If x -4-1- then 1-x2-4 x-7 belongs toA- -1-7- -B- - 1-7- -1-3- - -1-7- 1D- -1-7- 1-3-

The correct option is D [17,13]Here, x∈[ 4, 1] ⇒ x^2+4x+7 =(x+2)^2+3 It will be minimum at x= 2 and maximum at x= 4 ⇒ (x+2)^2+3∈[3,7] Since, (3)(7)0 So, nbsp ...

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Byju's Answer

Standard XII

Mathematics

Properties of Inequalities

If x ∈[-4,-1]...

Question

If $x \in [- 4, - 1]$ , then $\frac{1}{x^{2} + 4 x + 7}$ belongs to

[\frac{1}{7}, \infty)

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(- \infty, \frac{1}{7}] \cup [\frac{1}{3}, \infty)

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[\frac{1}{7}, 1)

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[\frac{1}{7}, \frac{1}{3}]

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Solution

The correct option is D $[\frac{1}{7}, \frac{1}{3}]$
Here, $x \in [- 4, - 1]$
$\Rightarrow x^{2} + 4 x + 7 = (x + 2)^{2} + 3$
It will be minimum at $x = - 2$ and maximum at $x = - 4$
$\Rightarrow (x + 2)^{2} + 3 \in [3, 7]$
Since, $(3) (7) > 0$
So, $\frac{1}{x^{2} + 4 x + 7} \in [\frac{1}{7}, \frac{1}{3}]$

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If x∈[−4,−1], then 1x2+4x+7 belongs to

The correct option is D [17,13]Here, x∈[−4,−1] ⇒x2+4x+7=(x+2)2+3 It will be minimum at x=−2 and maximum at x=−4 ⇒(x+2)2+3∈[3,7] Since, (3)(7)>0 So, 1x2+4x+7∈[17,13]

If $x \in [- 4, - 1]$ , then $\frac{1}{x^{2} + 4 x + 7}$ belongs to