ધારો કે a-2 b+c=1 છે . જો f(x)=left|begin{array}{lll}{x+a} & {x+2} &am

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3 and 4 .Determinants and Matrices

hard

ધારો કે $a-2 b+c=1$ છે . જો $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ હોય તો . . .

A

$f(-50)=501$

B

$f(-50)=-1$

C

$f(50)=1$

D

$f(50)=501$

(JEE MAIN-2020)

Solution

$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}$

$f(x)=\left|\begin{array}{ccc}{a+c-2 b} & {0} & {0} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|$

$=(a+c-2 b)\left((x+3)^{2}-(x+2)(x+4)\right)$

$=x^{2}+6 x+9-x^{2}-6 x-8=1$

$\Rightarrow f(x)=1 \Rightarrow f(50)=1$

Standard 12

Mathematics

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