Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.6k views
in Binomial Theorem by (91.3k points)
closed by
The largest value of `x` for which the fourth tem in the expansion `(5^2/3(log)_5sqrt(4^(x+44))+1/(5^(log)_5 2^((x-1)+7 3)))` is 336 is.

1 Answer

0 votes
by (94.1k points)
selected by
 
Best answer
Correct Answer - 4
`(5^(2/5sqrt(4^(2)+44))+(1)/(5^(log2sqrt(2^(2-1))+7)))^(8)`
`= ((sqrt(4^(4)+44)^(2//5))+((1)/(3sqrt(2^(x-1)+7))))^(8)`
`= ((4^(x)+44)^(1//5)+(1)/((2^(x-1)+7)^(1//3)))^(8)`
Now, `T_(4) = T_(3+1)=.^(8)C_(3)((4^(x)+44)^(1//5))^(8-3)(1)/((2^(x-1)+7)^(1//3))^(3)`
Given `336 = .^(8)C_(3)((4^(x) + 44)/(2^(x-1) + 7))`
Let `2^(x) = y`
`rArr 336 = .^(8)C_(3)((y^(2)+44)/((y//2)+7))`
or `336 = (8xx7xx6)/(3xx2xx1)((2(y^(2)+44))/(y+14))`
`rArr y^(2) -3y + 2 =0`
or `y = 0,2`
or `y = 2` or `x= 4`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...