Question

Assertion :If $n$ is an odd integer greater than 3 but not a multiple of 3, then $(x+1{)}^n-x^n-1$ is divisible by $x^3+x^2+x$ Reason: If $n$ is an odd integer greater than $3$ but not a multiple of $3$, we have $(1+{\omega }^n+{\omega }^{2n})=0$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect and Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$x^3+x^2+x=x(x^2+x+1)=x(x-\omega )(x-{\omega }^2)$

Hence the roots of $x^3+x^2+x=0$ are $x=0,x=\omega ,x={\omega }^2$

If $(x+1{)}^n-x^n-1$ has to be divisible by $x^3+x^2+x$, then $x=0,x=\omega$ and $x={\omega }^2$ are its factors.

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

$1-1=0$

Putting $x=\omega$,

$(\omega +1{)}^n-{\omega }^n-1=(-{\omega }^2{)}^n-{\omega }^n-1$ (As $1+\omega +{\omega }^2=0$)

$=-({\omega }^{2n}+{\omega }^n+1)$ ($n$ is odd)

$=-\left(0\right)=0$ ($n$ is an odd number greater than $3$ and not a multiple of $3$)

Putting $x={\omega }^2$,

$(1+{\omega }^2{)}^n-{\omega }^{2n}-1=-({\omega }^n+{\omega }^{2n}+1)=0$

Hence $(x+1{)}^n-x^n-1$ is divisible by $x^3+x^2+x$, if $n$ is an odd integer greater than $3$ but not a multiple of $3$.

$1+{\omega }^n+{\omega }^{2n}={\omega }^2+\omega +1=0$ always because $n$ is odd

Thus, the reason is correct and correct explanation for assertion.