JEE Advanced · Mathematics · 2. Quadratic Equations

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Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation, $x^{2}-x-1=0$, where $\alpha \neq \beta$. For $n=0,1,2, \ldots$, let $a_{n}=p \alpha^{n}+q \beta^{n}$
FACT: If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$.

Question:
If $a_{4}=28$, then $p+2 q=$

A $14$
B $7$
C $12$
D $21$

Verified Solution

Answer & Solution

Correct Answer

(C) $12$

Step-by-step Solution

Detailed explanation

α^{2} = α + 1

\Rightarrow α^{4} = {(α^{2})}^{2}

\Rightarrow α^{4} = {(α + 1)}^{2}

\Rightarrow α^{4} = α^{2} + 2 α + 1

\Rightarrow α^{4} = α + 1 + 2 α + 1

\Rightarrow α^{4} = 3 α + 2

A l s o α + β = 1

$\therefore a_4=28 \Rightarrow p \alpha^4$ $+~q \beta^4=p(3 \alpha+2)+q(3 \beta+2)=28$
$\Rightarrow p(3 \alpha+2)+q(3-3 \alpha+2)=28$
$\Rightarrow \alpha(3 p-3 q)+2 p+5 q=28 \quad$ (as $\alpha \in Q^c$ ) (as it is root of quadratic equation)
$\Rightarrow p=q, 2 p+5 q=28 \Rightarrow p=q=4$
$\therefore p+2 q=12$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let

p, q, r

be non-zero real numbers that are, respectively, the

10^{th}, 100^{th}

and

1000^{th}

terms of a harmonic progression. Consider the system of linear equations

x + y + z = 1

10 x + 100 y + 1000 z = 0

q r x + p r y + p q z = 0

	List- $I$		List- $II$
$(I)$	If $\frac{q}{r} = 10$ , then the system of linear equations has	$(P)$	$x = 0, y = \frac{10}{9}, z = - \frac{1}{9}$ as a solution
$(II)$	If $\frac{p}{r} \neq 100$ , then the system of linear equations has	$(Q)$	$x = \frac{10}{9}, y = - \frac{1}{9}, z = 0$ as a solution
$(III)$	If $\frac{p}{q} \neq 10$ , then the system of linear equations has	$(R)$	infinitely many solutions
$(IV)$	If $\frac{p}{q} = 10$ , then the system of linear equations has	$(S)$	no solution
		$(T)$	at least one solution

The correct option is:

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An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation in conductivity of these reactions is given in List II. Match List I with List II and select the correct using answer the code given below the lists :

List - I	List - II
a. $\underset{\quad\quad~~ X }{( C _2 H _5)_3 N}+\underset{ Y }{CH _3 COOH}$	p. Conductivity decreases and then increases
b. $\underset{\quad\quad~~ X }{KI (0.1 M )}+\underset{ Y }{AgNO _3(0.01 M )}$	q. Conductivity decreases and then does not change much
c. $\underset{\text X }{CH _3 COOH}+\underset{\text Y }{KOH}$	r. Conductivity increases and then does not change much
d. $\underset{\text X }{NaOH}+\underset{\text Y }{HI}$	s. Conductivity does not change much and then increases

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List - I	List - II
a. \(\underset{\quad\quad~~ X }{( C _2 H _5)_3 N}+\underset{ Y }{CH _3 COOH}\)	p. Conductivity decreases and then increases
b. \(\underset{\quad\quad~~ X }{KI (0.1 M )}+\underset{ Y }{AgNO _3(0.01 M )}\)	q. Conductivity decreases and then does not change much
c. \(\underset{\text X }{CH _3 COOH}+\underset{\text Y }{KOH}\)	r. Conductivity increases and then does not change much
d. \(\underset{\text X }{NaOH}+\underset{\text Y }{HI}\)	s. Conductivity does not change much and then increases