NDA 2021
Exam NDA
Year 2021
Subject Mathematics
Chapter Determinants NDA 2021 · Mathematics · Determinants
If f(x)=| array ccc 1 & x & x+1 2 x & x(x-1) & x(x+1) 3 x(x-1) &...
\text{ If } f(x)=\left| 1 x x + 1 2 x x ( x − 1 ) x ( x + 1 ) 3 x ( x − 1 ) 2 ( x − 1 ) ( x − 2 ) x ( x + 1 ) ( x − 1 ) \begin{array}{ccc}1 & x & x+1 \\2 x & x(x-1) & x(x+1) \\3 x(x-1) & 2(x-1)(x-2) & x(x+1)(x-1)
\end{array} 1 2 x 3 x ( x − 1 ) x x ( x − 1 ) 2 ( x − 1 ) ( x − 2 ) x + 1 x ( x + 1 ) x ( x + 1 ) ( x − 1 ) \right|
then what is
f ( − 1 ) + f ( 0 ) + f ( 1 ) f(-1)+f(0)+f(1) f ( − 1 ) + f ( 0 ) + f ( 1 ) equal to?
Options A 0 0 0 B 1 1 1 C 100 100 100 D − 100 -100 − 100
Step-by-step solution
f(x)=\left| 1 x x + 1 2 x x ( x − 1 ) x ( x + 1 ) 3 x ( x − 1 ) 2 ( x − 1 ) ( x − 2 ) x ( x + 1 ) ( x − 1 ) \begin{array}{ccc}1 & x & x+1 \\ 2 x & x(x-1) & x(x+1) \\ 3 x(x-1) & 2(x-1)(x-2) & x(x+1)(x-1)\end{array} 1 2 x 3 x ( x − 1 ) x x ( x − 1 ) 2 ( x − 1 ) ( x − 2 ) x + 1 x ( x + 1 ) x ( x + 1 ) ( x − 1 ) \right|
For
x = − 1 , C 3 → x=-1, C_3 \rightarrow x = − 1 , C 3 → \left[
− 1 + 1 − 1 ( − 1 + 1 ) − 1 ( − 1 + 1 ) ( − 1 − 1 ) \begin{array}{c}-1+1 \\ -1(-1+1) \\ -1(-1+1)(-1-1)\end{array} − 1 + 1 − 1 ( − 1 + 1 ) − 1 ( − 1 + 1 ) ( − 1 − 1 ) \right]=
\left[ 0 0 0 \begin{array}{l}0 \\ 0 \\ 0\end{array} 0 0 0 \right]
For
x = 1 x=1 x = 1 ,
R_3 \rightarrow\left[ 3 ( 1 − 1 ) 2 ( 1 − 1 ) ( 1 − 2 ) ( 1 + 1 ) ( 1 − 1 ) \begin{array}{lll}3(1-1) & 2(1-1)(1-2) & (1+1)(1-1)\end{array} 3 ( 1 − 1 ) 2 ( 1 − 1 ) ( 1 − 2 ) ( 1 + 1 ) ( 1 − 1 ) \right]=\left[ 0 0 0 \begin{array}{lll}0 & 0 & 0\end{array} 0 0 0 \right]\therefore f(-1)+f(0)+f(1)=0 .
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