The range of the function f x - x 2-2 x -15- is A- R -B- R -0C- -0- -D- R

The correct option is C [0,∞)Clearly, domain of f is ℝ. x^2 +2x 15=(x+1)^2 16 ⇒ 16 ≤ x^2 +2x 15 ∞ ⇒ 0≤ |x^2 +2x 15| ∞ Hence, range of nbsp; f is [0,∞) ...

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Standard XI

Mathematics

Range

The range of ...

Question

The range of the function $f (x) = | x^{2} + 2 x - 15 |$ is

R

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R^{+}

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R - {0}

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[0, \infty)

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Solution

The correct option is D $[0, \infty)$
Clearly, domain of $f$ is $R$ .
$x^{2} + 2 x - 15 = (x + 1)^{2} - 16$
$\Rightarrow - 16 \leq x^{2} + 2 x - 15 < \infty$
$\Rightarrow 0 \leq | x^{2} + 2 x - 15 | < \infty$

Hence, range of $f$ is $[0, \infty)$

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The range of the function f(x)=|x2+2x−15| is

The correct option is D [0,∞)Clearly, domain of f is R. x2+2x−15=(x+1)2−16 ⇒−16≤x2+2x−15<∞ ⇒0≤|x2+2x−15|<∞ Hence, range of f is [0,∞)

The range of the function $f (x) = | x^{2} + 2 x - 15 |$ is

The correct option is D $[0, \infty)$
Clearly, domain of $f$ is $R$ .
$x^{2} + 2 x - 15 = (x + 1)^{2} - 16$
$\Rightarrow - 16 \leq x^{2} + 2 x - 15 < \infty$
$\Rightarrow 0 \leq | x^{2} + 2 x - 15 | < \infty$

Hence, range of $f$ is $[0, \infty)$