Question

The 6th (from the beginning) term in the expansion of the [√2log(10−3x)+5√2(x−2)log3]m

is equal to 21. It is known that the binomial coefficient of the 2nd,3rd and 4th term in the expansion are in an A.P., then the value of x is/are

is equal to 21. It is known that the binomial coefficient of the 2nd,3rd and 4th term in the expansion are in an A.P., then the value of x is/are

- 0
- 1
- 2
- 3

Solution

The correct options are

**A** 0

**C** 2

The coefficients are

mC1,mC2 and mC3

They are T2,T3,T4 terms of an A.P.

∴2⋅mC2=mC1+mC3⇒m(m−1)=m+m(m−1)(m−2)6⇒m2−9m+14=0[∵m≠0]⇒m=2,7

There should be minimum 6 terms in the expasion, so

m≠2,m=7

Now,

⇒T6=7C5[√2log(10−3x)]7−5⋅[5√2(x−2)log3]5⇒21=21⋅2log(10−3x)⋅2(x−2)log3⇒1=2log(10−3x)+log3x−2⇒0=log(10−3x)+log3x−2⇒(10−3x)(3x−2)=1⇒(10−3x)3x=9⇒32x−10⋅3x+9=0⇒(3x−9)(3x−1)=0⇒3x=9,1⇒x=2,0

The coefficients are

mC1,mC2 and mC3

They are T2,T3,T4 terms of an A.P.

∴2⋅mC2=mC1+mC3⇒m(m−1)=m+m(m−1)(m−2)6⇒m2−9m+14=0[∵m≠0]⇒m=2,7

There should be minimum 6 terms in the expasion, so

m≠2,m=7

Now,

⇒T6=7C5[√2log(10−3x)]7−5⋅[5√2(x−2)log3]5⇒21=21⋅2log(10−3x)⋅2(x−2)log3⇒1=2log(10−3x)+log3x−2⇒0=log(10−3x)+log3x−2⇒(10−3x)(3x−2)=1⇒(10−3x)3x=9⇒32x−10⋅3x+9=0⇒(3x−9)(3x−1)=0⇒3x=9,1⇒x=2,0

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