ISI 2004 MIII

Question

$Q8$ If $\alpha_{1},\alpha_{2},\alpha_{3}, \dots , \alpha_{n}$ be the roots of $x^{n}+1=0$, then $\left ( 1-\alpha_{1} \right )\left ( 1-\alpha_{2} \right ) \dots \left ( 1-\alpha_{n} \right )$ is equal to

• A. $1$
• B. $0$
• C. $n$
• D. $2$

Answer 1

$x^{n}+1=0$ ------------$>(1)$

$\text{Case1:}$ If $n$ is odd $\text{(let n=3)}$

$x^{3}+1=0$

We can write like this

$x^{3}+0\cdot x^{2}+0\cdot x +1=0$ -----------$>(2)$

$(1-\alpha_{1})(1-\alpha_{2})(1-\alpha_{3})=1-\alpha_{1}-\alpha_{2}+\alpha_{1}\alpha_{2}-\alpha_{3}+\alpha_{1}\alpha_{3}-\alpha_{1}\alpha_{2}\alpha_{3}+\alpha_{2}\alpha_{3}$

                                                    $=1-\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{1}\alpha_{2}+\alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{3}-\alpha_{1}\alpha_{2}\alpha_{3}$

                                                   $=1-(\alpha_{1}+\alpha_{2}+\alpha_{3})+(\alpha_{1}\alpha_{2}+\alpha_{2}\alpha_{3}+\alpha_{1}\alpha_{3})-\alpha_{1}\alpha_{2}\alpha_{3}$ -----$>(3)$

                     $\text{From equation $(2)$}$

$$\alpha_{1}+\alpha_{2}+\alpha_{3}=\frac{-b}{a}=\frac{0}{1}=0$$

$$\alpha_{1}\alpha_{2}+\alpha_{2}\alpha_{3}+\alpha_{1}\alpha_{3}=\frac{c}{a}=\frac{0}{1}=0$$

$$\alpha_{1}\alpha_{2}\alpha_{3}=\frac{-d}{a}=\frac{-1}{1}=-1$$

Put these value into equation $(3)$ we get

$(1-\alpha_{1})(1-\alpha_{2})(1-\alpha_{3})=1-(0)+(0)-(-1)=1+1=2$ 

 

$\text{Case2:}$ If $n$ is even $\text{(let n=2)}$

$x^{2}+1=0$

We can write like this

$x^{2}+0\cdot x +1=0$ -----------$>(4)$

$(1-\alpha_{1})(1-\alpha_{2})=1-\alpha_{1}-\alpha_{2}+\alpha_{1}\alpha_{2}$

                                  $=1-(\alpha_{1}+\alpha_{2})+\alpha_{1}\alpha_{2}$ -------------$>(5)$

                            $\text{From equation $(4)$}$

$$\alpha_{1}+\alpha_{2}=\frac{-b}{a}=\frac{0}{1}=0$$

$$\alpha_{1}\alpha_{2}=\frac{c}{a}=\frac{1}{1}=1$$

Put these value into equation $(5)$ we get

$(1-\alpha_{1})(1-\alpha_{2})=1-(0)+1=2$

So$,$answer is $2$

Option $(D)$ is the correct choice$.$

Comments

You made it so complex :p

Since $\alpha_{1},\alpha _{2},\alpha _{3},........,\alpha _{n}$ are the roots of the equation $x^{n} + 1 = 0$

So, we can write $x^{n} + 1$ as :-

$x^{n} + 1 = (x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3}).......(x-\alpha _{n})$

Now, put $x=1$ both sides

$2= (1-\alpha _{1})(1-\alpha _{2})(1-\alpha _{3}).......(1-\alpha _{n})$

So, Answer is $(D)$