3 ooro k = -1

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3.
2x8y = 2x(23)y
2x8y = 2x23y
2x8y = 2x+3y

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4.
The remainder when x2+ax+b is divided by x+2 is -4:
(-2)2+a(-2)+b = -4
4-2a+b = -4
2a-b = 8 ...... (1)


The remainder when ax2+bx+1 is divided by x-2 is 9:
a(2)2+b(2)+1 = 9
4a+2b = 8
2a+b = 4 ...... (2)

(1)+(2):
4a = 12
a = 3

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5.
logx2=Log3x+1
logx2 = log3x+log10
logx2 = log(3x)(10)
logx2 = log30x
x2 = 30x
x2 - 30x = 0
(x-30)x = 0
x = 30 ooro x = 0 (rejected)

Ans: x = 30
+

1.
As f(x) can divisible by x+1, (when x+1=0,x=-1)
f(-1)= (-1)^3+ 2(-1)^2+a(-1)+b
= -1+2-a+b
= 1-a+b -----------------------(1)
As f(x) can divisible by x-2, (when x-2 =0, x=-2)
f(2) = (2)^3+2(2)^2+a(2)+b
= 8+8+4a+b
=16+4a+b ----------------------(2)
(2) - (1),
5a+15 = 0
a= -3
sub a= -3 in (1),
1-(-3)+b=0,
b= -4

2.
f(k)= k^2-k-3
As f(k) = k
thus, k^2-k-3= k
k^2-2k-3= 0
(k+1)(k-3)=0
k= -1 or k=3

3.
2^x times 8^y
= 2^x times 2^(3y) As 2^3=8
= 2^(x+3y)

4.
As x^2 +ax +b can divided by x+2 is -4, (when x+2=0, x= -2)
(-2)^2 + a(-2) +b= -4
4-2a+b = -4
-2a+b+8=0 ---------------------------(1)
As ax^2+bx+1 can divided by x-2 is 9, (when x-2 =0, x=2)
a{(2)^2} + b(2) +1= 9
4a+2b+1 =9
4a+2b-8 =0 ------------------------(2)
(1)* 2 - (2),
-8a +24 =0
a=3

5.
log x^2 =Log3x+1 -------(1)
log x ^2 = Log3x + log10 (as log10=1, log100=2 etc)
log x^2 = log (3x*10)
log x^2 =log(30x)
x^2 = 30x
x^2-30x =0
x(x-30) =0
x=0(discarded) or x=30
u may sub when x=0 in (1), it is not equal in both side.