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Question

Let p(x)=a2+bx,q(x)=lx2+mx+n. If p(1)q(1)=0,p(2)q(2)=1 and p(3)q(3)=4, then p(4)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 9
We 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+bxlx2mxn

h(x)=Ax2+Bx+C


Here, A=l,B=(bm),C=(a2n)


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 get

4A+2B+C=3A+B+(A+B+C)=3A+B=1 ........(iv) .....[From(i),(ii)]


Solving (iii), we get

9A+3B+C=3(3A+B)+C=3(1)+C=3+C=4 .....[From(iii),(iv)]

C=1


Substituting the value of C in (i), we get

A+B=1

We also have, from (iv), the relation, 3A+B=1

Subtracting the two, we get

(A+B)(3A+B)=2A=11=2

A=1

Substituting the value of a in (iv), we get

3A+B=3(1)+B=B+3=1

B=2


Thus, h(x)=p(x)q(x)=x22x+1=(x1)2

p(4)q(4)=h(4)=(41)2=9

p(4)q(4)=9


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