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VITEEE 2025 Paper

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Question : 84 of 125

Marks: +1, -0

If

x

\frac{x^2+2x+c}{x^2+4x+3c}

can take all real values, then

$0 \leq c \leq 1$
$2 < c<3$
$c>4$
None of these

Solution:

Let,

y=\;\frac{x^2+2x+c}{x^2+4x+3c}

or

(y-1)x^2+(4y-2)x+3cy-c=0

Now,

x

is real. Hence,

or

(2y-1)^2-(y-1)(3cy-c) \geq 0, \forall y \in R

or

\;\;(4-3c)y^2+(-4+c+3c)y+1-c \geq 0, \forall y \in R

or

\;\;4-3c>0

and

(4c-4)^2-4(4-3c)(1-c) \leq 0

or

c < \;\frac{4}{3}

and

4(c-1)^2-(4-3c)(1-c) \leq 0

or

\;\;c < \;\frac{4}{3}

and

(c-1) \times (4c-4+4-3c) \leq 0

or

\;\;c < \;\frac{4}{3}

and

(c-1)(c) \leq 0

or

\;\;c < \;\frac{4}{3}

and

0 \leq c \leq 1

or

\;\;0 \leq c \leq 1

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