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JEE Main 20209 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2020 · 9 January · Evening Shift · Mathematics · Determinants

Let a - 2b + c = 1 . If f(x) = | cases x + a & x + 2 & x + 1 x + b...

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

Options

  1. Af(−50)=501f\left(- 50\right) = 501
  2. Bf(−50)=−1f\left(- 50\right) = - 1
  3. Cf(50)=−501f\left(50\right) = - 501
  4. Df(50)=1f\left(50\right) = 1

Correct answer

D. f(50)=1f\left(50\right) = 1

Step-by-step solution

Apply R1=R1+R3−2R2{R}_{1} = {R}_{1} + {R}_{3} - 2{R}_{2}
⇒ f\left(x\right) = \left(a - 2b + c\right)\left|\begin{cases}1 & 0 & 0 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3\end{cases}\right|

⇒f(x)=1[(x+3)2−(x+2)(x+4)]⇒ f\left(x\right) = 1\left[{\left(x + 3\right)}^{2} - \left(x + 2\right)\left(x + 4\right)\right]

⇒f(x)=1⇒ f\left(x\right) = 1

⇒f(50)=1⇒ f\left(50\right) = 1

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