Let $a-2 b+c=1$. If $f(x)=\left|\begin{array}{ccc}x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3\end{array}\right|$, then:

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Determinants

Let

a - 2 b + c = 1

. If

f (x) = x + a x + b x + c x + 2 x + 3 x + 4 x + 1 x + 2 x + 3

, then:

Answer: (B)

Apply the row operation

R_{1} \to R_{1} + R_{3} - 2 R_{2}

. Because

a - 2 b + c = 1

, the first entry becomes

1

, while the other two become

(x + 2) + (x + 4) - 2 (x + 3) = 0

and

(x + 1) + (x + 3) - 2 (x + 2) = 0

. The determinant then expands along the first row to

1 \cdot x + 3 x + 4 x + 2 x + 3 = (x + 3)^{2} - (x + 2) (x + 4) = 1

. Thus

f (x) = 1

for every

x

, and in particular

f (50) = 1

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option