JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 7)
If $$\Delta $$ = $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{2x - 3} & {3x - 4} & {4x - 5} \cr
{3x - 5} & {5x - 8} & {10x - 17} \cr
} } \right|$$ =
Ax3 + Bx2 + Cx + D, then B + C is equal to :
Ax3 + Bx2 + Cx + D, then B + C is equal to :
-1
-3
9
1
Explanation
$$\Delta $$ = $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{2x - 3} & {3x - 4} & {4x - 5} \cr
{3x - 5} & {5x - 8} & {10x - 17} \cr
} } \right|$$
R2 $$ \to $$ R2 – R1
R3 $$ \to $$ R3 – R2
= $$\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {x - 1} & {x - 1} & {x - 1} \cr {x - 2} & {2\left( {x - 2} \right)} & {6\left( {x - 2} \right)} \cr } } \right|$$
= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr 1 & 1 & 1 \cr 1 & 2 & 6 \cr } } \right|$$
C1 $$ \to $$ C1 - C2
C2 $$ \to $$ C2 - C3
= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ { - x + 1} & { - x + 1} & {3x - 4} \cr 0 & 0 & 1 \cr { - 1} & { - 4} & 6 \cr } } \right|$$
= -(x - 1)(x - 2)[-4(1 - x) + 1(1 - x)]
= -(x2 - 3x + 2)[3x - 3]
= -3x3 + 9x2 - 6x + 3x2 - 9x + 6
= -3x3 + 12x2 - 15x + 6 = Ax3 + Bx2 + Cx + D
$$ \therefore $$ A = -3, B = 12, C = -15
$$ \therefore $$ B + C = 12 – 15 = – 3
R2 $$ \to $$ R2 – R1
R3 $$ \to $$ R3 – R2
= $$\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {x - 1} & {x - 1} & {x - 1} \cr {x - 2} & {2\left( {x - 2} \right)} & {6\left( {x - 2} \right)} \cr } } \right|$$
= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr 1 & 1 & 1 \cr 1 & 2 & 6 \cr } } \right|$$
C1 $$ \to $$ C1 - C2
C2 $$ \to $$ C2 - C3
= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ { - x + 1} & { - x + 1} & {3x - 4} \cr 0 & 0 & 1 \cr { - 1} & { - 4} & 6 \cr } } \right|$$
= -(x - 1)(x - 2)[-4(1 - x) + 1(1 - x)]
= -(x2 - 3x + 2)[3x - 3]
= -3x3 + 9x2 - 6x + 3x2 - 9x + 6
= -3x3 + 12x2 - 15x + 6 = Ax3 + Bx2 + Cx + D
$$ \therefore $$ A = -3, B = 12, C = -15
$$ \therefore $$ B + C = 12 – 15 = – 3
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