Question

Let f(x)=ax2+bx+c, g(x)=px?+qx+r such
that f(1)=g(1), F(2)=g(2) and f(3)-g(3)=2.
Then f(4)-g(4) is
Select one:
07
5
6
4.​

Answer 1

Step-by-step explanation:

ANSWER

Given,

f(x)=ax

2

+bx+c,g(x)=px

2

+qx+r

Since, f(1)=g(1)

⇒a+b+c=p+q+r ...... (i)

f(2)=g(2)

⇒4a+2b+c=4p+2q+r ....... (ii)

Subtracting Eq. (ii) from Eq. (i), we get

3a+b=3p+q ..... (iii)

f(3)−g(3)=2

⇒(9a+3b+c)−(9p+3q+r)=2

⇒3(3a+b)+c−3(3p+q)−r=2

⇒c−r=2 .... (∵3a+b=3p+q) (iv)

From Eq. (i),

(a−p)+(b−q)+(c−r)=0

⇒(a−p)+(b−q)+2=0 ..... (v)

From Eq. (ii)

4(a−p)+2(b−q)+c−r=0

⇒2(a−p)+(b−q)+1=0 ..... (vi)

Subtracting Eq. (v) from Eq. (vi), we get

(a−p)−1=0

a−p=1

∴ From Eq. (v), b−q=−3

Now,

f(4)−g(4)=(16a+4b+c)−(16p+4q+r)

=16(a−p)+4(b−q)+(c−r) ..... (vii)

Substituting the values of (a−p),(b−q) and (c−r) from above in Eq. (vii), we get

f(4)−g(4)=16×1+4(−3)+2

=16−12+2=6