Question

If ${}^7{C_r} + 3{}^7{C_{r + 1}} + 3{}^7{C_{r + 2}} + {}^7{C_{r + 3}} > {}^{10}{C_4}$, then the quadratic equations whose roots are $\alpha$, $\beta$, and ${\alpha ^{r - 1}}$, ${\beta ^{r - 1}}$ respectively, have
(a) No common roots
(b) Only one common root
(c) Two common roots
(d) None of these

Answer 1

Here, we need to find which of the options is true for the roots of the two quadratic equations. First, we will rewrite the given inequation. Then, we will use the property to simplify the expression until the left hand side becomes comparable to the right hand side. We will use this to find the value of $r$. Finally, we will substitute the value of $r$ in the roots of the second quadratic equation and check whether any of the roots of the quadratic equation are common or not, and thus, find the correct answer.
Formula Used: We will use the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$.

Complete step by step solution:
First, we will simplify the given inequation.
Rewriting $3{}^7{C_{r + 1}}$ as the sum of ${}^7{C_{r + 1}}$ and $2{}^7{C_{r + 1}}$, and $3{}^7{C_{r + 2}}$ as the sum of $2{}^7{C_{r + 2}}$ and ${}^7{C_{r + 2}}$, we get
$\Rightarrow {}^7{C_r} + {}^7{C_{r + 1}} + 2{}^7{C_{r + 1}} + 2{}^7{C_{r + 2}} + {}^7{C_{r + 2}} + {}^7{C_{r + 3}} > {}^{10}{C_4}$
Factoring out 2 from the terms, we get
$\Rightarrow {}^7{C_r} + {}^7{C_{r + 1}} + 2\left( {{}^7{C_{r + 1}} + {}^7{C_{r + 2}}} \right) + {}^7{C_{r + 2}} + {}^7{C_{r + 3}} > {}^{10}{C_4}$
Pairing the terms in the expression, we get
$\Rightarrow \left( {{}^7{C_r} + {}^7{C_{r + 1}}} \right) + 2\left( {{}^7{C_{r + 1}} + {}^7{C_{r + 2}}} \right) + \left( {{}^7{C_{r + 2}} + {}^7{C_{r + 3}}} \right) > {}^{10}{C_4}$
Now, we know that the sum of ${}^n{C_r}$ and ${}^n{C_{r + 1}}$ is given as ${}^{n + 1}{C_{r + 1}}$, that is ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$.
Substituting $n = 7$ in the formula, we get
$\Rightarrow {}^7{C_r} + {}^7{C_{r + 1}} = {}^{7 + 1}{C_{r + 1}}$
Therefore, we get
$\Rightarrow {}^7{C_r} + {}^7{C_{r + 1}} = {}^8{C_{r + 1}}$
Substituting $n = 7$ and $r + 1$ for $r$ in the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$, we get
$\Rightarrow {}^7{C_{r + 1}} + {}^7{C_{r + 1 + 1}} = {}^{7 + 1}{C_{r + 1 + 1}}$
Adding the terms in the expression, we get
$\Rightarrow {}^7{C_{r + 1}} + {}^7{C_{r + 2}} = {}^8{C_{r + 2}}$
Substituting $n = 7$ and $r + 2$ for $r$ in the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$, we get
$\Rightarrow {}^7{C_{r + 2}} + {}^7{C_{r + 2 + 1}} = {}^{7 + 1}{C_{r + 2 + 1}}$
Adding the terms in the expression, we get
$\Rightarrow {}^7{C_{r + 2}} + {}^7{C_{r + 3}} = {}^8{C_{r + 3}}$
Now, we can simplify the inequation further.
Substituting ${}^7{C_r} + {}^7{C_{r + 1}} = {}^8{C_{r + 1}}$, ${}^7{C_{r + 1}} + {}^7{C_{r + 2}} = {}^8{C_{r + 2}}$, and ${}^7{C_{r + 2}} + {}^7{C_{r + 3}} = {}^8{C_{r + 3}}$ in the inequation $\left( {{}^7{C_r} + {}^7{C_{r + 1}}} \right) + 2\left( {{}^7{C_{r + 1}} + {}^7{C_{r + 2}}} \right) + \left( {{}^7{C_{r + 2}} + {}^7{C_{r + 3}}} \right) > {}^{10}{C_4}$, we get
$\Rightarrow {}^8{C_{r + 1}} + 2{}^8{C_{r + 2}} + {}^8{C_{r + 3}} > {}^{10}{C_4}$
Rewriting $2{}^8{C_{r + 2}}$ as the sum of ${}^8{C_{r + 2}}$ and ${}^8{C_{r + 2}}$, we get
