Question

If y = sin^-1 `[(sqrt(1 + x) + sqrt(1 - x))/2]`, then `"dy"/"dx"` = ?

विकल्प

• `(1/4) 1/(sqrt(x^2 - 1))`

• `(-1/2) 1/(sqrt(x^2 - 1))`

• `(-1/2) 1/(sqrt(1 - x^2))`

• `(1/4) 1/(sqrt(1 - x^2))`

Answer 1

`(-1/2) 1/(sqrt(1 - x^2))`

Explanation:

We have, y = sin-1`[(sqrt(1 + x) + sqrt(1 - x))/2]`

Put, x = cos θ

∴ `"y" = sin^-1 [(sqrt(1 + cos 2 theta) + sqrt(1 - cos 2 theta))/2]`

`"y" = sin^-1 ((cos theta + sin theta)/sqrt2)`

`"y" = sin^-1 (sin (pi/4 + theta))`

y = `pi/4 + theta`

y = `pi/4 + 1/2 cos^-1 x`

`"dy"/"dx" = (- 1)/(2 sqrt(1 - x^2))`