If the middle term in the expansion of (x2+2)8is 1120; then x∈R is equal to
A
−2
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B
3
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C
−3
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D
2
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Solution
The correct options are A−2 D2 Here n is even ∴ Middle term is (n2+1)th⇒(4+1)th,i.e.,T5 ∴T5=8C4(x2)4⋅24=1120⇒8C4x4=8⋅7⋅6⋅51⋅2⋅3⋅4x4=1120 ⇒x4=112070=16 ⇒x4−16=0 ⇒(x2+4)(x2−4)=0 ∴x=±2 as x∈R