JEE Main 26 Feb 2021 Shift 1 Solved Paper

Given, √3cos2x=(√3−1)‌cos‌x+1,x∈[0,π∕2]
Let cos‌x=t, then
‌√3t2=(√3−1)t+1
⇒‌√3t2−√3t+t−1=0
⇒‌(√3t2−√3t)+(t−1)=0
⇒‌√3t(t−1)+1(t−1)=0
⇒‌(t−1)(√3t+1)=0
This gives t=1 and t=‌

−1

√3

Put, t=cos‌x, then
cos‌x=1 and cos‌x=‌

−1

√3

cos‌x=−1∕√3 is rejected as x∈[0,π∕2]
‌cos‌x=−1∕√3‌ is rejected as ‌x∈[0,π
∴cos‌x=1
‌ Since, ‌x∈[0,‌

π

2

],‌ then ‌cos‌x=cos‌0
This gives x=0 is only solution.
Therefore, number of solution when x∈[0,π∕2] is 1 .