JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 4)
Let a – 2b + c = 1.
If $$f(x)=\left| {\matrix{ {x + a} & {x + 2} & {x + 1} \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|$$, then:
If $$f(x)=\left| {\matrix{ {x + a} & {x + 2} & {x + 1} \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|$$, then:
ƒ(50) = 1
ƒ(–50) = –1
ƒ(50) = –501
ƒ(–50) = 501
Explanation
R1 $$ \to $$ R1 + R3 – 2R2
f(x) = $$\left| {\matrix{ {a + c - 2b} & 0 & 0 \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|$$
= (a + c – 2b) ((x + 3)2 – (x + 2)(x + 4))
= x2 + 6x + 9 – x2 – 6x – 8 = 1
$$ \therefore $$ f(x) = 1
$$ \Rightarrow $$ f(50) = 1
f(x) = $$\left| {\matrix{ {a + c - 2b} & 0 & 0 \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|$$
= (a + c – 2b) ((x + 3)2 – (x + 2)(x + 4))
= x2 + 6x + 9 – x2 – 6x – 8 = 1
$$ \therefore $$ f(x) = 1
$$ \Rightarrow $$ f(50) = 1
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