A) 0
B) 5
C) 6
D) 9
Correct Answer: D
Solution :
Given, \[p(x)={{a}^{2}}+bx,q(x)=l{{x}^{2}}+mx+n,\] \[\Rightarrow \] \[p(1)-q(1)=({{a}^{2}}+b)-1(l+m+n)=0\] ...(i) \[\Rightarrow \] \[p(2)-q(2)=({{a}^{2}}+2b)-(4l+2m+n)=1\] ...(ii) \[\Rightarrow \] \[p(3)-q(3)=({{a}^{2}}+3b)-(9l+3m+n)=4\] ...(iii) On subtract Eq. (ii) from Eq. (iii) \[b-5l-m=3\] ...(iv) On substract Eq. (i) from (ii) \[b-3l-m=1\] ...(v) On subtract Eq. (v) from Eq. (iv) \[-2l=2\Rightarrow l=-1\] \[\Rightarrow \]Put the value of I in Eq. (iv) \[b-m=-2\] ...(iv) From Eq. (i) \[{{a}^{2}}-n=1\] ... (vii) Now, \[P(4)-q(4)=({{a}^{2}}+4b)-(16l+4m+n)\] \[=({{a}^{2}}-n)+4(b-m)-16l\] \[=1+4(-2)-16(-1)\] \[=16-8+1\] \[=17-8=9\]
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