Let x 2lim( (x-2))(rx^2+(p-2)x-2p)(x-2)^2=5 for some r,p R.

Question

Let $\underset{x	o 2}{lim}\frac{(	an (x-2))(r{x}^{2}+(p-2)x-2p)}{(x-2{)}^{2}}=5$ for some $r,p\in R$. If the set of all possible values of q , such that the roots of the equation $r{x}^{2}-px+q=0$ lie in $(0,2)$, be the interval $(\alpha ,\beta ]$, then $4(\alpha +\beta )$ equals :

Accepted Answer

We are given

$\underset{x	o 2}{lim}\frac{	an (x-2)(r{x}^{2}+(p-2)x-2p)}{(x-2{)}^{2}}=5.$

We have to first find the relation between $r$ and $p$, and then determine all values of $q$ for which the roots of

$r{x}^{2}-px+q=0$

lie in $(0,2)$.

Simplify the limit: Let

$f(x)=r{x}^{2}+(p-2)x-2p.$

Then the given limit becomes

$\underset{x	o 2}{lim}\frac{	an (x-2)\, f(x)}{(x-2{)}^{2}}=5.$

Now, as $x	o 2$,

$	an (x-2)\sim x-2.$

So the limit becomes

$\underset{x	o 2}{lim}\frac{(x-2)f(x)}{(x-2{)}^{2}}=\underset{x	o 2}{lim}\frac{f(x)}{x-2}.$

For this limit to be finite, we must have

$f(2)=0.$

Now,

$f(2)=r(2{)}^{2}+(p-2)(2)-2p=4r+2p-4-2p=4r-4.$

So,

$4r-4=0\, \implies \, r=1.$

Then

$f(x)={x}^{2}+(p-2)x-2p.$

Since $f(2)=0$, factor $(x-2)$:

${x}^{2}+(p-2)x-2p=(x-2)(x+p).$

Hence the limit becomes

$\underset{x	o 2}{lim}\frac{	an (x-2)(x-2)(x+p)}{(x-2{)}^{2}}=\underset{x	o 2}{lim}\frac{	an (x-2)}{x-2}(x+p).$

Using

$\underset{t	o 0}{lim}\frac{	an t}{t}=1,$

we get

$1\cdot (2+p)=5.$

So,

$p+2=5\, \implies \, p=3.$

Thus,

$r=1,\, p=3.$

Form the quadratic equation: Now the equation becomes

${x}^{2}-3x+q=0.$

We want both roots to lie in $(0,2)$.

Let the roots be ${\alpha }_{1},{\alpha }_{2}$. Then

${\alpha }_{1}+{\alpha }_{2}=3,\, {\alpha }_{1}{\alpha }_{2}=q.$

Since both roots lie in $(0,2)$, let us find the condition on $q$.

Use condition for both roots in $(0,2)$: Let

$f(x)={x}^{2}-3x+q.$

Since coefficient of ${x}^{2}$ is positive, parabola opens upward.

For both roots to lie in $(0,2)$, we need:

1. Real roots
So discriminant must be non-negative:

$\Delta=9-4q\ge 0\, \implies \, q\le \frac{9}{4}.$

2. Both roots positive
Since product of roots is $q$,

$q>0.$

3. Both roots less than $2$
Now check value at $x=2$:

$f(2)=4-6+q=q-2.$

If both roots are less than $2$, then for an upward parabola, point $x=2$ lies to the right of both roots, so we need

$f(2)>0.$

Thus,

$q-2>0\, \implies \, q>2.$

Now combine:

$2So the set of all possible values of $q$ is

$(\alpha ,\beta ]=(2,\frac{9}{4}].$

Hence,

$\alpha =2,\, \beta =\frac{9}{4}.$

Therefore,

$4(\alpha +\beta )=4(2+\frac{9}{4})=4(\frac{17}{4})=17.$

Final Answer
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Solutions · 1