A) 4
B) 5
C) 6
D) 7
Correct Answer: C
Solution :
Since \[f(x)=a{{x}^{2}}+bx+c\]and \[g(x)=p{{x}^{2}}+qx+r\] Since \[f(1)=g(1)\Rightarrow a+b+c=p+q+r\] ....(i) Since \[f(2)=g(2)\Rightarrow 4a+2b+c=4p+2q+r\] ....(ii) From (i) - (ii), we have \[3a+b=3p+q\] ...(iii) Since \[f(3)-g(3)=2\Rightarrow (9a+3b+c)-(9p+3q+r)=2\] \[\Rightarrow \,\,\,3(3a+b)+c-3(3p+q)-r=2\] \[\Rightarrow \,\,c-r=2\] ?.(iv) [Using (iii)] From (i), \[(a-p)+(b-q)+(c-r)=0\] \[\Rightarrow \,\,\,(a-p)+(b-q)+2=0\] ?..(v) From (ii), \[4(a-p)+2(b-q)+(c-r)=0\] \[\Rightarrow \,\,2(a-p)+(b-q)+1=0\] ?.(vi) From (vi) - (v), we have \[(a-p)-1=0\] Now, \[f(4)-g(4)=(16a+4b+c)-(16p+4q+1)\] \[=16(a-p)+4(b-q)+(c-r)=16-12+2=6\]
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