Question

Statement 1: If \[x+\left( \dfrac{1}{x} \right)=1\] and \[p={{x}^{4000}}+\dfrac{1}{{{x}^{4000}}}\] and q is the digit at unit place in the number
\[{{2}^{{{2}^{n}}}}+1\] , \[n\in N\] and n > 1, then the value of \[p+q=8\] .
Statement 2: If \[\omega \] , \[{{\omega }^{2}}\] are the roots of \[x+\dfrac{1}{x}=-1\] , the \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}=-1\] , \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}=2\] .
(A) Both the statements are true, and statement 2 is the correct explanation for statement 1.
(B) Both the statements are true, but statement 2 is not the correct explanation for statement 1.
(C) Statement 1 is true and statement 2 is false.
(D) Statement 1 is false and statement 2 is true.

Answer 1

Hint: First of all, in statement 1, solve the equation \[x+\left( \dfrac{1}{x} \right)=1\] and get the value of x using the formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2as}\] . We know that \[\omega =\dfrac{-1+\sqrt{3}i}{2}\] and \[{{\omega }^{2}}=\dfrac{-1-\sqrt{3}i}{2}\] . Express the value of x in terms of \[\omega \] and \[{{\omega }^{2}}\] . Now, put \[x=-\omega \] in the equation \[p={{x}^{4000}}+\dfrac{1}{{{x}^{4000}}}\]and simplify it further using the property \[{{\omega }^{3n}}=1\] , and \[1+\omega +{{\omega }^{2}}=0\Rightarrow \omega +{{\omega }^{2}}=-1\] . Similarly, put \[x=-{{\omega }^{2}}\] in the equation \[p={{x}^{4000}}+\dfrac{1}{{{x}^{4000}}}\]and simplify it further using the property \[{{\omega }^{3n}}=1\] , and \[1+\omega +{{\omega }^{2}}=0\Rightarrow \omega +{{\omega }^{2}}=-1\] . Now, get the value of p. Put \[n=2\] and \[n=3\] in the equation \[{{2}^{{{2}^{n}}}}+1\] and get the unit place of the number \[{{2}^{{{2}^{n}}}}+1\] . ‘q’ is at the unit place of \[{{2}^{{{2}^{n}}}}+1\] . Now, calculate the value of \[p+q\] . For statement 2, solve the equation \[x+\dfrac{1}{x}=-1\] and get the value of x using the formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2as}\] . We know that \[\omega =\dfrac{-1+\sqrt{3}i}{2}\] and \[{{\omega }^{2}}=\dfrac{-1-\sqrt{3}i}{2}\] . Express the value of x in terms of \[\omega \] and \[{{\omega }^{2}}\] . Now, put \[x=\omega \] in \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}\] and get the value of \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}\] . Similarly, put \[x={{\omega }^{2}}\] in \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}\] and get the value of \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}\] . Put \[x=\omega \] and \[x={{\omega }^{2}}\] in \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}\] and calculate its value.

