wiz-icon
MyQuestionIcon
MyQuestionIcon
15
You visited us 15 times! Enjoying our articles? Unlock Full Access!
Question

If the equation cos4θ+sin4θ+λ=0 has real solutions for θ, then λ lies in the interval


A

(-12,-14]

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

-1,-12

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

-32,-54

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

-54,-1

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

The correct option is B

-1,-12


Explanation for correct option

Given equation is cos4θ+sin4θ+λ=0

λ=-cos4θ+sin4θλ=-cos2θ+sin2θ2-2cos2θsin2θa2+b2=a+b2-2abλ=-12+122cosθsinθ2cos2θ+sin2θ=1λ=12sin22θ-12cosθsinθ=sin2θ

We know that -1sinx1

Therefore 0sin2x1

So 12sin22θ0,12

for 12sin22θ=0λ=0-1=-1

for 12sin22θ=12λ=12-1=-12

λ-1,-12

Hence, Option(B) i.e. -1,-12 is correct


flag
Suggest Corrections
thumbs-up
49
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Range of Trigonometric Expressions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon