F - R - R - f x -3 x 2- mx - n - x 2-1- If the range of f x is -4-3- thenA- m -0B- n-4C- m-2 D- n -4

The correct options are A m=0 D n= 4y=3x^2+mx+nx^2+1 x^2(y 3) mx+y n = 0 ∵ x ∈ℝ⇒ D ≥ 0 ⇒ m^2 4(y 3)(y n) ≥ 0 ⇒ m^2 4(y^2 ny 3y+3n) ≥ 0 ⇒4y^2 4y (n+3)+12n m^ ...

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Standard XII

Mathematics

Range of Quadratic Expression

f : R → R , f...

Question

$f : R \to R,$ $f (x) = \frac{3 x^{2} + m x + n}{x^{2} + 1} .$ If the range of $f (x)$ is $[- 4, 3],$ then

m = 0

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n = 4

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m = 2

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n = - 4

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Solution

The correct options are
A $m = 0$
D $n = - 4$
$y = \frac{3 x^{2} + m x + n}{x^{2} + 1}$
$x^{2} (y - 3) - m x + y - n = 0$
$∵ x \in R \Rightarrow D \geq 0$
$\Rightarrow m^{2} - 4 (y - 3) (y - n) \geq 0$
$\Rightarrow m^{2} - 4 (y^{2} - n y - 3 y + 3 n) \geq 0$
$\Rightarrow 4 y^{2} - 4 y (n + 3) + 12 n - m^{2} \leq 0 \dots (1)$

Given range is $[- 4, 3]$
So, $y^{2} + y - 12 \leq 0 \dots (2)$

Compare $(1)$ and $(2),$ we get
$\frac{4}{1} = - \frac{4 (n + 3)}{1} = \frac{12 n - m^{2}}{- 12}$
$\Rightarrow m = 0$ and $n = - 4$

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Range of Quadratic Expression

Standard XII Mathematics

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f:R→R, f(x)=3x2+mx+nx2+1. If the range of f(x) is [−4,3], then

$f : R \to R,$ $f (x) = \frac{3 x^{2} + m x + n}{x^{2} + 1} .$ If the range of $f (x)$ is $[- 4, 3],$ then