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$If the 4^{th} term in the expansion of {(2 + \frac{3 x}{8})}^{10} has the maximum$ $n u m e r i c a l v a l u e, t h e n t h e r a n g e o f v a l u e s o f x f o r w h i c h t h i s w i l l b e t r u e i s$

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a

−6421

b

2164

c

−6421

d

−2164

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detailed solution

Correct option is A

T4 is numerically greatest term ⇒T4 T3>1,⇒T4 T5>1⇒T4 T3>1,T5 T4<1Now T4 T3=C3 10273x83C2 10283x82=x2,T5 T4=C4 10263x84C3 10273x83=21x64∴T4 T3>1⇒x2>1⇒|x|>2⇒x∈-∞,-2∪2,∞ →1and T5 T4<1⇒21x64<1⇒|x|<6421⇒x∈-6421,6421 →2By 1 and 2 we have x∈−6421,−2∪2,6421

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