Question

If, y = cos(2sin^-1x), then what is the value of (1 – x^2)y” – xy’ + 4y?

(a) –1

(b) 0

(c) 1

(d) Depends on the value of x

I had been asked this question in an interview for job.

Asked question is from Second Order Derivative topic in chapter Limits and Derivatives of Mathematics – Class 11

Answer 1

Right answer is (b) 0

Explanation: We have, y = cos(2sin^-1x)

Differentiating both the sides,

y’ = d/dxcos(2sin^-1x)

= -sin(2sin^-1x) * d/dx(2sin^-1x)

or, y’ = -sin(2sin^-1x)*2*1/√(1 – x^2)

or, y’*√(1 – x^2) = -2sin(2sin^-1x)

Squaring both sides,

or, (y’)^2 *(1 – x^2) = 4sin^2(2sin^-1x)

or, (y’)^2 *(1 – x^2) = 4(1 – y^2)

Differentiating again with respect to x, we get,

(1 – x^2)*2y’y” + (y’)^2 (-2x) = 4 * (-2yy’)

(1 – x^2)y” – xy’ = – 4y

=> (1 – x^2)y” – xy’ + 4y = 0