Question

# Let p(x)=a2+bx,q(x)=lx2+mx+n. If p(1)âˆ’q(1)=0,p(2)âˆ’q(2)=1 and p(3)âˆ’q(3)=4, then p(4)âˆ’q(4) equals

A
0
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B
5
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C
6
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D
9
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Solution

## The correct option is C 9We have 3 relations given, and have 5 unknown coefficients, a, b, l, m, n to be evaluated. This clearly does not have a unique solution. Thus, we need to modify the equation to be solved. In this question x is only an input variable and is not to be evaluated.We observe that the question only discusses p(x) - q(x), and not p(x) and q(x) individually. So it makes sense to solve for this quantity instead of the individual expressions. Let, h(x)=p(x)−q(x)⇒h(x)=a2+bx−lx2−mx−n⇒h(x)=Ax2+Bx+CHere, A=−l,B=(b−m),C=(a2−n)So we will solve for A, B , C thus reducing the number of variables.The first equation says h(1)=p(1)−q(1)=0⇒A+B+C=0 .......(i) The second relation is h(2)=p(2)−q(2)=1 ⇒4A+2B+C=1 ........(ii)The third relation give in the problem is h(3)=p(3)−q(3)=4⇒9A+3B+C=4 ........(iii)Solving (ii), we get4A+2B+C=3A+B+(A+B+C)=3A+B=1 ........(iv) .....[From(i),(ii)]Solving (iii), we get9A+3B+C=3(3A+B)+C=3(1)+C=3+C=4 .....[From(iii),(iv)]⇒C=1Substituting the value of C in (i), we getA+B=−1We also have, from (iv), the relation, 3A+B=1Subtracting the two, we get(A+B)−(3A+B)=−2A=−1−1=−2⇒A=1Substituting the value of a in (iv), we get3A+B=3(1)+B=B+3=1⇒B=−2Thus, h(x)=p(x)−q(x)=x2−2x+1=(x−1)2⇒p(4)−q(4)=h(4)=(4−1)2=9p(4)−q(4)=9

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