Question:
If a function \(f:\mathbb{R}-\{l\}\to \mathbb{R}-\{m\}\) defined by
\[
f(x)=\frac{x+3}{x-2}
\]
is a bijection, then \(3l+2m=\)
If a function \(f:\mathbb{R}-\{l\}\to \mathbb{R}-\{m\}\) defined by
\[
f(x)=\frac{x+3}{x-2}
\]
is a bijection, then \(3l+2m=\)
Show Hint
For a rational function of the form
\[
f(x)=\frac{ax+b}{cx+d},
\]
first exclude the value that makes the denominator zero from the domain, and then solve \(y=f(x)\) to find the value excluded from the range.
Updated On: Jun 24, 2026
- \(10\)
- \(12\)
- \(8\)
- \(14\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Find the value excluded from the domain.
The function is \[ f(x)=\frac{x+3}{x-2} \] For the function to be defined, the denominator must not be zero.
So, \[ x-2\neq 0 \] Therefore, \[ x\neq 2 \] Hence, \[ l=2 \]
Step 2: Find the value excluded from the range.
Let \[ y=\frac{x+3}{x-2} \] Now solve for \(x\): \[ y(x-2)=x+3 \] \[ xy-2y=x+3 \] \[ xy-x=2y+3 \] \[ x(y-1)=2y+3 \] \[ x=\frac{2y+3}{y-1} \] For \(x\) to be defined, \[ y-1\neq 0 \] So, \[ y\neq 1 \] Hence, the range is \[ \mathbb{R}-\{1\} \] Therefore, \[ m=1 \]
Step 3: Calculate \(3l+2m\).
Using \[ l=2,\quad m=1 \] we get \[ 3l+2m=3(2)+2(1) \] \[ 3l+2m=6+2 \] \[ 3l+2m=8 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{8} \]
The function is \[ f(x)=\frac{x+3}{x-2} \] For the function to be defined, the denominator must not be zero.
So, \[ x-2\neq 0 \] Therefore, \[ x\neq 2 \] Hence, \[ l=2 \]
Step 2: Find the value excluded from the range.
Let \[ y=\frac{x+3}{x-2} \] Now solve for \(x\): \[ y(x-2)=x+3 \] \[ xy-2y=x+3 \] \[ xy-x=2y+3 \] \[ x(y-1)=2y+3 \] \[ x=\frac{2y+3}{y-1} \] For \(x\) to be defined, \[ y-1\neq 0 \] So, \[ y\neq 1 \] Hence, the range is \[ \mathbb{R}-\{1\} \] Therefore, \[ m=1 \]
Step 3: Calculate \(3l+2m\).
Using \[ l=2,\quad m=1 \] we get \[ 3l+2m=3(2)+2(1) \] \[ 3l+2m=6+2 \] \[ 3l+2m=8 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{8} \]
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