The value of a+1a+2 a+2 1 a+2a+3 a+3 1 a+3a+4 a+4 1 is

Correct answer: (C) -2. R_2→R_2-R_1 and R_3→R_3-R_1 Δ= a+1a+2 a+2 1 a+2a+3-a-1 1 0 a^2+7a+12-a^2-3a-2 2 0 = a^2+3a+2 a+2 1 2a+2 1 0 4a+10 2 0 =4a+2-4 a-10 =4a+8-4a-10=-2

Q46 / 90 · JEE Main 2021 (26 Feb Shift 1) · Mathematics

The value of $|\begin{array}{l} (a + 1) (a + 2) & a + 2 & 1 \\ (a + 2) (a + 3) & a + 3 & 1 \\ (a + 3) (a + 4) & a + 4 & 1 \end{array}|$ is

A
$0$
B
$(a + 2) (a + 3) (a + 4)$
Correct answer
C
$- 2$
D
$(a + 1) (a + 2) (a + 3)$

Solution

R_{2} \to R_{2} - R_{1}

and

R_{3} \to R_{3} - R_{1}

$Δ = |\begin{matrix} (a + 1) (a + 2) & a + 2 & 1 \\ (a + 2) (a + 3 - a - 1) & 1 & 0 \\ a^{2} + 7 a + 12 - a^{2} - 3 a - 2 & 2 & 0 \end{matrix}|$

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

$= 4 (a + 2) - 4 a - 10$

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

Hence, the correct answer is option: 'C'.

The value of a+1a+2a+21a+2a+3a+31a+3a+4a+41 is

Solution

Relevant Questions

The value of $|\begin{array}{l} (a + 1) (a + 2) & a + 2 & 1 \\ (a + 2) (a + 3) & a + 3 & 1 \\ (a + 3) (a + 4) & a + 4 & 1 \end{array}|$ is