JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023If the solution of the equation logcosxcotx+4logsinxtanx=1,x∈(0,2π), is sin−1(2α+β), where α and β are integers, then α+β is equal to:
Choose the correct answer:
- A.
5
- B.
6
- C.
4
(Correct Answer) - D.
3
4
Explanation
Solution:
1. Equation simplification:
logcosxlog(cosx/sinx)+4logsinxlog(sinx/cosx)=1
logcosxlogcosx−logsinx+4(logsinxlogsinx−logcosx)=1
2. Variable substitution:
Let t=logcosxlogsinx.
(1−t)+4(1−t1)=1
1−t+4−t4=1
4−t−t4=0
t2−4t+4=0
(t−2)2=0⟹t=2
3. Value of sinx:
logcosxlogsinx=2⟹sinx=cos2x
sinx=1−sin2x
sin2x+sinx−1=0
sinx=2−1+12−4(1)(−1)=2−1+5
4. Final Calculation:
Comparing with sin−1(2α+β):
α=−1,β=5
α+β=−1+5=4
Correct Option: (3) 4
Explanation
Solution:
1. Equation simplification:
logcosxlog(cosx/sinx)+4logsinxlog(sinx/cosx)=1
logcosxlogcosx−logsinx+4(logsinxlogsinx−logcosx)=1
2. Variable substitution:
Let t=logcosxlogsinx.
(1−t)+4(1−t1)=1
1−t+4−t4=1
4−t−t4=0
t2−4t+4=0
(t−2)2=0⟹t=2
3. Value of sinx:
logcosxlogsinx=2⟹sinx=cos2x
sinx=1−sin2x
sin2x+sinx−1=0
sinx=2−1+12−4(1)(−1)=2−1+5
4. Final Calculation:
Comparing with sin−1(2α+β):
α=−1,β=5
α+β=−1+5=4
Correct Option: (3) 4