If Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x - 5 & 5x - 8 & 10x - 17 cases | = A x ^ 3 + B x ^ 2 + Cx + D , then B + C is equal to :

Correct answer: C. - 3 Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x - 5 & 5x - 8 & 10x - 17 cases | Applying R _ 2 → R _ 2 - R _ 1 , R _ 3 → R _ 3 - R _ 2 , we get Δ = | cases x - 2 & 2x - 3 & 3x - 4 x - 1 & x - 1 & x - 1 x - 2 & 2(x - 2) & 6(x - 2) cases | = (x - 1)(x - 2)| cases x - 2 & 2x - 3 & 3x - 4 1 & 1 & 1 1 & 2 & 6 cases | = - 3 x ^ 3 + 12 x ^ 2 - 15x + 6 ∴ B + C = 12 - 15 = - 3

JEE Main 20203 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2020 · 3 September · Morning Shift · Mathematics · Determinants

If Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x...

\Delta = \left|

\begin{cases}x - 2 & 2x - 3 & 3x - 4 \\ 2x - 3 & 3x - 4 & 4x - 5 \\ 3x - 5 & 5x - 8 & 10x - 17\end{cases}

\right| = A{x}^{3} + B{x}^{2} + Cx + D

, then

B + C

is equal to :

Options

A $- 1$
B $1$
C $- 3$
D $9$

Correct answer

C. $- 3$

Step-by-step solution

$\Delta = \left|$ \begin{cases}x - 2 & 2x - 3 & 3x - 4 \\ 2x - 3 & 3x - 4 & 4x - 5 \\ 3x - 5 & 5x - 8 & 10x - 17\end{cases} $\right|$

Applying ${R}_{2} \to {R}_{2} - {R}_{1}$ , ${R}_{3} \to {R}_{3} - {R}_{2}$ , we get

$\Delta = \left|$ \begin{cases}x - 2 & 2x - 3 & 3x - 4 \\ x - 1 & x - 1 & x - 1 \\ x - 2 & 2\left(x - 2\right) & 6\left(x - 2\right)\end{cases} $\right| = \left(x - 1\right)\left(x - 2\right)\left|$ \begin{cases}x - 2 & 2x - 3 & 3x - 4 \\ 1 & 1 & 1 \\ 1 & 2 & 6\end{cases} $\right|$

$= - 3{x}^{3} + 12{x}^{2} - 15x + 6$

$∴ B + C = 12 - 15 = - 3$

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