"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:"

Question

Accepted Answer

"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"

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:

Solution

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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:

Solution

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More Sets, Relations and Functions questions

Drill 25 more like these. Every day. Free.

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: