Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Let f(x)=absin⁡ x+b1−a2cos⁡ x+c, where |a|<1, b > 0 then

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

maximum value of f(x) is b if c = 0

b

difference of maximum and minimum values of f(x) is 2b

c

f(x)=c if x=−cos−1⁡ a

d

f(x)=c if x=cos−1⁡ a

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

f(x)=absin⁡x+b1−a2cos⁡x+c, where |a|<1,b<0f(x)=a2b2+b2−b2a2sin⁡(x+α)+c =bsin⁡(x+α)+c, where tan⁡α=b1−a2ab=1−a2a =bcos⁡(x−α)+c, where tan⁡α=abb1−a2=a1−a2f(x)max−f(x)min=c+b−(c−b)=2bf(x)=c if x+α=0or   x=−αor   x=−cos−1⁡ a
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon