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Let $p(x)=x^2-5 x+a$ and $q(x)=x^2-3 x+b$, where $a$ and $b$ are positive integers. Suppose HCF $(p(x), q(x))=x-1$ and $k(x)=1 cm (p(x), q(x))$ If the coefficient of the highest degree term of $k(x)$ is 1 , then sum of the roots of $(x-1)+k(x)$ is
$4$
$5$
$6$
$7$
Solution
(d)
We have,
$p(x)=x^2-5 x+a$ and $q(x)=x^2-3 x+b$
Given, $(x-1)$ is HCF of $p(x)$ and $q(x)$.
$\because p(1)=0$ and $q(1)=0$
$\because p(1)=0=1-5+a$ and $q(1)=0$
$=1-3+b$
$\Rightarrow \quad a=4$ and $b=2$
$\because p(x)=x^2-5 x+4$ and $q(x)=x^2-3 x+2$
$\Rightarrow p(x)=(x-1)(x-4)$
and $q(x)=(x-1)(x-2)$
LCM of $p(x)$ and $q(x)=k(x)$
$\because \quad k(x) =\frac{p(x) \cdot q(x)}{\text { HCF of } p(x) \text { and } q(x)}$
$k(x) =(x-1)(x-4) \cdot(x-1)(x-2)$
$k(x) =(x-1)(x-2)(x-4)$
Now, $x-1+k(x)$
$=x-1+(x-1)(x-2)(x-4)$
$=(x-1)\left(1+x^2-6 x+8\right)$
$=(x-1)(x-3)(x-3)$
$\because$ Roots of $x-1+k(x)$ are $1,3,3$.
Sum of roots are $1+3+3=7$
