JEE 2023 Mathematics PYQ — Let the sixth term in the binomial expansion of , in the increasi… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let the sixth term in the binomial expansion of (2log2(10−3x)+52(x−2)log23)m, in the increasing powers of 2(x−2)log23, be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is:
Choose the correct answer:
A.
4
(Correct Answer)
B.
5
C.
6
D.
7
Correct Answer:
4
Explanation
Solution
1. Simplify the Terms
First term: A=2log2(10−3x)=(10−3x)1/2
Second term: B=52(x−2)log23=(3x−2)1/5
2. Find m from Binomial Coefficients
The coefficients are (1m),(2m),(3m). We are told (1m), (2m), and (3m) are the 1st, 3rd, and 5th terms of an A.P.
Let the common difference be d.
a1=(1m)=m
a3=(2m)=2m(m−1)
a5=(3m)=6m(m−1)(m−2)
In an A.P., a5−a3=a3−a1 is not necessarily true because these are the 1st, 3rd, and 5th terms. However, a3 is the midpoint between a1 and a5:
2a3=a1+a5⟹22m(m−1)=m+6m(m−1)(m−2)
m2−m=m+6m(m−1)(m−2)
Dividing by m (since m=0):
m−1=1+6(m−1)(m−2)⟹6m−12=m2−3m+2
m2−9m+14=0⟹(m−7)(m−2)=0
Since we have a 6th term, m≥5, so m=7.
3. Solve for x using T6=21
T6=(57)A7−5B5=21
21⋅(10−3x)2⋅21⋅(3x−2)5⋅51=21
(10−3x)⋅(3x−2)=1
Let 3x=t:
(10−t)⋅9t=1⟹10t−t2=9
t2−10t+9=0⟹(t−9)(t−1)=0
So 3x=9⟹x=2 or 3x=1⟹x=0.
4. Final Sum
Sum of squares of x: 22+02=4.
Explanation
Solution
1. Simplify the Terms
First term: A=2log2(10−3x)=(10−3x)1/2
Second term: B=52(x−2)log23=(3x−2)1/5
2. Find m from Binomial Coefficients
The coefficients are (1m),(2m),(3m). We are told (1m), (2m), and (3m) are the 1st, 3rd, and 5th terms of an A.P.
Let the common difference be d.
a1=(1m)=m
a3=(2m)=2m(m−1)
a5=(3m)=6m(m−1)(m−2)
In an A.P., a5−a3=a3−a1 is not necessarily true because these are the 1st, 3rd, and 5th terms. However, a3 is the midpoint between a1 and a5: