Question

Let (\phi(x) = 2\tan^{-1} x + \sin^{-1}\left( \frac{2x}{1 + x^{2}} \right)). Then –

• A. (\phi'(2) = \phi'(3))
• B. (\phi'(2) = 0)
• C. (\phi'\left( \frac{1}{2} \right) = \frac{16}{5})
• D. (\phi'\left( \frac{1}{2} \right) = 0)

Answer 1

Answer: (A)

$\phi'(2) = \phi'(3)$

Answer 2

If x ≤ 1, sin^–1 $\frac{2x}{1 + x^{2}}$ = 2 tan^–1 x.

But, if x > 1, sin^–1 $\frac{2x}{1 + x^{2}}$ = π – 2tan^–1 x

∴ for x < 1, we have ƒ(x) = 4tan^–1x. So, ƒ′(x) = $\frac{4}{1 + x^{2}}$.

∴ ƒ′$\left( \frac{1}{2} \right)$= $\frac{4}{1 + \frac{1}{4}}$ = $\frac{16}{5}$.

and for x > 1, we have ƒ(x) = π ∴ f′(x) = 0.

Hence, ƒ′(2) = 0 and ƒ′(2) = ƒ′(3) = 0.