- Home
- Standard 12
- Mathematics
3 and 4 .Determinants and Matrices
hard
माना $a -2 b + c =1$ है। यदि $f(x)=\left|\begin{array}{lll}x+a & x+2 & x+1 \\x+b & x+3 & x+2 \\x+c & x+4 & x+3\end{array}\right|$ है, तो
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