Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

$a$ and $b$ are two integers such that $10a+b=5$ and $p(x)=x^2+ax+b$.The value of n such that $p(10)p(11)=p(n)$

$\begin{array}{1 1}(A)\;110&(B)\;100\\(C)\;115&(D)\;105\end{array} $

Can you answer this question?

1 Answer

0 votes
Let $\alpha$ and $\beta$ be the roots of $p(x)=0$
$\Rightarrow p(x)=(x-\alpha)(x-\beta)$
$\Rightarrow (10-\alpha)(11-\beta)(10-\beta)(11-\alpha)$
$\Rightarrow [110-10(\alpha+\beta)-\alpha+\beta\alpha][110-10(\alpha+\beta)-\beta+\beta\alpha]$
$\Rightarrow (110+10a+b-\alpha)(110+10a+b-\beta)$
$\Rightarrow (115-\alpha)(115-\beta)$
$\Rightarrow p(115)$
Hence (C) is the correct answer.
answered Apr 16, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App