Let p, q be integers and let alpha, beta be t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B, D) $(1)$ Since $\alpha$ and $\beta$ are roots of $x^2-x-1=0$,we have $\alpha^2 = \alpha+1$ and $\beta^2 = \beta+1$.Multiplying by $\alpha^{n-2}$ a

Vedclass · Accepted Answer

$A, D$

Vedclass · Answer

(B, D) $(1)$ Since $\alpha$ and $\beta$ are roots of $x^2-x-1=0$,we have $\alpha^2 = \alpha+1$ and $\beta^2 = \beta+1$.Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively: $\alpha^n = \alpha^{n-1} + \alpha^{n-2}$ and $\beta^n = \beta^{n-1} + \beta^{n-2}$.Multiplying by $p$ and $q$ and adding: $p\alpha^n + q\beta^n = (p\alpha^{n-1} + q\beta^{n-1}) + (p\alpha^{n-2} + q\beta^{n-2})$.Thus,$a_n = a_{n-1} + a_{n-2}$. For $n=12$,$a_{12} = a_{11} + a_{10}$. Option $B$ is correct.$(2)$ We have $a_0 = p+q$,$a_1 = p\alpha + q\beta$,$a_2 = p\alpha^2 + q\beta^2 = p(\alpha+1) + q(\beta+1) = a_1 + a_0$,$a_3 = a_2 + a_1 = 2a_1 + a_0$,$a_4 = a_3 + a_2 = 3a_1 + 2a_0$.Given $a_4 = 3(p\alpha + q\beta) + 2(p+q) = 28$.Since $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2}$,$3p(\frac{1+\sqrt{5}}{2}) + 3q(\frac{1-\sqrt{5}}{2}) + 2p + 2q = 28$.$\frac{3p+3p\sqrt{5}+3q-3q\sqrt{5}}{2} + 2p + 2q = 28$.$(3.5p + 3.5q) + \frac{3\sqrt{5}}{2}(p-q) = 28$.Since $28$ is rational,$p-q=0 \Rightarrow p=q$.$7p = 28 \Rightarrow p=4, q=4$.Then $p+2q = 4 + 2(4) = 12$. Option $D$ is correct.

Similar Questions

If the sum of two roots of the equation $x^4-2x^3+x^2+4x-6=0$ is zero,then the sum of the squares of the other two roots is

If $\alpha, \beta$, where $\alpha < \beta$, are the roots of the equation $\lambda x^{2} - (\lambda + 3)x + 3 = 0$ such that $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, then the sum of all possible values of $\lambda$ is:

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-4x+3=0$,then $\frac{2(\alpha^4+\beta^4)+3(\alpha^2+\beta^2)}{\alpha+\beta} = $

The harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module