First slide

Functions (XII)

Question

For real $x$ , the function $f (x)$ $= \frac{(x - a) (x - b)}{(x - c)}$ will be onto function provided

Moderate

A

$a \geq b \geq c$
B

$a \leq b \leq c$
C

$a \leq c \leq b$ or $a \geq c \geq b$
D

None of these

Solution

$y = f (x)$ $= \frac{(x - a) (x - b)}{(x - c)}$ is onto
$\Rightarrow x^{2} - (y + a + b) x + (a b + c y) = 0$ …(1) has real roots for all $y \in R$

$\Rightarrow {(y + a + b)}^{2} - 4 (a b + c y) \geq 0 \forall y \in R$

$\Rightarrow y^{2} + 2 (a + b - 2 c) y + {(a - b)}^{2} \geq 0$ $\forall y \in R$

$\Rightarrow 4 {(a + b - 2 c)}^{2} + 4 {(a - b)}^{2} \leq 0$

$\Rightarrow (c - a) (c - b) \leq 0$

$\Rightarrow$ $c$ lies between $a$ and $b$

$\Rightarrow a \leq c \leq b$ or $a \geq c \geq b$ .

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