wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Values of x for which the sixth term of the expansion of
E=(3log3 9|x2|+7(15)log7 [(4).3|x2|9])7 is 567, are


A

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

3


Put y=3log3p|x2|log3 y=log3 9|x2|
y=9|x2|=3|x2|
Next, put z=(715log7 [(4).3x29])
log7 z=15 log7[(4)3|x2|9]=log7[(4)3|x2|9]15 z=[(4)3|x2|9]15
Now, E=(y+z)7 and 6th term is given by
t6=7C5 y75 z5=21(3|x2|)2{(4)3|x2|9} 567=21{32|x2|}{(4)3|x2|9} 27=(4)33|x2|(9)(32|x2|) 27=(4)33|x2|(9)(32|x2|) 4u39u227=0 where u=3|x2|
Note that u=3 satisfies this equation
3|x2|=3|x2|=1x2=±1x=2±1=3 or 1


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
What is Binomial Expansion?
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon