JEE Main 24 June 2022 Shift 1 Solved Paper

(tan−1x)3+(cot−1x)3=kπ3
Let f(t)=t3+(‌

π

2

−t)3
Where t=tan−1x;x∈(−‌

π

2

,‌

π

2

)
=t3+(‌

π

2

)3−‌

3π2t

4

+‌

3π

2

t2−t3
f(t)=‌

3π

2

t2−‌

3π2

4

⋅t+‌

π3

8

This is a quadratic equation of t.
Here, coefficient of t2 term is ‌

3π

2

which is >0.
∴ It is a upward parabola.
Now, f′(t)=3πt−‌

3π2

4

f′′(t)=3π>0
∴3πt−‌

3π2

4

=0
⇒t=‌

π

4

‌ (minima) ‌
∴ vertex of graph at ‌

π

4

∴ Minimum value at ‌

π

4

and maximum value at −‌

π

2

.
∴f(‌

π

4

)=‌

π3

64

+(‌

π

2

−‌

π

4

)3=‌

π3

32

f(−‌

π

2

)=−‌

π3

8

+π3
=‌

7π3

8

∴kπ3∈[‌

π3

32

,‌

7π3

8

)
⇒k∈[‌

1

32

,‌

7

8

)