JEE Main 17 March 2021 Shift 2 Solved Paper

Given, sin−1[x2+‌

1

3

]+cos−1[x2−‌

2

3

]=x2
(∵[x+n]=[x]+n,n∈I)
∵[x2−‌

2

3

] gives always integral values.
∴[x2−‌

2

3

]=0,−1 are possible values for x∈[−1,1]
[∵‌‌−1≤x≤1
∴‌‌0≤x2≤1
⇒‌‌−0.66≤x2−‌

2

3

≤0.33
∴‌‌[x2−‌

2

3

]=−1,0 are possibilities. ]
Case I If [x2−‌

2

3

]=0
Then, Eq. (i) becomes,
sin−1(1)+cos−1(0)=x2
⇒‌‌x2=π⇒x=±√π
But at this value of
x2,[x2−‌

2

3

]≠0
∴‌‌x2=π  (Rejected)
Case II If [x2−‌

2

3

]=−1
Then, Eq. (i) becomes,
sin−1(0)+cos−1(−1)=x2⇒x2=π
⇒‌‌x=±√π
But at this value of x2,[x2−‌

2

3

]≠−1
∴‌‌x2=π
Hence, there is no solution for Eq. (i).
∴ Total number of solution (s)=0