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JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Determinants

If

Δ = x - 2 2 x - 3 3 x - 5 2 x - 3 3 x - 4 5 x - 8 3 x - 4 4 x - 5 10 x - 17 =

A x^{3} + B x^{2} + C x + D

, then

B + C

is equal to:

(A)-3✓ Correct
(B)9
(C)-1
(D)1

Answer: (A)

Solution

Using

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

and

R_{3} \to R_{3} - R_{2}

, the second row becomes

(x - 1, x - 1, x - 1)

and the third

(x - 2, 2 (x - 2), 6 (x - 2))

. Taking out

(x - 1)

and

(x - 2)

,

Δ = (x - 1) (x - 2) x - 2 11 2 x - 3 12 3 x - 4 16

, and expanding the remaining determinant gives

- 3 (x - 1)

. Hence

Δ = - 3 (x - 1)^{2} (x - 2) = - 3 x^{3} + 12 x^{2} - 15 x + 6

, so

B = 12

,

C = - 15

and

B + C = - 3

.

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option

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