Let $f: R -\{2,6\} \rightarrow R$ be real valued function defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$ Then range of $f$ is
Let $f: R -\{2,6\} \rightarrow R$ be real valued function defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$ Then range of $f$ is
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- $\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$
- $\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$
- $\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)$
- $\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)$
The Correct Option is B
Approach Solution - 1
By cross multiplying
Case
Case
So, can be 1
Hence
So, the correct option is (B) : $\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$
Approach Solution -2
Let \( y = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}. \) By cross-multiplying: \[ y(x^2 - 8x + 12) = x^2 + 2x + 1. \] Simplifying the equation: \[ yx^2 - 8xy + 12y = x^2 + 2x + 1, \] \[ yx^2 - x^2 - 8xy + 12y - 2x - 1 = 0. \] Case 1: Assume \( y \neq 1 \). \[ x^2(y - 1) - x(8y + 2) + (12y - 1) = 0. \] The discriminant condition for real solutions is \( D \geq 0 \). Simplifying: \[ (8y + 2)^2 - 4(y - 1)(12y - 1) \geq 0. \] Step 1: Solving this inequality results in the range for \( y \), which is \[ y \in \left( -\infty, \frac{-21}{4} \right] \cup [0, \infty). \] Case 2: Assume \( y = 1 \). Substitute into the equation: \[ x^2 + 2x + 1 = x^2 - 8x + 12. \] Simplifying: \[ 10x = 11 \quad \Rightarrow \quad x = \frac{11}{10}. \] Thus, \( y \) can be 1.
Step 2: Combining the solutions, the range of \( f(x) \) is \[ \left( -\infty, \frac{-21}{4} \right] \cup [0, \infty). \]
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions