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Functions question

2020 · 9 Jan · Shift 2 · Q24
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Functions question

2020 · 9 Jan · Shift 2 · Q24

JEE MainMathematicsFunctionsMCQ+4 / −1
Let a – 2b + c = 1. If f(x)=∣x+ax+2x+1x+bx+3x+2x+cx+4x+3∣f(x)=\left| {\begin{matrix} {x + a} & {x + 2} & {x + 1} \\ {x + b} & {x + 3} & {x + 2} \\ {x + c} & {x + 4} & {x + 3} \\ \end{matrix} } \right|f(x)=​x+ax+bx+c​x+2x+3x+4​x+1x+2x+3​​, then:
  1. A
    ƒ(50) = 1
  2. B
    ƒ(–50) = –1
  3. C
    ƒ(50) = –501
  4. D
    ƒ(–50) = 501
View written solutionFree

Correct answer: A

  1. We need evaluate
x+a & x+2 & x+1\\ x+b & x+3 & x+2\\ x+c & x+4 & x+3 \end{vmatrix}$$ given $$a-2b+c=1.$$ --- 2. Write the columns explicitly: Let $$C_1=\begin{pmatrix}x+a\\x+b\\x+c\end{pmatrix},\quad C_2=\begin{pmatrix}x+2\\x+3\\x+4\end{pmatrix},\quad C_3=\begin{pmatrix}x+1\\x+2\\x+3\end{pmatrix}.$$ Notice that $$C_2-C_3=\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ Also, $$C_1-C_3=\begin{pmatrix}a-1\\b-2\\c-3\end{pmatrix}.$$ Now perform column operations: $$C_1\to C_1-C_3,\qquad C_2\to C_2-C_3.$$ These do not change the determinant. Hence $$f(x)=\begin{vmatrix} a-1 & 1 & x+1\\ b-2 & 1 & x+2\\ c-3 & 1 & x+3 \end{vmatrix}.$$ --- 3. Now do row operations: $$R_2\to R_2-R_1,\qquad R_3\to R_3-R_2 \text{ (original consecutive difference idea)}$$ More directly from the original simplified matrix, $$R_2-R_1=(b-a-1,0,1),$$ $$R_3-R_2=(c-b-1,0,1).$$ So it is easier to expand along the second column of $$\begin{vmatrix} a-1 & 1 & x+1\\ b-2 & 1 & x+2\\ c-3 & 1 & x+3 \end{vmatrix}.$$ Expanding along column 2: $$f(x)=(-1)^{1+2}\begin{vmatrix}b-2 & x+2\\ c-3 & x+3\end{vmatrix} +(-1)^{2+2}\begin{vmatrix}a-1 & x+1\\ c-3 & x+3\end{vmatrix} +(-1)^{3+2}\begin{vmatrix}a-1 & x+1\\ b-2 & x+2\end{vmatrix}.$$ Thus $$f(x)= -\big((b-2)(x+3)-(c-3)(x+2)\big) +\big((a-1)(x+3)-(c-3)(x+1)\big) -\big((a-1)(x+2)-(b-2)(x+1)\big).$$ --- 4. Simplify each bracket: First, $$ (b-2)(x+3)-(c-3)(x+2) = (b-c+1)x + (3b-2c). $$ Second, $$ (a-1)(x+3)-(c-3)(x+1) = (a-c+2)x + (3a-c). $$ Third, $$ (a-1)(x+2)-(b-2)(x+1) = (a-b+1)x + (2a-b). $$ Therefore $$f(x)= -\big((b-c+1)x + (3b-2c)\big) +\big((a-c+2)x + (3a-c)\big) -\big((a-b+1)x + (2a-b)\big).$$ Collect coefficients of $x$: $$-(b-c+1)+(a-c+2)-(a-b+1)=0.$$ So $f(x)$ is a constant, independent of $x$. Constant term: $$-(3b-2c)+(3a-c)-(2a-b)=a-2b+c.$$ Given $$a-2b+c=1,$$ we get $$f(x)=1 \quad \text{for all } x.$$ --- 5. Check options: - **A:** $f(50)=1$ ✅ - **B:** $f(-50)=-1$ ❌ since $f(-50)=1$ - **C:** $f(50)=-501$ ❌ - **D:** $f(-50)=501$ ❌ So the correct option is: $$\boxed{A}$$
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