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B
(−∞,17]∪[13,∞)
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C
[17,1)
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D
[17,13]
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Solution
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]