Complete step by step answer:
According to the question, we have two statements and we have to first check whether both of the options are correct or not.
 In statement 1, we have
If \[x+\left( \dfrac{1}{x} \right)=1\] and \[p={{x}^{4000}}+\dfrac{1}{{{x}^{4000}}}\] and q is the digit at the unit place in the number
\[{{2}^{2n}}+1\] , \[n\in N\] and n>1, then the value of \[p+q=8\] .
First of all, let us check it.
\[\Rightarrow x+\left( \dfrac{1}{x} \right)=1\]
\[\Rightarrow {{x}^{2}}-x+1=0\] ……………………………….(1)
We know the formula that for the roots of the quadratic equation \[a{{x}^{2}}+bx+c\] is
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] ………………………………………….(2)
Now, from equation (1) and equation (2), we get
\[\Rightarrow x=\dfrac{1\pm \sqrt{1-4}}{2}\]
\[x=\dfrac{1-\sqrt{3}i}{2}\] or \[x=\dfrac{1+\sqrt{3}i}{2}\] …………………………………….(3)
We also know that \[\omega =\dfrac{-1+\sqrt{3}i}{2}\] and \[{{\omega }^{2}}=\dfrac{-1-\sqrt{3}i}{2}\] ……………………………………………(4)
Now, from equation (3) and equation (4), we get
\[x=-\left( \dfrac{-1+\sqrt{3}i}{2} \right)=-\omega \] or \[x=-\left( \dfrac{-1-\sqrt{3}i}{2} \right)=-{{\omega }^{2}}\] …………………………………………(5)
In the statement, we also have,
\[p={{x}^{4000}}+\dfrac{1}{{{x}^{4000}}}\] ………………………………………..(6)
Now, on putting \[x=-\omega \] in equation (6), we get
\[\Rightarrow p={{\left( -\omega \right)}^{4000}}+\dfrac{1}{{{\left( -\omega \right)}^{4000}}}\]
 \[\Rightarrow p={{\omega }^{4000}}+\dfrac{1}{{{\omega }^{4000}}}\]
\[\Rightarrow p={{\omega }^{3\times 1333}}\times \omega +\dfrac{1}{{{\omega }^{3\times 1333}}\times \omega }\] ……………………………………………………..(7)
We know the property, \[{{\omega }^{3n}}=1\] …………………………………(8)
Now, from equation (7) and equation (8), we get
\[\begin{align}
  & \Rightarrow p=1\times \omega +\dfrac{1}{1\times \omega } \\
 & \Rightarrow p=\omega +\dfrac{1}{\omega } \\
\end{align}\]
\[\Rightarrow p=\omega +{{\omega }^{2}}\] …………………………………………..(9)
We also know the property that \[1+\omega +{{\omega }^{2}}=0\Rightarrow \omega +{{\omega }^{2}}=-1\] …………………………………………….(10)
From equation (9) and equation (10), we get
\[\Rightarrow p=-1\] ……………………………………….(11)
Similarly, on putting \[x=-{{\omega }^{2}}\] in equation (6), we get
\[\Rightarrow p={{\left( -{{\omega }^{2}} \right)}^{4000}}+\dfrac{1}{{{\left( -{{\omega }^{2}} \right)}^{4000}}}\]
 \[\Rightarrow p={{\omega }^{8000}}+\dfrac{1}{{{\omega }^{8s000}}}\]
\[\Rightarrow p={{\omega }^{3\times 2666}}\times {{\omega }^{2}}+\dfrac{1}{{{\omega }^{3\times 2666}}\times {{\omega }^{2}}}\] ……………………………………………………..(12)
Now, from equation (8) and equation (12), we get
\[\begin{align}
  & \Rightarrow p=1\times {{\omega }^{2}}+\dfrac{1}{1\times {{\omega }^{2}}} \\
 & \Rightarrow p={{\omega }^{2}}+\dfrac{1}{{{\omega }^{2}}} \\
\end{align}\]
\[\Rightarrow p={{\omega }^{2}}+\omega \] …………………………………………..(13)
From equation (10) and equation (13), we get
\[\Rightarrow p=-1\] ……………………………………….(14)
From equation (11) and equation (14), we have the value of p which is equal to -1.
So, \[p=-1\] ……………………………………..(15)
Also, in statement 1, we have
q is the digit at the unit place in the number \[{{2}^{{{2}^{n}}}}+1\] , \[n\in N\] and n>1, and the value of \[p+q=8\] .
On putting \[n=2\] in \[{{2}^{{{2}^{n}}}}+1\] , we get
\[\begin{align}
  & ={{2}^{{{2}^{2}}}}+1 \\
 & ={{2}^{4}}+1 \\
\end{align}\]
\[=17\] ………………………………………..(16)
Similarly, on putting \[n=3\] in \[{{2}^{{{2}^{n}}}}+1\] , we get
\[\begin{align}
  & ={{2}^{{{2}^{3}}}}+1 \\
 & ={{2}^{8}}+1 \\
\end{align}\]
\[=257\] …………………………………………(17)
