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Question

Given that $P(x)=F(x)-g(x)$ has only one root $-1$

$\Rightarrow p(x)=\left(a-a_{1}\right) x^{2}+\left(b-b_{1}\right) x+\left(c-c_{1}\right)$ has

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$18$

Vedclass · Answer

Given that $P(x)=F(x)-g(x)$ has only one root $-1$

$\Rightarrow p(x)=\left(a-a_{1}ight) x^{2}+\left(b-b_{1}ight) x+\left(c-c_{1}ight)$ has

one root, $-1$ only

$\Rightarrow p^{\prime}(x)$ will also has root as $-1$ $\Rightarrow p^{\prime}(x)=0$ at $x=-1$

$\Rightarrow 2\left(a-a_{1}ight) x+\left(b-b_{1}ight)=0$ at $x=-1$

$\Rightarrow-2\left(a-a_{1}ight)+\left(b-b_{1}ight)=0$

$\Rightarrow \quad \frac{-\left(b-b_{1}ight)}{\left(a-a_{1}ight)}=-2$         ......$(i)$

Now, $p(x)=\left(a-a_{1}ight) x^{2}+\left(b-b_{1}ight) x+\left(c-c_{1}ight)$

$\Rightarrow \quad \frac{p(x)}{a-a_{1}}=x^{2}+\frac{b-b_{1}}{a-a_{1}} x+\frac{\left(c-c_{1}ight)}{a-a_{1}}$

$\because \quad p(-1)=0$

$	herefore \quad 0=(-1)^{2}-\frac{\left(b-b_{1}ight)}{a-a_{1}}+\frac{\left(c-c_{1}ight)}{a-a_{1}}$

$\Rightarrow \quad 0=1-2+\frac{\left(c-c_{1}ight)}{a-a_{1}} $ $\quad [{m{ using Eq}}{m{.}}\left( i ight)]$

$\Rightarrow \frac{\mathrm{c}-\mathrm{c}_{1}}{\mathrm{a}-\mathrm{a}_{1}}=1$         ......$(ii)$

Also, given that $p(-2)=2$ 

$\Rightarrow 4\left(a-a_{1}ight)-2\left(b-b_{1}ight)+\left(c-c_{1}ight)=2 \quad \ldots$ $(iii)$

From Eqs. $(i), (ii)$ and $(iii),$ we have

${4\left(a-a_{1}ight)-4\left(a-a_{1}ight)+\left(a-a_{1}ight)=2} $

$\Rightarrow$  ${a-a_{1}=2}$

On substituting $a-a_{1}=2$ in Eq. $(ii),$ we get

$c-c_{1}=2$

On substituting $a-a_{1}=2$ in Eq. $(i),$ we get

$b-b_{1}=4$

$ 	ext { Now, } p(2) =4\left(a-a_{1}ight)+2\left(b-b_{1}ight)+\left(c-c_{1}ight) $

$=4 	imes 2+2 	imes 4+2 $

$=8+8+2=18$

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