The value of | cases (a + 1)(a + 2) & a + 2 & 1 (a + 2)(a + 3) & a + 3 & 1 (a + 3)(a + 4) & a + 4 & 1 cases | is

Correct answer: C. - 2 R _ 2 → R _ 2 - R _ 1 and R _ 3 → R _ 3 - R _ 1 Δ = | cases (a + 1)(a + 2) & a + 2 & 1 (a + 2)(a + 3 - a - 1) & 1 & 0 a ^ 2 + 7a + 12 - a ^ 2 - 3a - 2 & 2 & 0 cases | = | cases a ^ 2 + 3a + 2 & a + 2 & 1 2(a + 2) & 1 & 0 4a + 10 & 2 & 0 cases | = 4(a + 2) - 4a - 10 = 4a + 8 - 4a - 10 = - 2

JEE Main 202126 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Morning Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2021 · 26 February · Morning Shift · Mathematics · Determinants

The value of | cases (a + 1)(a + 2) & a + 2 & 1 (a + 2)(a + 3) & a...

The value of

\left|

\begin{cases}\left(a + 1\right)\left(a + 2\right) & a + 2 & 1 \\ \left(a + 2\right)\left(a + 3\right) & a + 3 & 1 \\ \left(a + 3\right)\left(a + 4\right) & a + 4 & 1\end{cases}

\right|

Options

A $0$
B $\left(a + 2\right)\left(a + 3\right)\left(a + 4\right)$
C $- 2$
D $\left(a + 1\right)\left(a + 2\right)\left(a + 3\right)$

Correct answer

C. $- 2$

Step-by-step solution

{R}_{2} \to {R}_{2} - {R}_{1}

and

{R}_{3} \to {R}_{3} - {R}_{1}

$\Delta = \left|$ \begin{cases}\left(a + 1\right)\left(a + 2\right) & a + 2 & 1 \\ \left(a + 2\right)\left(a + 3 - a - 1\right) & 1 & 0 \\ {a}^{2} + 7a + 12 - {a}^{2} - 3a - 2 & 2 & 0\end{cases} $\right|$

$= \left|$ \begin{cases}{a}^{2} + 3a + 2 & a + 2 & 1 \\ 2\left(a + 2\right) & 1 & 0 \\ 4a + 10 & 2 & 0\end{cases} $\right|$

$= 4\left(a + 2\right) - 4a - 10$

$= 4a + 8 - 4a - 10 = - 2$

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