$\Rightarrow {}^8{C_{r + 1}} + {}^8{C_{r + 2}} + {}^8{C_{r + 2}} + {}^8{C_{r + 3}} > {}^{10}{C_4}$
Pairing the terms in the expression, we get
$\Rightarrow \left( {{}^8{C_{r + 1}} + {}^8{C_{r + 2}}} \right) + \left( {{}^8{C_{r + 2}} + {}^8{C_{r + 3}}} \right) > {}^{10}{C_4}$
Substituting $n = 8$ and $r + 1$ for $r$ in the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$, we get
$\Rightarrow {}^8{C_{r + 1}} + {}^8{C_{r + 1 + 1}} = {}^{8 + 1}{C_{r + 1 + 1}}$
Adding the terms in the expression, we get
$\Rightarrow {}^8{C_{r + 1}} + {}^8{C_{r + 2}} = {}^9{C_{r + 2}}$
Substituting $n = 8$ and $r + 2$ for $r$ in the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$, we get
$\Rightarrow {}^8{C_{r + 2}} + {}^8{C_{r + 2 + 1}} = {}^{8 + 1}{C_{r + 2 + 1}}$
Adding the terms in the expression, we get
$\Rightarrow {}^8{C_{r + 2}} + {}^8{C_{r + 3}} = {}^9{C_{r + 3}}$
Now, we can simplify the inequation further.
Substituting ${}^8{C_{r + 1}} + {}^8{C_{r + 2}} = {}^9{C_{r + 2}}$ and ${}^8{C_{r + 2}} + {}^8{C_{r + 3}} = {}^9{C_{r + 3}}$ in the inequation $\left( {{}^8{C_{r + 1}} + {}^8{C_{r + 2}}} \right) + \left( {{}^8{C_{r + 2}} + {}^8{C_{r + 3}}} \right) > {}^{10}{C_4}$, we get
$\Rightarrow {}^9{C_{r + 2}} + {}^9{C_{r + 3}} > {}^{10}{C_4}$
Substituting $n = 9$ and $r + 2$ for $r$ in the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$, we get
$\Rightarrow {}^9{C_{r + 2}} + {}^9{C_{r + 2 + 1}} = {}^{9 + 1}{C_{r + 2 + 1}}$
Adding the terms in the expression, we get
$\Rightarrow {}^9{C_{r + 2}} + {}^9{C_{r + 3}} = {}^{10}{C_{r + 3}}$
Substituting ${}^9{C_{r + 2}} + {}^9{C_{r + 3}} = {}^{10}{C_{r + 3}}$ in the inequation ${}^9{C_{r + 2}} + {}^9{C_{r + 3}} > {}^{10}{C_4}$, we get
$\Rightarrow {}^{10}{C_{r + 3}} > {}^{10}{C_4}$
Thus, we get
$\Rightarrow r + 3 = 5$
Subtracting 3 from both sides of the equation, we get
$\begin{array}{l} \Rightarrow r + 3 - 3 = 5 - 3\\\ \Rightarrow r = 2\end{array}$
$\therefore$ The value of $r$ is 2.
Now, we will find the roots of the two quadratic equations.
The roots of the first quadratic equation are $\alpha$, $\beta$.
The roots of the second quadratic equation are ${\alpha ^{r - 1}}$, ${\beta ^{r - 1}}$.
Substituting $r = 2$ in the roots of the second quadratic equation, we get
${\alpha ^{2 - 1}} = {\alpha ^1} = \alpha$
${\beta ^{2 - 1}} = {\beta ^1} = \beta$
Therefore, the roots of the second quadratic equation are also $\alpha$, $\beta$.
The roots of both the quadratic equations are equal/common.
Thus, the quadratic equations have two common roots.

The correct option is option (c).

Note:
We obtained the equation $r + 3 = 5$ from the inequation ${}^{10}{C_{r + 3}} > {}^{10}{C_4}$.
We know that ${}^{10}{C_6} = {}^{10}{C_4}$, ${}^{10}{C_7} = {}^{10}{C_3}$, ${}^{10}{C_8} = {}^{10}{C_2}$, ${}^{10}{C_9} = {}^{10}{C_1}$, and ${}^{10}{C_{10}} = {}^{10}{C_0}$.
Since ${}^{10}{C_0}$, ${}^{10}{C_1}$, ${}^{10}{C_2}$, and ${}^{10}{C_3}$ are less than ${}^{10}{C_4}$, therefore ${}^{10}{C_6}$, ${}^{10}{C_7}$, ${}^{10}{C_8}$, ${}^{10}{C_9}$, and ${}^{10}{C_{10}}$ are also less than ${}^{10}{C_4}$.
Therefore, the only possible value of $r + 3$ for which the inequation ${}^{10}{C_{r + 3}} > {}^{10}{C_4}$ is true is 5.