It is given that,, f(x) = x 2 - 4x + 5 Let y = x 2 - 4x + 5 y = x 2 - 4x + 4 + 1 = (x - 2) 2 + 1 = y - 1 x - 2 = x = 2 + y - 1 0,y 1 Range = [1, )

Relations and Functions — Class 12 Maths Solution

exemplar objective MCQ 2,\infty ) \to R$ be the function defined by $f(x) = {x^2} - 4x + 5$, then the range of $f$ is \begin{enumerate} \item $R$ \item $[1,\infty )$ \item $[4,\infty )$ \item $[5,\infty )$ \end{enumerate} [NCERT Exemp.Q.44,Pa

Question

If $f:$

(a) $R$
(b) $[1,\infty )$ ✓ Correct
(c) $[4,\infty )$
(d) $[5,\infty )$

Step-by-step Solution

It is given that,, $f(x) = {x^2} - 4x + 5$

Let $y = {x^2} - 4x + 5$

$\Rightarrow$ $y = {x^2} - 4x + 4 + 1 = {(x - 2)^2} + 1$

$\Rightarrow$ ${(x - 2)^2} = y - 1 \Rightarrow x - 2 = \sqrt {y - 1}$

$\Rightarrow$ $x = 2 + \sqrt {y - 1}$

$\therefore y - 1 \ge 0,y \ge 1$

Range $= [1,\infty )$

← All Relations and Functions solutions Practice tests →

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Relations and Functions. Curated by Sachin Sharma. Free for all students.