BITSAT 2024 Shift 1 Paper

Given the function y=tan−1(‌

√x−x

1+x‌ 3 2

).
We can rewrite this expression using an identity for inverse tangent:
y=tan−1(‌

√x−x

1+√x⋅x

)
This can be expressed as:
y=tan−1(√x)−tan−1(x)
This transformation uses the identity:
tan−1(‌

a−b

1+ab

)=tan−1(a)−tan−1(b)
To find the derivative y′ with respect to x, we use the differentiation of inverse tangent functions:
y′=‌

d

dx

[tan−1(√x)−tan−1(x)]
The derivative is calculated as follows:
y′=‌

1

1+(√x)2

⋅‌

1

2√x

−‌

1

1+x2

Now, evaluate y′(1) :
y′(1)=‌

1

1+1

⋅‌

1

2⋅1

−‌

1

1+1

=‌

1

2

⋅‌

1

2

−‌

1

2

=‌

1

4

−‌

1

2

=−‌

1

4

Thus, the value of y′(1) is −‌

1

4

.