JEE Advanced 2022 — Paper 1 — MATH-III solution

JEE Advanced2022Paper 1MATH-III

Let $p, q, r$ be non-zero real numbers that are, respectively, the $1 0^{th}$ , $10 0^{th}$ and $100 0^{th}$ terms of a harmonic progression. Consider the system of linear equations

$x + y + z = 1$
$10 x + 100 y + 1000 z = 0$
$q r x + p r y + pq z = 0$

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:

A
(I) $\to$ (T); (II) $\to$ (R); (III) $\to$ (S); (IV) $\to$ (T)
B
(I) $\to$ (Q); (II) $\to$ (S); (III) $\to$ (S); (IV) $\to$ (R)
C
(I) $\to$ (Q); (II) $\to$ (R); (III) $\to$ (P); (IV) $\to$ (R)
D
(I) $\to$ (T); (II) $\to$ (S); (III) $\to$ (P); (IV) $\to$ (T)

Solution

Note that Your institute's published answer key marks the correct option as (B). The reasoning below confirms it.

Let the HP terms have reciprocals in AP with first term $a$ and common difference $d$ : $\frac{1}{p} = a + 9 d$ , $\frac{1}{q} = a + 99 d$ , $\frac{1}{r} = a + 999 d$ . The third equation $q r x + p r y + pq z = 0$ divided by $pq r$ becomes $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$ , i.e., $(a + 9 d) x + (a + 99 d) y + (a + 999 d) z = 0$ .

The coefficient determinant is

$Δ = 111101001000 a + 9 d a + 99 d a + 999 d .$

Applying $C_{2} \to C_{2} - C_{1}$ , $C_{3} \to C_{3} - C_{1}$ on the third row: $(a + 99 d) - (a + 9 d) = 90 d$ , $(a + 999 d) - (a + 9 d) = 990 d$ . So

$Δ = 1001090990 a + 9 d 90 d 990 d = 90 \cdot 990 \cdot d - 990 \cdot 90 \cdot d = 0.$

So $Δ = 0$ unconditionally, and the system is either inconsistent or has infinitely many solutions.

(I) $\frac{q}{r} = 10$ : $\frac{a + 999 d}{a + 99 d} = 10 \Rightarrow a + 999 d = 10 a + 990 d \Rightarrow 9 d = 9 a \Rightarrow a = d$ . With $a = d$ , the third equation becomes $a (1 + 9) x + a (1 + 99) y + a (1 + 999) z = 0$ , i.e., $10 x + 100 y + 1000 z = 0$ (after dividing by $a$ ). This is identical to equation (2). The system reduces to equations (1) and (2), which represent two distinct, non-parallel planes intersecting in a line $\Rightarrow$ infinitely many solutions. So (I) $\to$ (R), and this answer also satisfies (T). One particular solution is $x = \frac{10}{9}, y = - \frac{1}{9}, z = 0$ (substitute and verify), so (I) $\to$ (Q) as well.

(II) $\frac{p}{r} \neq = 100$ : This means $\frac{a + 999 d}{a + 9 d} \neq = 100$ , i.e., $a + 999 d \neq = 100 a + 900 d$ , so $99 (a - d) \neq = 0$ , i.e., $a \neq = d$ . Now $Δ = 0$ but the augmented matrix has rank $3$ (since $Δ_{x}, Δ_{y}, Δ_{z}$ are not all zero when $a \neq = d$ ). So no solution. (II) $\to$ (S).

(III) $\frac{p}{q} \neq = 10$ : This means $a + 99 d \neq = 10 (a + 9 d) \Rightarrow 9 d \neq = 9 a \Rightarrow a \neq = d$ . By the same reasoning as (II), no solution. (III) $\to$ (S).

(IV) $\frac{p}{q} = 10$ : Gives $a = d$ , identical to case (I). Infinitely many solutions. (IV) $\to$ (R), and the particular solution $(0, \frac{10}{9}, - \frac{1}{9})$ checks out. So (IV) $\to$ (P).

Combining best matches: (I) $\to$ (Q); (II) $\to$ (S); (III) $\to$ (S); (IV) $\to$ (R), which is option (B).

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