Now, from equation (16) and equation (17), we have the unit place of \[{{2}^{{{2}^{n}}}}+1\] which is equal to 7 and for every natural value of n, n>1, the unit place of \[{{2}^{{{2}^{n}}}}+1\] is 7.
So, the value of q is 7 ……………………………………..(18)
Now, from equation (15) and equation (17), we have
\[\Rightarrow p+q=-1+7=6\] ……………………………………..(19)
But, in statement 1 we are given that \[p+q=8\] ………………………………………….(20)
Using equation (19) and equation (20), we can say that statement 1 is not true ………………………………………..(21)
Now, in statement 2, we have
If \[\omega \] , \[{{\omega }^{2}}\] are the roots of \[x+\dfrac{1}{x}=-1\] , the \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}=-1\] , \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}=2\] ……………………………………….(22)
First of all, let us solve \[x+\dfrac{1}{x}=-1\]
\[\Rightarrow x+\dfrac{1}{x}=-1\]
\[\Rightarrow {{x}^{2}}+x+1=0\] ………………………………………………..(23)
From equation (2) and equation (23), we get
\[\Rightarrow x=\dfrac{-1\pm \sqrt{1-4}}{2}\]
\[x=\dfrac{-1+\sqrt{3}i}{2}\] or \[x=\dfrac{-1-\sqrt{3}i}{2}\] …………………………………….(24)
Using equation (4) and equation (24), we have
\[x=\left( \dfrac{-1+\sqrt{3}i}{2} \right)=\omega \] or \[x=\left( \dfrac{-1-\sqrt{3}i}{2} \right)={{\omega }^{2}}\] ………………………………….(25)
So, \[\omega \] , \[{{\omega }^{2}}\] are the roots of \[x+\dfrac{1}{x}=-1\] .
Now, on putting \[x=\omega \] in \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}\] , we get
\[={{\omega }^{2}}+\dfrac{1}{{{\omega }^{2}}}\]
\[={{\omega }^{2}}+\omega \] …………………………………(26)
Now, from equation (10) and equation (26), we get
 \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}=1\] ……………………………………………(27)
Similarly, on putting \[x={{\omega }^{2}}\] in \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}\] , we get
\[={{\omega }^{4}}+\dfrac{1}{{{\omega }^{4}}}\]
\[={{\omega }^{3}}\times \omega +\dfrac{1}{{{\omega }^{3}}\times \omega }\] …………………………………(28)
Now, from equation (8) and equation (28), we get
\[=1\times \omega +\dfrac{1}{1\times \omega }\]
\[=\omega +{{\omega }^{2}}\] ………………………………………….(29)
Now, from equation (10) and equation (29), we get
So, \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}=1\] ……………………………………………(30)
Now, on putting \[x=\omega \] in \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}\] , we get
\[={{\omega }^{3}}+\dfrac{1}{{{\omega }^{3}}}\] ………………………………(31)
From equation (8) and equation (31), we get
\[\begin{align}
  & =1+\dfrac{1}{1} \\
 & =2 \\
\end{align}\]
Similarly, on putting \[x={{\omega }^{2}}\] in \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}\] , we get
\[={{\omega }^{6}}+\dfrac{1}{{{\omega }^{6}}}\] ………………………………(32)
From equation (8) and equation (32), we get
\[\begin{align}
  & =1+\dfrac{1}{1} \\
 & =2 \\
\end{align}\]
So, \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}=2\] ……………………………(33)
Using equation (22), equation (25), equation (30), and equation (33), we can say that statement (2) is true ……………………………………….(34)
Now, from equation (21) and equation (34), we have statement 1 as a false statement and statement 2 as a true statement.

So, the correct answer is “Option D”.

Note: For this type of question, one might get confused about how to approach it. So, whenever this type of question appears, always try to solve statement-wise , first solve the information provided in statement 1 and then check whether it is true or false. Similarly, solve statement 2 and check whether it is true or false. At last, verify whether statement 2 is the correct explanation of statement 1 or not.