Question

If y = `2/(sqrt(a^2 - b^2))tan^-1[sqrt((a - b)/(a + b))  tan  x/2], "then" (d^2y)/dx^2|_{x = pi/2}` = ______ 

Options

• `b/(2a^2)`

• `b/a^2`

• `(2b)/a`

• `b^2/(2a)`

Answer 1

If y = `2/(sqrt(a^2 - b^2))tan^-1[sqrt((a - b)/(a + b))  tan  x/2], "then" (d^2y)/dx^2|_{x = pi/2}` = `underline(b/a^2)`.

Explanation:

y = `2/(sqrt(a^2 - b^2))tan^-1(sqrt((a - b)/(a + b))  tan  x/2)`

∴ `dy/dx = 2/sqrt(a^2 - b^2) 1/(1 + ((a - b)/(a + b))tan^2  x/2) xx sqrt((a - b)/(a + b)) sec^2  x/2(1/2)`

= `1/(a + b) xx (sec^2  x/2)/(1 + ((a - b)/(a + b)) tan^2  x/2)`

⇒ `dy/dx = (sec^2  x/2)/((a + b)(a - b)tan^2  x/2)`

`[(a + b) + (a - b)tan^2  x/2](sec  x/2 sec  x/2 tan  x/2)`

∴ `(d^2y)/(dx^2) = (-sec^2  x/2[(a - b) tan  x/2 sec^2  x/2])/[[(a + b) + (a - b) tan^2  x/2]`

(a + b + a - b)`(sqrt2 xx sqrt2 xx 1)`

`((d^2y)/dx^2)_{(x = pi/2)} = (-(sqrt2)^2[(a - b)(sqrt2)^2])/(a + b + a - b)^2`

= `(4a - 4(a - b))/(4a^2)`

= `(4b)/(4a^2) = b/a^2`