CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

lf cosα is a root of 25x2+5x12=0 and (1<x<0), then the value of sin2α=

A
1225
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1225
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2025
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2425
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 2425
We can write the equation as (5x+4)(5x3)=0
Now x lies between 0 and 1 so cosα=45
Hence, sinα=35
sin(2α)=2(45)(35)=2425
Option D is correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Properties of Modulus
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon