© examsiri.com
Question : 5 of 160
Marks:
+1,
-0
Solution:
Given,
=2x2+4x+=7−2+x‌, also ‌90∘<θ<180∘ We know that,
cos‌θ=‌ cos‌θ=‌| (2x2+4x+)⋅(7−2+x) |
| √4x4+16x2+1⋅√49+4+x2 |
cos‌θ=‌| 14x2−8x+x |
| √4x4+16x2+1⋅√53+x2 |
=‌| 14x2−7x |
| √4x4+16x2+1⋅√53+x2 |
cos‌θ=‌| 7x(2x−1) |
| √4x4+16x2+1⋅√53+x2 |
∵θ‌ lies between ‌(90∘,180∘) ‌ ie, ‌cos‌θ‌ is negative in IInd quadrant. ‌ ‌ So, RHS is also negative ie, ‌ ‌| 7x(2x−1) |
| √4x4+16x2+1⋅√53+x2 |
<0 7x(2x−1)<0 So,
© examsiri.com
Go to Question: