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JEE Advanced Mathematics Determinants 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Mathematics Question (2022) — Solution

Question

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  q r = 10 , then the system of linear equations has P x = 0 , y = 10 9 , z = - 1 9  as a solution II If  p r ≠ 100 , then the system of linear equations has Q x = 10 9 , y = - 1 9 , z = 0  as a solution III If  p q ≠ 10 , then the system of linear equations has R infinitely many solutions IV If  p q = 10 , then the system of linear equations has S no solution     T at least one solution The correct option is:

Options

  1. A. I → T ; II → R ; III → S ; IV → T
  2. B. I → Q ; II → S ; III → S ; IV → R
  3. C. I → Q ; II → R ; III → P ; IV → R
  4. D. I → T ; II → S ; III → P ; IV → T

Answer

B. I → Q ; II → S ; III → S ; IV → R

Step-by-step solution

Given, x + y + z = 1             ⋯ 1 10 x + 100 y + 1000 z = 0             ⋯ 2 q r x + p r y + p q z = 0               ⋯ 3 Now equation 3  can be re-written as x p + y q + z r = 0         ∵   p , q , r ≠ 0 Now given  p ,   q   &   r  are 10 t h , 100 t h  & 1000 t h  term of an h.p , So, let  p = 1 a + 9 d ,   q = 1 a + 99 d   &   r = 1 a + 999 d Now, equation  3  will be a + 9 d x + a + 99 d y + a + 999 d z = 0 Now from equation  1 ,   2   &   3  we get, Δ = 1 1 1 10 100 1000 a + 9 d a + 99 d a + 999 d = 0 Δ x = 1 1 1 0 100 1000 0 a + 99 d a + 999 d = 900 d - a Δ y = 1 1 1 10 0 1000 a + 9 d 0 a + 999 d = 990 a - d Δ z = 1 1 1 10 100 0 a + 9 d a + 99 d 0 = 90 d - a Option I: If  q r = 10 ⇒ a = d Δ = Δ x = Δ y = Δ z = 0 And eq.  1  and eq.  2  represents non-parallel planes eq.  2  and eq.  3  represents same plane ⇒  Infinitely many solutions Now finding solution by taking  z = λ  so from equation  1   &   2  we get, x + y = 1 - λ  and  x + 10 y = - 100 λ ⇒ x = 10 9 + 10 λ ,   y = - 1 9 - 11 λ ⇒ x , y , z ∈ 10 9 + 10 λ ,   - 1 9 - 11 λ ,   λ So,  P  is not valid for any value of  λ  rest are valid. So, option (i) → Q , R , T Option II:  p r ≠ 100 ⇒ a ≠ d Δ = 0   &   Δ x , Δ y , Δ z ≠ 0 So, no solution Option (ii) → S Option (iii): If  p q ≠ 10 ⇒ a ≠ d  then  Δ z ≠ 0 So, no solution Option (iii)  → S Now option (iv): If  p q = 10 ⇒ a = d  then  Δ z = 0 ⇒ Δ x = Δ y = 0 So, infinitely many solutions Option (iv) → Q , R , T   similar to option (i)

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